Chapter 10: Magnetic Fields due to Electric Current
In 1820, in Copenhagen, the Danish scientist Hans Christian Oersted was working with a simple arrangement: a wire connected to a battery, and an ordinary magnetic compass lying close by. Until that moment, electricity and magnetism were treated as two completely separate subjects. Electric charges attracted or repelled other charges. Magnets attracted iron and pointed north. Nobody expected the two to have anything to do with each other.
When Oersted allowed current to flow through the wire, the compass needle moved. That alone was surprising. What was stranger still was the direction in which it moved. The needle did not swing along the wire, as one might expect if the wire had simply become “charged”. It settled at right angles to the wire, as though the wire were surrounded by invisible circles of magnetism.
That single observation joined two branches of physics into one. A current — nothing but charges in motion — was producing a magnetic field around itself. Within a few years, this idea grew into the theory that today runs electric motors, generators, loudspeakers, MRI scanners and particle accelerators.
Diagram
Fig. 10.0 – A straight wire connected to a battery with a compass needle beneath it, the needle turning perpendicular to the wire, with dashed circular field lines around the wire.
Keep these questions in mind as you read. Each one of them is answered somewhere in this chapter.
10.1 Introduction
In this chapter you will study how magnetic fields are produced by an electric current. Along the way, an important foundation will be laid for the developments that follow in later chapters.
The story begins with Hans Christian Oersted, who first discovered that a magnetic field is produced by an electric current passing through a wire. Later, Gauss, Henry, Faraday and others showed that the magnetic field is an important partner of the electric field — the two are not independent subjects but two sides of a single description of nature.
Maxwell’s theoretical work highlighted this close relationship between electric and magnetic fields. That understanding has resulted in several practical applications in day-to-day life, for example electrical motors, generators, communication systems and computers.
From Charges at Rest to Charges in Motion
In electrostatics, we considered static charges — charges at rest — and the force exerted by them on another charge or on a test charge. A test charge is simply a very small positive charge used to probe a field without disturbing the arrangement that produces it.
We now take the next step and consider forces between charges in motion. This single change — from charges at rest to charges moving — is what brings magnetism into the picture.
A magnetic field is the region around a magnet or a current in which a moving charge experiences a force. Like the electric field, it is a vector quantity: at every point it has both a magnitude and a direction. It is written as \(\vec{B}\), and its SI unit will be defined carefully in the next section.
The Magnetic Field Circles the Wire
You have learnt in Std. X that if a magnetic needle is held in close proximity to a current-carrying wire, it shows the direction of the magnetic field circling around the wire. This is worth noticing carefully: the field does not point along the wire, and it does not point straight outward from the wire. It wraps around the wire in closed circles.
To find which way these circles run, imagine that the current-carrying wire is grabbed with your right hand, with the thumb pointing in the direction of the current. Your fingers then curl around in the direction of the magnetic field, as shown in Fig. 10.2.
Diagram
Fig. 10.2 – Right hand thumb rule: a right hand gripping a vertical wire, thumb pointing along the current, fingers curling around the wire to show the circular magnetic field.
Note that the rule is stated for the conventional current — the direction in which positive charge would flow. Inside a metal wire the actual carriers are electrons, and they drift in the direction opposite to the conventional current. This does not create any confusion as long as we consistently use conventional current in the rule.
Two Wires That Push and Pull
A magnetic field is not merely something to be detected with a compass. It exerts real, measurable forces — and the clearest demonstration uses nothing more than two wires.
The force in this experiment is certainly not of electrostatic origin, even though the current is due to the electrons flowing in the wires. The reason is that the overall charge neutrality is maintained throughout the wire: for every mobile electron there is a fixed positive ion in the metal lattice, so the wire as a whole carries no net charge. With no net charge on either wire, electrostatic forces are ruled out.
Accounting for the Force on the Neighbouring Wire
How can one account for the force on the neighbouring current-carrying wire? Consider the arrangement in Fig. 10.3, with wire 1 and wire 2 placed side by side.
The magnetic field due to the current in wire 1 is, at any point on wire 2, directed into the plane of the paper. Wire 2 therefore does not sit in empty space — it sits inside the magnetic field created by wire 1.
Diagram
Fig. 10.3 – Two parallel wires 1 and 2 carrying currents in the same direction, with the field of wire 1 shown going into the plane of the paper at wire 2, and the force \(\vec{F}\) on wire 2 directed towards wire 1.
Inside wire 2, the electrons flow in a direction opposite to the conventional current. These moving electrons, sitting in the magnetic field of wire 1, experience a magnetic force. Adding up the force on all of them, the wire 2 as a whole experiences a force \(\vec{F}\) directed towards wire 1 — an attraction, exactly as the hanging-wire experiment showed for parallel currents.
This is the central idea that the rest of the chapter builds on: a current creates a magnetic field, and a magnetic field exerts a force on moving charges. Everything that follows — the force law on a single charge, the circular and helical paths of particles, the torque on a coil, motors, galvanometers, solenoids and toroids — is worked out from these two statements.
10.2 Magnetic Force
From the discussion in the previous section and from Fig. 10.3, you must have realised something about the three directions involved: the velocity of the charge, the magnetic field, and the force produced. The force does not point along the velocity, and it does not point along the field either. It points perpendicular to both. This is exactly the behaviour of a vector cross product.
Recall from Std. XI that the cross product of two vectors \(\vec{A}\) and \(\vec{B}\) is itself a vector, whose magnitude is \(AB\sin\theta\) and whose direction is perpendicular to the plane containing \(\vec{A}\) and \(\vec{B}\). That perpendicular direction is what makes the cross product the right mathematical tool here.
The Magnetic Force on a Moving Charge
The magnetic force \(\vec{F}_m\) on an electron carrying a charge \(-e\), moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\), is
\[\vec{F}_m = -e\left(\vec{v} \times \vec{B}\right) \qquad \text{--- (10.1)}\]
In general, for a charge \(q\), the magnetic force will be
\[\vec{F}_m = q\left(\vec{v} \times \vec{B}\right) \qquad \text{--- (10.2)}\]
Here \(q\) is the charge in coulomb (C), \(\vec{v}\) is its velocity in metre per second (m s\(^{-1}\)), and \(\vec{B}\) is the magnetic field, whose SI unit will be fixed later in this section.
Notice that the sign of the charge is built into the formula. For a positive charge, \(\vec{F}_m\) points along \(\vec{v} \times \vec{B}\); for a negative charge such as an electron, the minus sign in Eq. (10.1) reverses it.
The Lorentz Force
A charge does not always find itself in a magnetic field alone. If both an electric field \(\vec{E}\) and a magnetic field \(\vec{B}\) are present, the net force on a charge \(q\) moving with velocity \(\vec{v}\) is
\[\vec{F} = q\left[\vec{E} + \left(\vec{v} \times \vec{B}\right)\right] \qquad \text{--- (10.3)}\]
\[= q\vec{E} + q\left(\vec{v} \times \vec{B}\right) = \vec{F}_e + \vec{F}_m \qquad \text{--- (10.4)}\]
Justification for this law can be found in experiments such as the one described in Fig. 10.1 (a) and (b). The force described in Eq. (10.4) is known as the Lorentz force. Here \(\vec{F}_e\) is the force due to the electric field and \(\vec{F}_m\) is the force due to the magnetic field.
From Eq. (10.4) it may be observed that the two parts behave quite differently. The force on the charge due to the electric field depends only on the strength of the electric field and the magnitude of the charge — a charge sitting still feels it just as much as a charge in motion. However, the magnetic force depends on the velocity of the charge, on the cross product of the velocity vector \(\vec{v}\) with the magnetic field vector \(\vec{B}\), and on the charge \(q\).
The Direction of the Magnetic Force
Consider the vectors \(\vec{v}\) and \(\vec{B}\) with a certain angle between them. Then \(\vec{v} \times \vec{B}\) will be a vector perpendicular to the plane containing the vectors \(\vec{v}\) and \(\vec{B}\), as shown in Fig. 10.4.
Diagram
Fig. 10.4 – Vectors \(\vec{v}\) and \(\vec{B}\) drawn in a plane with angle \(\theta\) between them, and the unit vector \(\hat{n}\) along \(\vec{v} \times \vec{B}\) standing perpendicular to that plane.
The cross product is in the direction of the unit vector \(\hat{n}\), perpendicular to both \(\vec{v}\) and \(\vec{B}\).
There are interesting consequences of the Lorentz force law.
If the velocity \(\vec{v}\) of a charged particle is parallel to the magnetic field \(\vec{B}\), the magnetic force is zero. This follows directly from the cross product: when the two vectors are parallel, the angle between them is zero, and \(\sin 0^\circ = 0\).
If the charge is stationary, \(\vec{v} = 0\), the force \(= 0\), even if \(\vec{B} \neq 0\). A magnetic field, however strong, has nothing to act on unless the charge is moving.
Why a Magnetic Force Never Does Work
Since \(\vec{F}\) is along \(\vec{v} \times \vec{B}\), the vectors \(\vec{v}\) and \(\vec{F}\) are always perpendicular to each other. Hence
\[\vec{F} \cdot \vec{v} = 0\]
for any magnetic field \(\vec{B}\). The magnetic force \(\vec{F}_m\) is thus perpendicular to the displacement, and hence the magnetic force never does any work on moving charges.
This deserves a moment’s thought. Work is done only by the component of a force along the direction of motion. The magnetic force has no such component — it is entirely sideways. So it can pull the particle off its straight path, but it can never add energy to it or take energy away.
The magnetic forces may change the direction of motion of a charged particle, but they can never affect the speed.
Defining the Unit of Magnetic Field
Interestingly, Eq. (10.2) leads to the definition of the unit of \(\vec{B}\). From Eq. (10.2), the magnitude of the force is
\[\vec{F} = q\left|\vec{v} \times \vec{B}\right| = qvB\sin\theta\ \hat{n} \qquad \text{--- (10.5)}\]
where \(\theta\) is the angle between \(\vec{v}\) and \(\vec{B}\), and \(\hat{n}\) is the unit vector in the direction of the force.
If the force \(F\) is 1 N acting on a charge of 1 C moving with a speed of 1 m s\(^{-1}\) perpendicular to \(\vec{B}\), then we can define the unit of \(B\). Rearranging,
\[B = \frac{F}{qv}\]
\(\therefore\) unit of \(B\) is \(\dfrac{\text{N} \cdot \text{s}}{\text{C} \cdot \text{m}}\)
Dimensionally,
\[[B] = \left[\frac{F}{qv}\right]\]
This SI unit is called the tesla (T).
\(1\ \text{T} = 10^4\) gauss. Gauss is not an SI unit, but is used as a convenient unit.
Because one tesla is such a large field, the gauss survives in everyday use — the field values you meet for the Earth and for ordinary magnets are far below one tesla, and are more comfortably quoted in gauss.
10.3 Cyclotron Motion
In a magnetic field, a charged particle typically undergoes circular motion. To see why, recall the two results established in the previous section: the magnetic force is always perpendicular to the velocity, and it never changes the speed. A force of constant magnitude that is always at right angles to the motion is precisely the condition for a circular path.
Figure 10.5 shows a uniform magnetic field directed perpendicularly into the plane of the paper, that is, parallel to the negative \(z\) axis.
Diagram
Fig. 10.5 – A charged particle moving in a circle of radius \(R\) in a uniform magnetic field \(\vec{B}\) directed into the plane of the paper, with \(\vec{v}\) tangential and \(\vec{F}_m\) pointing towards the centre.
Setting Up the Circular Path
Figure 10.5 shows a particle with charge \(q\) moving with a speed \(v\), and a uniform magnetic field \(\vec{B}\) directed into the plane of the paper. According to the Lorentz force law, the magnetic force on the particle will act towards the centre of a circle of radius \(R\), and this force will provide the centripetal force needed to sustain a uniform circular motion.
Notice how the direction works out. At every instant the force points at right angles to the velocity, towards one fixed point. As the particle moves, its velocity turns, and the force turns with it — always aimed at that same centre. The particle therefore keeps curving around, retracing the same circle.
The Cyclotron Formula
Since the velocity here is perpendicular to the field, \(\sin\theta = 1\) and the magnitude of the magnetic force is simply \(qvB\). Equating this to the centripetal force \(mv^2/R\) required for a circle of radius \(R\),
\[qvB = \frac{mv^2}{R} \qquad \text{--- (10.6)}\]
Cancelling one factor of \(v\) from both sides and multiplying through by \(R\),
\[\therefore mv = p = qBR \qquad \text{--- (10.7)}\]
Here \(m\) is the mass of the particle in kilogram (kg), \(v\) its speed in m s\(^{-1}\), \(p = mv\) its linear momentum in kg m s\(^{-1}\), \(q\) the charge in coulomb (C), \(B\) the magnetic field in tesla (T), and \(R\) the radius of the circular path in metre (m).
Equation (10.7) represents what is known as the cyclotron formula. It describes the circular motion of a charged particle in a particle accelerator, the cyclotron.
Reading the Formula Physically
Written as \(R = \dfrac{mv}{qB}\), the relation says something useful at a glance. A faster or heavier particle has more momentum, resists being turned, and therefore sweeps out a wider circle. A stronger field or a larger charge bends it more sharply into a tighter circle.
This is exactly why the formula is so valuable in the laboratory. If a particle’s track is photographed in a known magnetic field, measuring the radius of that track immediately gives its momentum. This is the standard way charged particles are identified in detectors.
10.3.1 Cyclotron Accelerator
Particle accelerators have played a key role in providing high energy particle beams, in the range of MeV to GeV, useful in studying particle–matter interactions. Some of these machines are also useful in the medical treatment of certain tumours and diseases.
The idea behind an accelerator is simple to state and hard to build: give a charged particle a push, then another, then another, until it carries an enormous amount of energy. The cyclotron solves this problem with an elegant trick — it makes the particle come back for its next push, again and again, using a magnetic field.
The cyclotron is a charged particle accelerator that accelerates charged particles to high energies.
Crossed Fields: the Two Halves of the Idea
Both electric as well as magnetic fields are used in a cyclotron, in combination. These are applied in directions perpendicular to each other, and hence they are called crossed fields.
Each field has its own job. The magnetic field puts the particle (ion) into a circular path, and a high frequency electric field accelerates it. Neither field alone would do — the magnetic field can only steer, never speed up, while the electric field can speed up but cannot bring the particle back for another push.
The whole design rests on one remarkable fact, which you can read off Eq. (10.9) below: the frequency of revolution of a charged particle in a magnetic field is independent of its energy. This fact is used in this machine.
Construction of the Cyclotron
A cyclotron consists of two semicircular disc-like metal chambers, \(D_1\) and \(D_2\), called the dees (Ds). Figure 10.6 shows a schematic diagram of a cyclotron.
Diagram
Fig. 10.6 – Schematic diagram of a cyclotron showing the two hollow dees \(D_1\) and \(D_2\) separated by a narrow gap, the ion source at P, the a.c. supply across the dees, the spiral path of the ion widening outward, the exit radius \(R_{exit}\), and the magnetic field \(\vec{B}\) coming out of the plane of the paper.
A uniform magnetic field \(B\) is applied perpendicular to the plane of the Ds. This magnetic field is produced using an electromagnet producing a field up to 1.5 T. An alternating voltage up to 10000 V at a high frequency of 10 MHz (\(f_a\)) is applied between the two Ds.
Positive ions are produced by a gas ionizing source kept at the point O in between the two Ds. The electric field provides acceleration to the charged particle (ion). The entire assembly is placed in a vacuum chamber, so that the ions are not slowed down or scattered by collisions with air molecules, and the ions are injected into the D at point P.
How the Ion Gains Energy
Once the ion is emitted, it accelerates due to the negative voltage of a D and performs a semicircular motion within the D. Whenever the ion moves from one D to the other D, it accelerates due to the potential difference between the two Ds, and again performs semicircular motion in the other D.
Thus the ion is acted upon by the electric field every time it moves from one D to the other D. As the electric field is alternating, its sign is changed in accordance with the circular motion of the ion.
The timing is everything here. If the polarity did not flip, the ion would be pushed forward on one crossing and pulled back on the next, gaining nothing. Because the polarity reverses exactly in step with the ion’s motion, every single crossing of the gap is an accelerating one.
Hence the ion is always accelerated, its energy increases, and the radius of its circular path also increases, making the entire path a spiral, as seen in Fig. 10.6.
The Period and the Cyclotron Frequency
Consider an ion source placed at P. An ion moves in a semicircular path in one of the Ds and reaches the gap between the two Ds in a time interval \(T/2\), \(T\) being the period of a full revolution.
Using the cyclotron formula, Eq. (10.7),
\[mv = qBR,\]
where \(q\) is the charge on the ion. The period of one full revolution is the circumference of the circle divided by the speed, that is \(T = 2\pi R/v\). Substituting \(v = qBR/m\) from the cyclotron formula,
\[\therefore\ T = \frac{2\pi R}{v} = \frac{m\,2\pi R}{qBR} = \frac{2\pi m}{qB} \qquad \text{--- (10.8)}\]
Look closely at what happened in the last step: the radius \(R\) cancelled completely. The time taken for one revolution does not depend on how big the circle is.
The frequency of revolution (cyclotron frequency) is
\[f_c = \frac{1}{T} = \frac{qB}{2\pi m} \qquad \text{--- (10.9)}\]
where \(q\) is the charge of the ion in coulomb (C), \(B\) the magnetic field in tesla (T), \(m\) the mass of the ion in kilogram (kg), and \(f_c\) the frequency in hertz (Hz).
The Resonance Condition
The frequency of the applied voltage (\(f_a\)) between the two Ds is adjusted so that the polarity of the two Ds is reversed as the ion arrives at the gap after completing one semicircle. This condition,
\[f_a = f_c\]
is the resonance condition. If this condition fails — for example if the field drifts, or if the ion becomes so fast that its mass changes appreciably — the ion arrives at the gap at the wrong moment and is decelerated instead of accelerated.
The Energy of the Extracted Beam
The ions do not experience any electric field while they travel within the D. Their kinetic energy increases by \(eV\) every time they cross over from one D to the other. Here \(V\) is the voltage difference across the gap.
The ions move in a circular path with successively larger and larger radius, up to a maximum radius at which they are deflected by a magnetic field so that they can be extracted through an exit slit.
From Eq. (10.7),
\[v = \frac{qBR_{exit}}{m},\]
where \(R_{exit}\) is the radius of the path at the exit. This is simply the cyclotron formula applied at the last, outermost turn — the point at which the ion leaves the machine.
The kinetic energy of the ions or protons will be
\[\text{K.E.} = \frac{1}{2}mv^2 = \frac{q^2B^2R^2_{exit}}{2m} \qquad \text{--- (10.10)}\]
obtained by substituting the expression for \(v\) into \(\tfrac{1}{2}mv^{2}\).
Thus the final energy is proportional to the square of the radius of the outermost circular path (\(R_{exit}\)).
This tells the designer exactly what to do to build a more powerful machine: make the magnet stronger, or make it wider. Doubling the exit radius quadruples the energy of the beam. It also explains why high-energy accelerators are such enormous structures.
10.4 Helical Motion
So far it has been assumed that the charged particle moves in a plane perpendicular to the magnetic field \(\vec{B}\). That is a special case. In general, a particle may enter the field at any angle, and then its velocity has one part along the field and another part across it.
If such a particle has some component of velocity parallel to \(\vec{B}\), written as \(\vec{v}_{\parallel}\), then it leads to helical motion. The reason lies in the cross product. Since the component \(\vec{v}_{\parallel}\) is parallel to \(\vec{B}\), the magnetic force \(\vec{F}_m\) on account of this component will be
\[\vec{F}_m = \vec{v}_{\parallel} \times \vec{B} = v \cdot B \sin(0^\circ) = 0 \qquad \text{--- (10.11)}\]
Thus, \(\vec{v}_{\parallel}\) will not be affected and the particle will move along the direction of \(\vec{B}\) with this component of velocity unchanged, exactly as if the field were absent.
At the same time, the perpendicular component of the velocity (\(\vec{v}_{\perp}\)) leads to circular motion, as stated above in Section 10.3. This component alone would carry the particle round and round a circle of radius \(R = mv_{\perp}/qB\).
As a result, the particle moves parallel to the field \(\vec{B}\) while simultaneously moving along a circular path perpendicular to \(\vec{B}\). Combining a steady forward drift with a circular motion gives a coiled path. Thus the path becomes a helix, as shown in Fig. 10.7.
Diagram
Fig. 10.7 – A charged particle following a helical path about the direction of a uniform magnetic field \(\vec{B}\) along the \(x\) axis, with the velocity resolved into \(\vec{v}_{\perp}\) and \(\vec{v}_{\parallel}\) at the point of entry.
Note also that the speed of the particle is still unchanged throughout this motion. The magnetic force continues to do no work; it only redistributes the direction of the velocity, never its magnitude.
10.5 Magnetic Force on a Wire Carrying a Current
We have seen earlier the Lorentz force law, Eq. (10.4). It tells us the force on a single moving charge. From this equation, we can obtain the force on a current-carrying wire, which is nothing but a very large number of charges moving together in a confined path.
This is an important step, because in practice we rarely handle free charges one at a time. What we can handle easily is a wire with a current in it — and every motor, loudspeaker and moving-coil meter depends on the force such a wire feels in a magnetic field.
Straight Wire
Consider a straight wire of length \(L\) as shown in Fig. 10.8. An external magnetic field \(\vec{B}\) is applied perpendicular to the wire, coming out of the plane of the paper. Let a current \(I\) flow through the wire under an applied potential difference.
Diagram
Fig. 10.8 – A straight wire of length \(L\) carrying a current \(I\) with the field \(\vec{B}\) coming out of the plane of the paper; electrons with drift velocity \(\vec{v}_d\) shown moving opposite to the current, and the force \(\vec{F}_m\) shown acting upwards.
Inside the metal, the conduction electrons do not travel in a straight rush; they drift slowly along the wire, colliding repeatedly with the ions of the lattice. Their small average velocity along the wire is called the drift velocity, written \(\vec{v}_d\).
If \(\vec{v}_d\) is the drift velocity of the conduction electrons in the part of length \(L\) of the wire, then an electron takes a time \(t = L/v_d\) to travel through this portion. The charge \(q\) flowing across the plane \(pp'\) in time \(t\) will therefore be
\[q = I\,t\]
\[q = \frac{IL}{v_d} \qquad \text{--- (10.12)}\]
The magnetic force \(\vec{F}_m\) on this charge, according to Eq. (10.2), due to the applied magnetic field \(\vec{B}\) is given by
\[\vec{F}_m = q\left(\vec{v}_d \times \vec{B}\right) = \frac{IL}{v_d}\left(v_d\,B\,\sin 90^\circ\right)\hat{n} = I\,L\,B\,\hat{n},\]
where \(\hat{n}\) is a unit vector perpendicular to both \(\vec{B}\) and \(\vec{v}_d\), in the direction of \(\vec{F}_m\).
\[\vec{F}_m = ILB \qquad \text{--- (10.13)}\]
Notice what happened in the middle step: the drift velocity \(v_d\) appears once in the numerator and once in the denominator, and cancels out completely. This is a satisfying result. The force does not depend on how fast the individual electrons crawl along — only on the current, the length and the field.
Here \(I\) is the current in ampere (A), \(L\) the length of the wire in metre (m), and \(B\) the magnetic field in tesla (T), so that \(F_m\) comes out in newton (N).
This is, therefore, the magnetic force acting on the portion of the straight wire having length \(L\).
If \(\vec{B}\) is not perpendicular to the wire, then the above Eq. (10.13) takes the form
\[\vec{F}_m = I\,\vec{L} \times \vec{B}, \qquad \text{--- (10.14)}\]
where \(\vec{L}\) is the length vector directed along the portion of the wire of length \(L\). Its direction is the direction of the current in that portion.
Arbitrarily Shaped Wire
In the previous section we considered a straight wire. Equation (10.14) can be extended to a wire of arbitrary shape as shown in Fig. 10.9.
Diagram
Fig. 10.9 – A wire of arbitrary curved shape carrying a current \(I\), with one infinitesimal element \(d\vec{l}\) marked along it and the field \(\vec{B}\) coming out of the plane of the paper.
The difficulty with a bent wire is that \(\vec{L}\) no longer has a single direction. The way around this is standard in physics: cut the wire into pieces so short that each piece is effectively straight, apply the result we already have to each piece, and then add up.
Consider a segment of infinitesimal length \(dl\) along the wire. If \(I\) is the current flowing, then using Eq. (10.14), the magnetic force on this segment due to the perpendicular magnetic field \(\vec{B}\) (coming out of the plane of the paper) is given by
\[d\vec{F}_m = I\,d\vec{l} \times \vec{B} \qquad \text{--- (10.15)}\]
The force on the total length of the wire is thus obtained by adding — that is, integrating — the contributions of all such segments:
\[\vec{F}_m = \int d\vec{F}_m = I\int d\vec{l} \times \vec{B} \qquad \text{--- (10.16)}\]
If \(\vec{B}\) is uniform over the whole wire, it is the same constant vector for every segment and can be taken outside the integral:
\[\vec{F}_m = I\left(\int d\vec{l}\right) \times \vec{B} \qquad \text{--- (10.17)}\]
10.6 Force on a Closed Circuit in a Magnetic Field \(\vec{B}\)
Equation (10.17) was written for a wire running from one end point to another. It can be extended to a closed wire circuit \(C\) — a loop whose two ends join up:
\[\vec{F}_m = I\oint_C d\vec{l} \times \vec{B} \qquad \text{--- (10.18)}\]
Here, the integral is over the closed circuit \(C\). The circle drawn on the integral sign is simply a reminder that the path closes on itself.
For uniform \(\vec{B}\), the field is the same at every element of the loop and can be taken outside:
\[\vec{F}_m = I\left[\oint_C d\vec{l}\right] \times \vec{B} \qquad \text{--- (10.19)}\]
The situation is the one sketched in Fig. 10.9A, where the forces on opposite portions of the loop point in opposite senses.
Diagram
Fig. 10.9A – A closed current-carrying loop of arbitrary shape lying in a uniform magnetic field, with the forces on opposite portions of the loop drawn as arrows that cancel in pairs.
The term in the bracket in Eq. (10.19) is the sum of vectors along a closed circuit. Adding these vectors head to tail carries you all the way round the loop and back to the starting point, so the resultant displacement is zero. Hence it must be zero.
\[\therefore \vec{F}_m = 0 \quad (\vec{B}\ \text{uniform}) \qquad \text{--- (10.20)}\]
This result is also a time-saver in problem solving. If a question places a closed loop of any shape — square, circular or irregular — in a uniform field and asks for the resultant force on it, the answer is zero without any calculation. What still has to be worked out in such problems is the force on each individual arm, and the torque about an axis.
10.7 Torque on a Current Loop
It will be very interesting to apply the results of the above sections to a current-carrying loop of wire. The previous section showed that the net force on such a loop in a uniform field is zero. What remains to be examined is the turning effect, and that turns out to be the more useful of the two.
You have learnt about an electric motor in Std. X. An electric motor works on the principle you have studied in the preceding sections, i.e., the magnetic force on a current-carrying wire due to a magnetic field.
Figure 10.10 shows a current-carrying loop (abcd) in a uniform magnetic field. There will, therefore, be magnetic forces \(\vec{F}_m\) acting in opposite directions on the segments of the loop ab and cd. This results in the rotation of the loop about its central axis.
Diagram
Fig. 10.10 – A rectangular current loop abcd in a uniform magnetic field, with the forces \(\vec{F}_m\) on the segments ab and cd shown pointing in opposite directions and turning the loop about its central axis.
Without going into the details of the contact carbon brushes and the external circuit, we can visualise the rotating action of a motor from this diagram alone.
The current-carrying wire loop is of rectangular shape and is placed in the uniform magnetic field in such a way that the segments ab and cd of the loop are perpendicular to the field \(\vec{B}\). We can use Eq. (10.14) to find out the direction of the magnetic force \(\vec{F}_m\) on each segment.
Setting Up the Calculation
Let us now look at the action of rotation in detail. For this purpose, consider Fig. 10.11, showing the rectangular loop abcd placed in a uniform magnetic field \(\vec{B}\) such that the sides ab and cd are perpendicular to the magnetic field \(\vec{B}\), but the sides bc and da are not.
Diagram
Fig. 10.11 – The rectangular loop abcd of sides \(l_1\) and \(l_2\) in a uniform magnetic field, with the sides ab and cd perpendicular to \(\vec{B}\) and the sides bc and da inclined to it; the normal \(\hat{n}\) to the plane of the loop is marked.
To describe the tilt of the loop, we use the normal to its plane. The unit vector \(\hat{n}\) is drawn perpendicular to the plane of the loop, and the angle \(\theta\) is measured between this normal and the direction of \(\vec{B}\). When \(\theta = 0\), the plane of the loop is perpendicular to the field; when \(\theta = 90^\circ\), the plane of the loop contains the field.
Now we can calculate the net force and the net torque on the loop in the situation depicted in Fig. 10.11 and Fig. 10.12. Let us obtain the forces on all sides of the loop.
Diagram
Fig. 10.12 – Side view of the loop abcd tilted at an angle \(\theta\), showing the forces \(\vec{F}_1\) and \(\vec{F}_3\) acting on the two arms in opposite directions and the moment arm \(\tfrac{1}{2}l_2\sin\theta\) about the central axis.
Forces on the Slanted Sides
The force \(\vec{F}_4\) on side 4 (bc) is
\[\vec{F}_4 = I\,l_2\,B \sin(90 - \theta), \qquad \text{--- (10.21)}\]
where \(\theta\) is the angle between the normal (\(\hat{n}\)) to the plane of the loop and the direction of the magnetic field \(\vec{B}\). Here \(I\) is the current in ampere (A), \(l_2\) the length of that side in metre (m), and \(B\) the field in tesla (T).
The force \(\vec{F}_2\) on side 2 (da) is equal and opposite to \(\vec{F}_4\), and both act along the same line. Thus, \(\vec{F}_2\) and \(\vec{F}_4\) cancel out each other. They contribute neither a net force nor a torque, and may be set aside for the rest of the calculation.
Forces on the Perpendicular Sides
The magnitude of the forces \(\vec{F}_1\) and \(\vec{F}_3\) on sides 3 (ab) and 1 (cd) is \(I\,l_1 B \sin 90^\circ\), i.e., \(I\,l_1 B\). These sides stay perpendicular to the field as the loop turns, so their forces keep the same magnitude throughout the rotation.
These two forces do not act along the same line, and hence they produce a net torque. This torque results in the rotation of the loop, so that the loop turns towards the orientation in which its plane is perpendicular to the direction of \(\vec{B}\), the magnetic field.
The moment arm is \(\dfrac{1}{2}\left(l_2 \sin\theta\right)\) about the central axis of the loop. The reasoning behind this is worth spelling out: each of these two sides is at a perpendicular distance \(\tfrac{1}{2}l_2\) from the central axis when the loop is edge-on to the field, and as the loop turns through an angle \(\theta\), the effective perpendicular distance from the line of the force to the axis shrinks by a factor \(\sin\theta\).
The torque \(\tau\) due to the forces \(\vec{F}_1\) and \(\vec{F}_3\) is the sum of the moments of the two forces, each equal to force \(\times\) moment arm:
\[\tau = \left(I\,l_1 B\,\frac{1}{2}l_2\sin\theta\right) + \left(I\,l_1 B\,\frac{1}{2}l_2\sin\theta\right)\]
\[= I\,l_1\,l_2\,B\sin\theta \qquad \text{--- (10.22)}\]
The two halves add rather than cancel, because both forces turn the loop the same way round the axis.
Torque on a Coil of N Turns
If the current-carrying loop is made up of multiple turns \(N\), in the form of a flat coil, then each turn contributes the same torque and the total torque becomes
\[\tau' = N\tau = N I\,l_1\,l_2\,B\sin\theta\]
\[\tau' = (NIA)B\sin\theta \qquad \text{--- (10.23)}\]
where
\[A = l_1 l_2\]
Here \(A\) is the area enclosed by the loop, in square metre (m\(^2\)), \(N\) is the number of turns, and \(\tau'\) is the torque in newton metre (N m).
The above equation holds good for all flat (planar) coils irrespective of their shape, in a uniform magnetic field. A circular coil, a triangular coil or an irregular flat coil of the same area and the same current experiences the same torque — only the enclosed area matters, not the outline.
Numerical questions on this section usually give \(N\), \(I\), \(A\) and \(B\) and ask for the torque at a stated orientation. The commonest error is to read the given angle as the angle between the plane of the coil and the field, when Eq. (10.23) requires the angle between the normal to the coil and the field. The two are complementary, so mixing them up replaces \(\sin\theta\) by \(\cos\theta\) and gives a wrong answer.
10.7.1 Moving Coil Galvanometer
A current in a circuit, or the voltage of a battery, can be measured in terms of the torque exerted by a magnetic field on a current-carrying coil. Analog voltmeters and ammeters work on this principle.
The idea is to convert something invisible — a current — into something you can see and read off a scale, namely the angle through which a pointer turns. Figure 10.13 shows a cross-sectional diagram of a galvanometer.
Diagram
Fig. 10.13 – Cross-sectional diagram of a moving coil galvanometer showing the permanent magnet with N and S poles, the soft iron cylindrical core, the coil wound around it, the pivot, the spring and the pointer moving over a scale, with the uniform radial field marked between the pole pieces and the core.
Construction
It consists of a coil of several turns, mounted — suspended or pivoted — in such a way that it can freely rotate about a fixed axis, in a radial uniform magnetic field. A soft iron cylindrical core makes the field radial and strong.
The word radial is the key to the whole design. The pole pieces are curved and the soft iron core is cylindrical, so that the field lines in the gap always point towards the axis of the cylinder. Whatever position the coil turns to, the field there still lies in the plane of the coil.
Working
The coil rotates due to a torque acting on it as the current flows through it. This torque is given by Eq. (10.23),
\[\tau = N\,I\,A \cdot B,\]
where \(A\) is the area of the coil in square metre (m\(^2\)), \(B\) the strength of the magnetic field in tesla (T), \(N\) the number of turns of the coil, and \(I\) the current in the coil in ampere (A).
Here, \(\sin\theta = 1\), as the field is radial — the plane of the coil will always be parallel to the field. This is why the radial field matters so much: without it, the torque would fall off as \(\sin\theta\) while the coil turned, and the instrument would need an awkward, crowded, non-uniform scale.
However, this torque is counterbalanced by a torque due to a spring fitted as shown in Fig. 10.13.
The Deflection Equation
This counter torque balances the magnetic torque, so that a fixed steady current \(I\) in the coil produces a steady angular deflection \(\phi\).
The larger the current is, the larger is the deflection, and the larger is the torque due to the spring. If the deflection is \(\phi\), the restoring torque due to the spring is equal to \(K\phi\), where \(K\) is the torsional constant of the spring — the torque required to twist it through unit angle, measured in N m per radian or per degree.
At equilibrium the two torques are equal. Thus,
\[K\phi = NIAB,\]
and the deflection
\[\phi = \left(\frac{NAB}{K}\right)I \qquad \text{--- (10.24)}\]
Thus the deflection \(\phi\) is proportional to the current \(I\).
This proportionality is exactly what makes the instrument usable. Because \(\phi\) and \(I\) rise together in fixed proportion, the divisions on the scale can be drawn evenly spaced, and the reading can be converted into a current by a single multiplying factor.
In numerical problems from this section you are typically given \(N\), \(A\), \(B\) and \(K\) and asked for the current that produces a stated deflection, or the reverse. Keep the units of \(K\) consistent with the units in which the deflection is measured — if \(K\) is given per degree, the deflection must be substituted in degrees, not in radians.
10.8 Magnetic Dipole Moment
In the preceding section, we have dealt with a current-carrying coil. Notice how often the same combination of quantities kept appearing in the torque expression: the number of turns, the current, and the area. It is convenient to bundle these three into a single quantity that describes the coil once and for all.
This current-carrying coil can be described with a vector \(\vec{\mu}\), its magnetic dipole moment. If \(\hat{n}\) is a unit vector normal to the plane of the coil, the direction of \(\vec{\mu}\) is the direction of \(\hat{n}\), shown in Fig. 10.12 (b).
Diagram
Fig. 10.12 (b) – The rectangular current loop viewed edge-on, with the unit normal \(\hat{n}\) and the magnetic dipole moment vector \(\vec{\mu}\) drawn along it, perpendicular to the plane of the coil.
We can then define the magnitude of \(\vec{\mu}\) as
\[\mu = NIA, \qquad \text{--- (10.25)}\]
where \(N\) is the number of turns of the coil, \(I\) the current passing through the coil in ampere (A), and \(A\) the area enclosed by each turn of the coil in square metre (m\(^2\)). The SI unit of magnetic dipole moment is therefore A m\(^2\), which is the same as J/T.
The Torque, Rewritten
If held in a uniform magnetic field \(\vec{B}\), the torque responsible for the rotation of the coil, according to Eq. (10.23), will be
\[\tau = \mu B \sin\theta,\]
\(\theta\) being the angle between \(\vec{\mu}\) (i.e., \(\hat{n}\)) and \(\vec{B}\). This is simply Eq. (10.23) with \(NIA\) replaced by \(\mu\) — nothing new has been assumed, only the notation has been compressed.
Since the torque is perpendicular to both \(\vec{\mu}\) and \(\vec{B}\), and has magnitude \(\mu B\sin\theta\), it can be written compactly as a cross product:
\[\therefore \vec{\tau} = \vec{\mu} \times \vec{B} \qquad \text{--- (10.26)}\]
The Electrostatic Parallel
You have learnt in Std. XI about the torque on an electric dipole exerted by an electric field \(\vec{E}\):
\[\therefore \vec{\tau} = \vec{p} \times \vec{E} \qquad \text{--- (10.27)}\]
Here \(\vec{p}\) is the electric dipole moment.
The two expressions, Eq. (10.26) and Eq. (10.27), are analogous to each other. A current loop in a magnetic field behaves in the same mathematical way as a pair of equal and opposite charges in an electric field: both are turned until their dipole moment lines up with the field.
This analogy is not a coincidence, and it is worth holding on to. Several results that follow later in this chapter — the potential energy of a dipole, and the field of a current loop at large distances — can be written down almost directly from their electrostatic counterparts by replacing \(\vec{p}\) with \(\vec{\mu}\) and \(\vec{E}\) with \(\vec{B}\).
10.9 Magnetic Potential Energy of a Dipole
A magnetic dipole freely suspended in a magnetic field possesses magnetic potential energy because of its orientation in the field.
The reason is that the field exerts a torque on the dipole, as we saw in Eq. (10.26). To hold the dipole at an orientation other than the one it naturally settles into, work has to be done against that torque — and work done against a restoring effect is stored as potential energy. Turn the dipole to a different angle and its stored energy changes.
You have learnt about an electric dipole in Chapter 8. Electrical potential energy is associated with an electric dipole on account of its orientation in an electric field. It has been shown that the potential energy \(U\) of an electric dipole \(\vec{p}\) in an electric field \(\vec{E}\) is given by
\[U = -\vec{p} \cdot \vec{E} \qquad \text{--- (10.28)}\]
Analogously, the magnetic potential energy of a magnetic dipole \(\vec{\mu}\) in a magnetic field \(\vec{B}\) is given by
\[U = -\vec{\mu} \cdot \vec{B} \qquad \text{--- (10.29)}\]
\[= -\mu B\cos\theta, \qquad \text{--- (10.30)}\]
where \(\theta\) is the angle between \(\vec{\mu}\) and \(\vec{B}\). Here \(\mu\) is in A m\(^2\), \(B\) is in tesla (T), and \(U\) comes out in joule (J).
The dot product in Eq. (10.29) is what produces the \(\cos\theta\) in Eq. (10.30). Note the minus sign carefully — it is not decoration. It is what makes the energy lowest when the dipole is lined up with the field, which is the orientation the dipole tends to move towards.
The Two Extreme Orientations
Since \(\cos\theta\) ranges between \(+1\) and \(-1\), the energy has a definite lowest and highest value, reached at the two ends of that range. These are shown in Fig. 10.14.
Diagram
Fig. 10.14 – Two sketches side by side: Case (i) with \(\vec{\mu}\) parallel to \(\vec{B}\) marked as minimum potential energy, and Case (ii) with \(\vec{\mu}\) antiparallel to \(\vec{B}\) marked as maximum potential energy.
Case (i): If \(\theta = 0\), \(U = -\mu B\cos(0^\circ) = -\mu B\)
This is the minimum potential energy of a magnetic dipole in a magnetic field, i.e., when \(\vec{\mu}\) and \(\vec{B}\) are parallel to each other.
Case (ii): If \(\theta = 180^\circ\), \(U = -\mu \cdot B \cdot \cos(180^\circ) = \mu B\).
This is the maximum potential energy of a magnetic dipole in a magnetic field, i.e., when \(\vec{\mu}\) and \(\vec{B}\) are antiparallel to each other.
Stable and Unstable Positions
These two cases are not merely the largest and smallest numbers the formula can produce; they describe two physically different situations.
At \(\theta = 0\) the dipole sits at the bottom of its energy curve. Disturb it slightly and the torque pushes it back, so this is a position of stable equilibrium. At \(\theta = 180^\circ\) the dipole sits at the top. The torque here is also zero, so it can in principle balance there, but the smallest disturbance sends it swinging around towards the aligned position. This is unstable equilibrium.
10.10 Magnetic Field due to a Current : Biot-Savart Law
In Sections 10.1 and 10.2, we have seen that a magnetic field is produced by a current-carrying wire. So far we have only spoken of that field qualitatively — that it circles the wire, and which way it points. Can we calculate this magnetic field?
Everything so far in this chapter has dealt with what a magnetic field does to a current. From here on we turn the question around and ask what a current does to make a magnetic field.
The Field of a Small Piece of Current
Figure 10.15 shows an arbitrarily shaped wire carrying a current \(I\). Here \(d\vec{l}\) is a length element along the wire, and the current in this element is in the direction of the length vector \(d\vec{l}\).
Diagram
Fig. 10.15 – A current-carrying wire of arbitrary shape with a small length element \(d\vec{l}\) marked, the position vector \(\vec{r}\) from the element to a point P, and the field \(d\vec{B}\) at P shown going into the plane of the paper.
Let us calculate the differential field \(d\vec{B}\) at the point P, produced by the current \(I\) through the length element \(dl\). The strategy is the same one used earlier for the force on a bent wire: no single formula can handle a wire of arbitrary shape, so we cut it into tiny pieces, find the contribution of each, and add.
The net magnetic field at the point P can be obtained by superimposition of the magnetic fields \(d\vec{B}\) at that point due to the different length elements along the wire. This can be done by integrating, i.e., summing up the magnetic fields \(d\vec{B}\) from these length elements.
Experimentally, the magnetic field \(d\vec{B}\) produced by the current \(I\) in the length element \(d\vec{l}\) is
\[dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^2} \qquad \text{--- (10.31)}\]
Here, \(\theta\) is the angle between the directions of \(d\vec{l}\) and \(\vec{r}\), where \(\vec{r}\) is the vector from the length element to the point P, and \(r\) is its magnitude in metre (m). The current \(I\) is in ampere (A), \(dl\) in metre (m), and \(dB\) comes out in tesla (T).
The constant \(\mu_0\) is called the permeability constant, given by
\[\mu_0 = 4\pi \times 10^{-7}\ \text{T. m/A} \qquad \text{--- (10.32)}\]
\[\approx 1.26 \times 10^{-6}\ \text{T. m/A} \qquad \text{--- (10.33)}\]
The direction of \(d\vec{B}\) is dictated by the cross product \(d\vec{l} \times \vec{r}\). Vectorially,
\[d\vec{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\vec{l} \times \vec{r}}{r^3} \qquad \text{--- (10.34)}\]
Equation (10.31) and Eq. (10.34) are known as the Biot and Savart law. This inverse square law is experimentally deduced. It may be noted that this is still an inverse square law, as \(\vec{r}\) appears in the numerator and \(r^3\) in the denominator — one power of \(r\) in the numerator is used up in supplying the direction, leaving \(1/r^2\) for the magnitude.
Using the Biot-Savart law, we can calculate the magnetic field produced by various distributions of currents, as discussed below.
Current in a Straight, Long Wire
You are aware of the right hand thumb rule, which gives the direction of the magnetic field produced by a current flowing in a wire. What we now want is the magnitude.
Figure 10.16 shows a long wire of length \(l\). We want to calculate the magnetic field \(\vec{B}\) at a point P which is at a perpendicular distance \(R\) from the wire.
Diagram
Fig. 10.16 – A long straight wire carrying current \(I\) with a length element \(dl\) at a distance \(l\) from the foot of the perpendicular, the point P at perpendicular distance \(R\), the distance \(r\) from the element to P, and the angle \(\theta\) marked; \(d\vec{B}\) at P is directed into the plane of the paper.
Let us consider a current length element of infinitesimal length \(d\vec{l}\) of the wire. Let \(I\) be the current passing through the wire, situated at a distance \(r\) from the point P. The product \(I\,d\vec{l}\) is called the current element, \(I\) being the current passing through the wire.
Using Eq. (10.31), the magnetic field \(d\vec{B}\) produced at P due to the current length element \(I\,d\vec{l}\) becomes
\[dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^2} \qquad \text{--- (10.35)}\]
Here, the direction of \(d\vec{B}\) is given by the cross product \(d\vec{l} \times \vec{r}\), see Eq. (10.34), and hence it is into the plane of the paper. This is true for every element of the wire, above P as well as below it, which is important: all the contributions point the same way and therefore add up rather than cancelling.
Adding Up the Whole Wire
We now calculate the magnitude of the magnetic field produced at P by all the current length elements in the upper half of the infinitely long wire. We do this by integrating Eq. (10.35) from zero to infinity.
Let us now calculate the magnitude of the magnetic field produced at P by a similar current length element in the lower half of the wire. By symmetry, its magnitude and direction are the same as that due to the current length element in the upper half of the wire. The magnetic field is directed into the plane of the paper.
Adding both the contributions (a) and (b), the total magnetic field \(B\) at point P is
\[B = \int dB = 2 \times \frac{\mu_0}{4\pi}\int_{0}^{\infty}\frac{I\sin\theta\,dl}{r^{2}} \qquad \text{--- (10.36)}\]
The factor 2 in front is exactly the statement that the lower half contributes as much as the upper half.
Before this can be integrated, both \(r\) and \(\sin\theta\) must be expressed in terms of the one variable \(l\), which is what changes as we move along the wire. From the geometry of Fig. 10.16,
\[r = \sqrt{l^{2}+R^{2}}\quad\text{and}\quad \sin\theta = \sin(\pi-\theta) = \frac{R}{r} = \frac{R}{\sqrt{l^{2}+R^{2}}} \qquad \text{--- (10.37)}\]
Substituting these into Eq. (10.36),
\[B = \frac{\mu_0}{2\pi}\int_{0}^{\infty}\frac{I R\,dl}{\left(l^{2}+R^{2}\right)^{3/2}} = \frac{\mu_0 I R}{2\pi}\int_{0}^{\infty}\frac{dl}{\left(l^{2}+R^{2}\right)^{3/2}}\]
The integral works out to \(1/R^{2}\), as evaluated in the note below. Hence
\[B = \frac{\mu_0 I R}{2\pi}\cdot\frac{1}{R^{2}} = \frac{\mu_0 I}{2\pi R} \qquad \text{--- (10.38)}\]
From Eq. (10.38), this is the magnetic field at a point P at a perpendicular distance \(R\) from the infinitely long straight wire. This is due to both the upper semi-infinite part and the lower semi-infinite part of the wire.
Thus, the magnetic field \(B\) due to a semi-infinite straight wire is just half of this:
\[B = \frac{\mu_0 I}{4\pi R} \qquad \text{--- (10.39)}\]
In Eq. (10.38) and Eq. (10.39), the field is inversely proportional to the distance from the wire.
In examinations, this result is used constantly — often combined with the right hand thumb rule to fix the direction, and often with two or more wires whose fields have to be added as vectors at a point. Take care to use the perpendicular distance \(R\) from the wire, not the distance from some other reference point.
10.11 Force of Attraction between two Long Parallel Wires
As an application of the result obtained in the last section, let us obtain the force of attraction between two long, parallel wires separated by a distance \(d\), as shown in Fig. 10.17. Let the currents in the two wires be \(I_1\) and \(I_2\).
Diagram
Fig. 10.17 – Two long parallel wires a distance \(d\) apart, carrying currents \(I_1\) and \(I_2\) in the same direction, with the field of wire 1 shown at the position of wire 2 and the resulting force on wire 2 directed towards wire 1.
This is the experiment described right at the start of the chapter, where two hanging wires were seen to move towards or away from each other. We are now in a position to explain it completely, and to calculate exactly how large the force is.
Two Steps: Field, Then Force
The calculation is done in two stages, using two results already established. First, wire 1 sets up a magnetic field in the space around it. Second, wire 2 sits in that field and, being a current-carrying wire, feels a force.
The magnetic field at the second wire due to the current \(I_1\) in the first one, according to Eq. (10.38), is
\[B = \frac{\mu_0 I_1}{2\pi d} \qquad \text{--- (10.40)}\]
By the right hand rule, the direction of this field is into the plane of the paper.
We now apply the Lorentz force law. Accordingly, the force on wire 2, because of the current \(I_2\) and the magnetic field \(B\) due to the current in wire 1, is given by Eq. (10.13):
\[F = I_2\int dl \cdot B = \frac{\mu_0 I_1 I_2}{2\pi d}\int dl \qquad \text{--- (10.41)}\]
The direction of this force is towards wire 1, i.e., it will be an attractive force.
For infinitely long wires, this force will be infinite — because an infinite length of wire is being acted upon. The sensible quantity to quote is therefore not the total force but the force on each metre of the wire.
Force per unit length of the wire will be
\[\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \qquad \text{--- (10.42)}\]
Here \(I_1\) and \(I_2\) are the currents in ampere (A), \(d\) the separation in metre (m), and \(F/L\) the force per unit length in newton per metre (N/m).
If the currents \(I_1\) and \(I_2\) are antiparallel, the force will be repulsive.
The Force on a Finite Section
Let us consider a section of length \(L\) of wire 2. The force on this section due to the current in wire 1 is given by
\[F = I_2\,B\,L \qquad \text{--- (10.43)}\]
Substituting for \(B\) from Eq. (10.40),
\[= \frac{\mu_0 I_1 I_2}{2\pi d}L = F_{21} \qquad \text{--- (10.44)}\]
We will denote this force by \(F_{21}\), i.e., the force on a section of length \(L\) of wire 2 due to the current in wire 1. Similarly, a section of the same length \(L\) of wire 1 will experience a force due to the current in wire 2. This force we denote as \(F_{12}\), which is equal and opposite to \(F_{21}\).
\[\therefore F_{21} = -F_{12} \qquad \text{--- (10.44 A)}\]
The force of attraction per unit length is then, from Eq. (10.44),
\[\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \qquad \text{--- (10.45)}\]
If the currents \(I_1\) and \(I_2\) are flowing in opposite directions, then there is a force of repulsion on the section of length \(L\) of each of the wires. The magnitude of the repulsive force per unit length of the wire is also given by
\[\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \qquad \text{--- (10.46)}\]
Note that the magnitude is the same in both cases; only the sense of the force changes. Reversing one of the currents reverses the direction of the field at the other wire, and therefore reverses the direction of the force, without altering its size.
We can summarise these results as: parallel currents attract, antiparallel currents repel.
Defining the Ampere
This force gives a way of fixing the size of the unit of current, without reference to any other electrical quantity — the force between two wires can be measured with an ordinary balance.
This definition of the ampere was adopted a few decades ago. Note how small the force is — two wires a metre apart, each carrying a current of one ampere, pull on each other with only about a ten-millionth of a newton per metre. Measuring such a force accurately is delicate work, which is one reason the effect is not obvious in everyday wiring.
10.12 Magnetic Field Produced by a Current in a Circular Arc of a Wire
After considering straight parallel wires, let us obtain the magnetic field at a point produced by a current in a circular arc of a wire.
A circular arc is a particularly convenient shape to work with, because every point of the arc is at the same distance from the centre, and every length element along it is perpendicular to the line joining it to the centre. Both of these facts simplify the Biot-Savart law considerably.
Figure 10.18 depicts a circular arc of a wire (AB), carrying a current \(I\). We can first obtain the magnetic field produced by one current-length element of the arc, and then integrate over the entire arc length.
Diagram
Fig. 10.18 – A circular arc AB of radius \(r\) carrying a current \(I\), subtending an angle \(\theta\) at the centre O, with a length element \(d\vec{l}\) on the arc, the radius vector \(\vec{r}\) from the element to O, and \(d\vec{B}\) at O directed into the plane of the paper.
The circular arc AB subtends an angle \(\theta\) at the centre O of the circle of which the arc is a part, and \(r\) is its radius.
Applying the Biot-Savart Law
Using the Biot-Savart law, Eq. (10.34), the magnetic field produced at O is
\[d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l} \times \vec{r}}{r^3}\]
\[dB = \frac{\mu_0}{4\pi}I\,\frac{dl \cdot r \cdot \sin 90^\circ}{r^3} = \frac{\mu_0}{4\pi}\frac{I\,dl}{r^2} \qquad \text{--- (10.47)}\]
The angle appearing here is \(90^\circ\) because the length element \(d\vec{l}\) along the arc is always perpendicular to the radius \(\vec{r}\) drawn to the centre — a tangent to a circle is always at right angles to the radius at that point. Since \(\sin 90^\circ = 1\) and one factor of \(r\) cancels, the expression reduces to the simple form on the right.
Equation (10.47) gives the magnitude of the field. The direction of the field is given by the right hand rule: aligning the thumb in the direction of the current, the field direction is indicated by the curling fingers. Thus, the direction of each of the \(d\vec{B}\) is into the plane of the paper.
Because every element gives a contribution in the same direction, the vector sum again reduces to an ordinary sum of magnitudes. The total field at O is therefore
\[B = \int dB = \frac{\mu_0}{4\pi}I\int_A^B\frac{dl}{r^2}\]
To carry out this integration, the arc length is written in terms of the angle: an element subtending an angle \(d\theta\) at the centre has length \(dl = r\,d\theta\). Hence
\[= \frac{\mu_0}{4\pi}I\int_0^{\theta}\frac{r}{r^2}\,d\theta = \frac{\mu_0}{4\pi}\frac{I}{r}\theta, \qquad \text{--- (10.48)}\]
where the angle \(\theta\) is in radians. Here \(I\) is in ampere (A), \(r\) in metre (m), and \(B\) in tesla (T).
The Full Circle
The result for a complete loop follows at once, because a full circle is simply an arc that subtends \(2\pi\) radians at the centre.
For a full circular wire carrying a current \(I\), the magnetic field at the centre of the circle, using Eq. (10.48), is
\[B = \frac{\mu_0}{4\pi}\frac{I}{r}\,2\pi\]
\[B = \frac{\mu_0 I}{2r} \qquad \text{--- (10.49)}\]
10.13 Axial Magnetic Field Produced by Current in a Circular Loop
In the previous section we found the field at the centre of a circular loop. Here we shall obtain the magnetic field, due to a current in a circular loop, at different points along its axis. We assume that the current is steady.
The axis of the loop is the straight line through its centre, perpendicular to the plane of the loop. Points on this line are the easiest ones to handle, because the arrangement looks exactly the same from every direction around the axis, and this symmetry does much of the work for us.
Figure 10.19 shows a circular loop of wire carrying a current \(I\). The loop itself is in the \(x\)-\(y\) plane with its centre at the origin O. The radius of the loop, carrying a steady current \(I\), is \(R\). We need to calculate the magnetic field at a point P on the \(z\)-axis, at a distance \(\vec{r}\) from the element \(d\vec{l}\) on the loop.
Diagram
Fig. 10.19 – A circular loop of radius \(R\) in the \(x\)-\(y\) plane carrying a current \(I\), with a length element \(d\vec{l}\) on the loop, the vector \(\vec{r}\) from the element to the point P on the \(z\)-axis, and the field \(d\vec{B}\) at P resolved into the axial component \(dB_z\) and the perpendicular component \(dB_\perp\).
The Field of One Element
Using the Biot-Savart law, the magnitude of the magnetic field \(dB\) is given by
\[dB = \frac{\mu_0}{4\pi}\,I\,\frac{\left|d\vec{l} \times \vec{r}\right|}{r^3} \qquad \text{--- (10.50)}\]
From the right-angled triangle formed by the radius of the loop, the axial distance and the line to P, we have
\[r^2 = R^2 + z^2\]
Any element \(d\vec{l}\) will always be perpendicular to the vector \(\vec{r}\) from the element to the point P. The element \(d\vec{l}\) is in the \(x\)-\(y\) plane, while the vector \(\vec{r}\) is in the \(y\)-\(z\) plane. Hence
\[\left|d\vec{l} \times \vec{r}\right| = dl \cdot r\]
\[\therefore dB = \frac{\mu_0}{4\pi}\frac{I\,dl}{r^2} \qquad \text{--- (10.51)}\]
\[= \frac{\mu_0}{4\pi}\frac{I\,dl}{\left(z^2 + R^2\right)} \qquad \text{--- (10.52)}\]
Why Only the Axial Component Survives
Unlike the case of the centre of the loop, the contributions from different elements now point in different directions, so they cannot simply be added as numbers.
The direction of \(d\vec{B}\) is perpendicular to the plane formed by \(d\vec{l}\) and \(\vec{r}\). Its \(z\) component is \(dB_z\), and the component perpendicular to the \(z\)-axis is \(dB_\perp\).
The components \(dB_\perp\), when summed over the whole loop, yield zero, as they cancel out due to symmetry. This can be easily seen from the diametrically opposite element \(d\vec{l}\), which gives a \(dB_\perp\) opposite to that due to the first element. Hence, only the \(z\) component remains.
\(\therefore\) The net contribution along the \(z\)-axis is obtained by integrating \(dB_z = dB\cos\theta\) over the entire loop.
From Fig. 10.19,
\[\cos\theta = \frac{R}{r} = \frac{R}{\sqrt{z^2 + R^2}}\]
Note that \(\cos\theta\) here involves the radius of the loop, not the axial distance, because the angle is measured between \(d\vec{B}\) and the axis.
Carrying Out the Integration
\[B = \int dB_z = \oint \frac{\mu_0}{4\pi}\frac{I\,dl}{\left(z^2 + R^2\right)}\cos\theta = \frac{\mu_0}{4\pi}\oint\frac{I\,R\,dl}{\left(z^2 + R^2\right)^{3/2}}\]
Everything in this expression except \(dl\) is the same for every element of the loop, so these constants come outside the integral, and \(\oint dl\) is simply the circumference \(2\pi R\):
\[= \frac{\mu_0}{4\pi}\frac{I\,R}{\left(z^2 + R^2\right)^{3/2}}\,2\pi R\]
\[B = \frac{\mu_0 I R^2}{2\left(z^2 + R^2\right)^{3/2}} \qquad \text{--- (10.53)}\]
This is the magnitude of the magnetic field due to a current \(I\) in a loop of radius \(R\), at a point P on the \(z\)-axis of the loop. Here \(R\) and \(z\) are in metre (m), \(I\) in ampere (A), and \(B\) in tesla (T).
The field is therefore strongest at the centre and falls off as one moves away along the axis in either direction. Far from the loop, when \(z\) is much greater than \(R\), the term \(R^2\) in the bracket becomes negligible and the field falls off roughly as \(1/z^3\) — a much faster decrease than for a straight wire. This limiting case is examined more closely in the next section.
In numerical work, note that \(z\) is measured from the centre of the loop along the axis, and that the bracket carries the power \(3/2\) — a common slip is to forget the square root implied by that fractional power.
10.14 Magnetic Lines for a Current Loop
We know that the magnetic field at a point P on the axis is given by Eq. (10.53) as
\[B_z = \frac{\mu_0 I R^2}{2\left(z^2 + R^2\right)^{3/2}}\]
Field at the Centre, and for a Coil of N Turns
As a special case, the field at the centre of the loop is obtained from the above equation by letting \(z = 0\):
\[B_0 = \frac{\mu_0 I}{2R} \qquad \text{--- (10.54)}\]
For a coil of \(N\) turns,
\[B = \frac{\mu_0 N I}{2R} \qquad \text{--- (10.55)}\]
The reason for the factor \(N\) is straightforward. A coil of \(N\) turns is simply \(N\) loops of the same radius stacked closely together, each carrying the same current, and each contributing the same field in the same direction. The contributions therefore add, and the field is \(N\) times that of a single loop.
The Pattern of the Field Lines
The magnetic field lines from a circular loop are depicted in Fig. 10.20. Inside the loop they are crowded together and nearly straight, showing a strong field there; outside, they spread out and curve around, returning to the other face of the loop.
Diagram
Fig. 10.20 – Magnetic field lines for a circular current loop, shown in a plane containing the axis, crowded and parallel near the centre and looping around outside, with the current direction marked on the loop.
The direction of the field is as per the right hand thumb rule: curl the palm of your right hand along the circular wire with the fingers in the direction of the current. The stretched right hand thumb then gives the direction of the magnetic field, as illustrated in Fig. 10.21.
Diagram
Fig. 10.21 – The right hand thumb rule for a current loop: fingers curled along the direction of the current in the loop, thumb pointing along the axis in the direction of the magnetic field.
Notice that this is the same rule as for a straight wire, applied the other way round. For a straight wire the thumb followed the current and the fingers gave the field; for a loop the fingers follow the current and the thumb gives the field.
Thus, the upper part of the loop seen in Fig. 10.20 may be regarded as the North pole and the lower part as the South pole of a bar magnet.
Circular Loop Carrying a Current as a Magnetic Dipole
The behaviour of the magnetic field due to a circular current loop, at large distances, is very similar to that due to the electric field of an electric dipole.
From the above equation for \(B_z\), at a large distance \(z\) from the loop along its axis,
\[z \gg R\]
so that \(R^2\) may be neglected in comparison with \(z^2\) inside the bracket, giving
\[\therefore B_z = \frac{\mu_0 I R^2}{2z^3}\]
The area of the loop is \(A = \pi R^2\), so \(R^{2} = A/\pi\), and substituting this,
\[\therefore B_z = \frac{\mu_0 I A}{2\pi z^3}\quad \text{at } z \gg R \qquad \text{--- (10.56)}\]
The magnetic moment \(\vec{m}\) of a circular loop is defined as \(\vec{m} = I\vec{A}\), where \(\vec{A}\) is a vector of magnitude \(A\) and direction perpendicular to the plane of the loop. This is the same quantity as the magnetic dipole moment introduced in Section 10.8, written here for a single turn.
Using Eq. (10.56),
\[\therefore B_z = \frac{\mu_0}{2\pi}\frac{m}{z^3}\]
\[\vec{B}_z = \frac{\mu_0}{4\pi}\frac{2\vec{m}}{z^3} \qquad \text{--- (10.57)}\]
Note that \(\vec{B}_z\) and \(\vec{m}\) are in the same direction, perpendicular to the plane of the loop.
Using the electrostatic analogue,
\[\vec{E} = \frac{2\vec{p}}{4\pi\varepsilon_0 z^3},\]
which is the electric field at an axial point of an electric dipole.
The two expressions have exactly the same form. Replace \(\vec{p}\) by \(\vec{m}\) and \(\dfrac{1}{4\pi\varepsilon_0}\) by \(\dfrac{\mu_0}{4\pi}\), and one turns into the other. Both fall off as the inverse cube of the distance, which is faster than the inverse square law of a single charge, because the two “ends” of a dipole nearly cancel each other when viewed from far away.
10.15 Ampere’s Law
We know that if a distribution of charges is given, one can obtain the electric field by using the inverse square law. If the distribution of charges is planar, or has spherical or cylindrical symmetry, then with the help of Gauss’ law we can find the net electric field with relative ease.
On a similar note, we can obtain the magnetic field produced by a distribution of currents — not charges. The Biot-Savart law will always give the right answer, but the integrals involved can be long and awkward, as the derivations of the last few sections showed.
Again, if the distribution of currents has some symmetry, then we can use Ampere’s law to find out the magnetic field with fair ease, as you will see below. You have studied the Biot-Savart law and its consequences. Ampere’s law can be derived from the Biot-Savart law. The law is due to André Marie Ampere (1775–1836), after whom the SI unit of current is named.
Statement of the Law
Ampere’s law is:
\[\oint \vec{B} \cdot d\vec{s} = \mu_0 I \qquad \text{--- (10.58)}\]
The \(\oint\) sign indicates that the integral is to be evaluated over a closed loop, called an Amperian loop. The current \(I\) on the right hand side is the net current encircled by the Amperian loop.
An Amperian loop is not a physical object. It is an imaginary closed path that we are free to draw wherever it is convenient — and choosing it cleverly is the whole art of using this law.
In the example shown in Fig. 10.22, cross-sections of four long straight wires carrying currents \(I_1\), \(I_2\), \(I_3\) and \(I_4\) into or out of the plane of the paper are shown. An Amperian loop is drawn to encircle three of the current-carrying wires and not the fourth one.
Diagram
Fig. 10.22 – An irregular closed Amperian loop drawn in the plane of the paper, enclosing three wire cross-sections carrying currents \(I_1\) and \(I_2\) out of the page and \(I_3\) into the page, with a fourth wire carrying \(I_4\) lying outside the loop; a length element \(d\vec{l}\) and the field \(\vec{B}\) are marked on the loop.
As the current goes perpendicular to the plane of the paper, \(\vec{B}\) is in the plane of the paper, even if its direction is unknown. The length element on the Amperian loop is \(d\vec{l}\), also in the plane of the paper.
\[\vec{B} \cdot d\vec{l} = B\,dl\cos\theta,\]
and from Eq. (10.58),
\[\oint \vec{B} \cdot d\vec{l} = \oint B\,dl\cos\theta = \mu_0 I \qquad \text{--- (10.59)}\]
Thus the integration is over the product of the length \(dl\) of the Amperian loop and the component of the magnetic field \(B\cos\theta\) tangent to the loop. Only the tangential component counts; wherever the field runs across the loop rather than along it, it contributes nothing to the integral.
Signs of the Enclosed Currents
We use the curled-palm right hand rule so that we can mark the currents with a positive or a negative sign. Curl the right hand palm along the Amperian loop, with the fingers in the direction of integration. Then a current in the direction of the stretched thumb is assigned a positive sign, and a current in the direction opposite to the stretched thumb is assigned a negative sign.
For the distribution of currents as shown in Fig. 10.22, \(I_1\) and \(I_2\) are coming out of the paper (\(\odot\)), parallel to the stretched thumb. Hence these are positive. \(I_3\), on the other hand, is going into the plane of the paper (\(\otimes\)). Thus, it is negative.
\[\therefore \oint B\cos\theta\,dl = \mu_0\left(I_1 + I_2 - I_3\right) \qquad \text{--- (10.60)}\]
The current \(I_4\) is not within the Amperian loop. As the integration is over a full loop, the contributions of \(I_4\) to \(B\) cancel out.
This last point deserves emphasis. The wire carrying \(I_4\) certainly does produce a magnetic field at every point of the loop, so the field appearing in the integral is not the field of the enclosed currents alone. But when the trip around the closed loop is completed, the positive and negative contributions of the outside current cancel exactly, and only the enclosed currents survive on the right hand side.
Equation (10.58) represents Ampere’s law, or Ampere’s circuital law.
Applying the Law to a Long Straight Wire
As an application of Ampere’s law, let us consider a long straight wire carrying a current \(I\), as in Fig. 10.23. The natural Amperian loop here is a circle centred on the wire, drawn in a plane perpendicular to it, because the field of a straight wire is known to run in exactly such circles.
Diagram
Fig. 10.23 – A long straight wire seen end-on, carrying a current \(I\), with a circular Amperian loop of radius \(r\) around it; the field \(\vec{B}\) and the element \(d\vec{l}\) are tangential to the circle and the angle \(d\theta\) subtended at the centre is marked.
Here \(\vec{B}\) and \(d\vec{l}\) are tangential to the Amperian loop, which is a circle. Since the two vectors are parallel at every point, \(\cos\theta = 1\) and
\[\therefore \vec{B} \cdot d\vec{l} = B\,dl = B\,r\,d\theta,\]
writing the arc length of the circle as \(r\,d\theta\).
The field \(B\) at a distance \(r\) from the wire is given by
\[B = \frac{\mu_0 I}{2\pi r} \qquad \text{--- (10.61)}\]
By symmetry, \(B\) has the same value at every point of this circle, so it comes outside the integral:
\[\therefore \oint B\,dl = \frac{\mu_0 I}{2\pi r}\int_0^{2\pi} r\,d\theta = \mu_0 I \qquad \text{--- (10.62)}\]
This is in agreement with Ampere’s law. Equation (10.61) shows that the magnetic field \(B\) of an infinitely long wire is proportional to the current \(I\) but inversely proportional to the distance from the wire, as seen earlier.
Note what has happened here: a result that took a substitution and a full integration to obtain from the Biot-Savart law in Section 10.10 has been recovered in two lines.
The Place of Ampere’s Law in the Subject
You have studied Gauss’ law in electrostatics as well as in magnetism. The above example shows that if the distribution of currents has a high degree of symmetry — such as the cylindrical symmetry in the case of a long wire — then the magnetic field for the given distribution of currents can be easily calculated. It then becomes unnecessary to solve the integrals which appear in the Biot-Savart law.
We note here that the Biot-Savart law plays a role in magnetostatics that Coulomb’s law plays in electrostatics. On parallel lines, we can say that the role Gauss’ law plays in electrostatics is played by Ampere’s law in magnetostatics.
10.16 Magnetic Field of a Solenoid and a Toroid
Ampere’s law comes into its own when the arrangement of currents has a high degree of symmetry. Two such arrangements are of great practical importance, and both are examined here.
Solenoid
You have learnt about a solenoid in Std. XI qualitatively. Consider a long, closely wound helical coil of conducting wire. We assume that the diameter of the coil is much smaller than its length — this is what makes the coil “long” in the sense required here.
Figure 10.24 shows the schematic diagram of a cross-section of a current-carrying solenoid.
Diagram
Fig. 10.24 – Schematic cross-section of a current-carrying solenoid showing the wire turns as dots (current out of the page) along the top and crosses (current into the page) along the bottom, with closely spaced straight field lines inside and widely spaced curved lines outside.
The density of the magnetic field lines along the axis of the solenoid within the solenoid, and at a certain distance away from the wire, is uniform. Hence the magnetic field \(B\) is parallel to the axis of the solenoid. The lines are widely spaced outside the solenoid, and hence the magnetic field is weak there.
The reason for this pattern is worth seeing. Each turn of the coil is a circular loop, and along the axis their fields all point the same way and reinforce one another. Outside the coil, the returning field lines of the many turns spread through a large volume, and the field there becomes correspondingly feeble.
For a real solenoid of finite length, the magnetic field is uniform and has good strength at the centre, and is comparatively weak outside the coil.
Applying Ampere’s Law to the Solenoid
Let us consider an ideal solenoid as shown in Fig. 10.25. For the application of Ampere’s law, an Amperian loop is drawn as shown in the figure — a rectangle abcd with one side lying along the axis inside the solenoid and the opposite side lying entirely outside it.
Diagram
Fig. 10.25 – Ampere’s law applied to part of a long ideal solenoid: a rectangular Amperian loop abcd with side ab of length \(L\) inside the solenoid parallel to the axis and side cd outside; the dots (.) show the current coming out of the plane of the paper and the crosses (x) show the current going into the plane of the paper, both in the coil of square cross-section wire.
From Eq. (10.58),
\[\oint \vec{B} \cdot d\vec{l} = \mu_0 I\]
Over the rectangular loop abcd, the above integral takes the form
\[\int_a^b \vec{B} \cdot d\vec{l} + \int_b^c \vec{B} \cdot d\vec{l} + \int_c^d \vec{B} \cdot d\vec{l} + \int_d^a \vec{B} \cdot d\vec{l} = \mu_0 I \qquad \text{--- (10.63)}\]
In the above equation, \(I\) is the net current encircled by the loop. Taking the four sides one at a time,
\[\therefore B\!\cdot\!L + 0 + 0 + 0 = \mu_0 I \qquad \text{--- (10.64)}\]
The second and fourth integrals are zero because \(\vec{B}\) and \(d\vec{l}\) are perpendicular to each other along those sides — the field runs along the axis while the path runs across it, so the dot product vanishes. The third integral is zero because outside the solenoid, \(B = 0\). Only the side of length \(L\) inside the solenoid contributes, and along it \(\vec{B}\) is constant and parallel to \(d\vec{l}\).
We can now obtain the net current \(I\) encircled by this loop. If the number of turns is \(n\) per unit length of the solenoid and the current flowing through the wire is \(i\), then the length \(L\) of the loop encloses \(nL\) turns, and the net current coming out of the plane of the paper is
\[I = nLi\]
\(\therefore\) From Eq. (10.64),
\[BL = \mu_0 n L i\]
\[\therefore B = \mu_0 n i \qquad \text{--- (10.65)}\]
Here \(n\) is the number of turns per unit length in m\(^{-1}\), \(i\) the current in ampere (A), and \(B\) the field in tesla (T).
Notice that neither the radius of the solenoid nor the position of the point inside it appears in this result. The field is the same everywhere in the interior — which is exactly what “uniform” means.
Although the above result for \(B\) is obtained for an ideal solenoid, it is also valid for a realistic solenoid, particularly when applied to points in the middle of it, but certainly not to points near the ends. Near an end the field lines have already begun to spread outwards, and the field there falls to roughly half its central value.
Thus, a solenoid can be designed for a specific value of \(B\) by a choice of \(i\) and \(n\).
Toroid
A toroid is a solenoid of finite length bent into a hollow circular tube-like structure, similar to a pressurised rubber tube inside the tyre of a vehicle. In effect, the solenoid is joined end to end so that it has no ends at all.
Schematic of a cross-section of a toroid is shown in Fig. 10.26. By applying Ampere’s law and taking into account the symmetry of this structure, we can obtain the magnetic field along the central axis of the tube in terms of the current. We construct a circular Amperian loop along the central axis of the tube, as shown in the figure.
Diagram
Fig. 10.26 – A toroid with wire wound around a circular tube, showing the circular Amperian loop of radius \(R\) drawn along the central axis of the tube and the magnetic field \(\vec{B}\) tangential to it.
The magnetic field lines are concentric circles in the toroid. The direction of the field is dictated by the direction of the current \(i\) in the coil around the toroid.
Again, by Ampere’s law,
\[\oint \vec{B} \cdot d\vec{l} = \mu_0 I,\]
where \(I\) is the net current encircled by the loop. The field is tangential to the Amperian loop at every point and has the same magnitude all round it, so the integral is simply \(B\) times the circumference \(2\pi R\). Every one of the \(N\) turns of the winding passes through this loop once, so the enclosed current is \(iN\).
\[B\,.\,2\pi R = \mu_0 i N \qquad \text{--- (10.66)}\]
Here \(N\) is the total number of turns in the toroid, as the integration is over the full length of the loop, \(2\pi R\).
\[\therefore B = \frac{\mu_0 i N}{2\pi R} \qquad \text{--- (10.67)}\]
From Eq. (10.67), \(B\) is inversely proportional to \(R\). Thus, unlike the solenoid, the magnetic field is not constant over the cross-section of the toroid.
10.17 Summary
Magnetic Force on Charges
A magnetic field is produced by moving charges, that is, by an electric current. This was first discovered by Oersted, and the close partnership of electric and magnetic fields was later established by Gauss, Henry, Faraday and Maxwell.
The direction of the magnetic field around a straight current-carrying wire is given by the right hand thumb rule: with the thumb along the current, the curled fingers give the direction of the field, which forms closed circles around the wire.
The magnetic force on a charge \(q\) moving with velocity \(\vec{v}\) in a magnetic field \(\vec{B}\) is \(\vec{F}_m = q(\vec{v} \times \vec{B})\); for an electron of charge \(-e\) it is \(\vec{F}_m = -e(\vec{v} \times \vec{B})\).
When both an electric field and a magnetic field are present, the total force is the Lorentz force, \(\vec{F} = q\left[\vec{E} + (\vec{v} \times \vec{B})\right] = \vec{F}_e + \vec{F}_m\).
Consequences of this law: the force is zero if \(\vec{v}\) is parallel to \(\vec{B}\), and zero if the charge is stationary, however strong \(\vec{B}\) may be.
Since \(\vec{F} \cdot \vec{v} = 0\) always, the magnetic force never does work on a moving charge. It can change the direction of motion but never the speed.
The magnitude of the force is \(F = q\left|\vec{v} \times \vec{B}\right| = qvB\sin\theta\), where \(\theta\) is the angle between \(\vec{v}\) and \(\vec{B}\).
The SI unit of magnetic field is the tesla (T), defined through \(B = F/qv\), with unit N s/(C m). Also, \(1\ \text{T} = 10^{4}\) gauss; the gauss is not an SI unit but is a convenient one.
Motion of a Charged Particle in a Magnetic Field
A charge moving perpendicular to a uniform field travels in a circle, the magnetic force supplying the centripetal force: \(qvB = \dfrac{mv^{2}}{R}\), giving the cyclotron formula \(mv = p = qBR\).
If the velocity has a component parallel to \(\vec{B}\), that component is unaffected, since \(\vec{F}_m = \vec{v}_{\parallel} \times \vec{B} = vB\sin(0^\circ) = 0\). Combined with the circular motion of the perpendicular component, the path becomes a helix.
In a cyclotron, crossed electric and magnetic fields are used: the magnetic field makes the ion circle, and a high-frequency alternating voltage across the two dees accelerates it at every crossing of the gap.
The period of revolution is \(T = \dfrac{2\pi m}{qB}\) and the cyclotron frequency is \(f_c = \dfrac{1}{T} = \dfrac{qB}{2\pi m}\). Both are independent of the speed, radius and energy of the particle.
The resonance condition for a cyclotron is \(f_a = f_c\), where \(f_a\) is the frequency of the applied voltage.
The kinetic energy of the extracted particles is \(\text{K.E.} = \dfrac{1}{2}mv^{2} = \dfrac{q^{2}B^{2}R_{exit}^{2}}{2m}\), proportional to the square of the exit radius.
Force and Torque on Current-Carrying Conductors
The force on a straight wire of length \(L\) carrying a current \(I\), placed perpendicular to a field \(B\), is \(F_m = ILB\). In general, \(\vec{F}_m = I\vec{L} \times \vec{B}\).
For a wire of arbitrary shape, \(d\vec{F}_m = I\,d\vec{l} \times \vec{B}\), so that \(\vec{F}_m = I\displaystyle\int d\vec{l} \times \vec{B}\), and for a uniform field \(\vec{F}_m = I\left(\displaystyle\int d\vec{l}\right) \times \vec{B}\).
For a closed circuit in a uniform field, \(\oint_C d\vec{l} = 0\), and therefore \(\vec{F}_m = 0\). The net force on any closed circuit in a uniform magnetic field is zero.
A rectangular loop of sides \(l_1\) and \(l_2\) in a uniform field experiences a torque \(\tau = Il_1l_2B\sin\theta\), where \(\theta\) is the angle between the normal to the plane of the loop and \(\vec{B}\). The forces on the two slanted sides are equal, opposite and collinear, and cancel.
For a flat coil of \(N\) turns and area \(A = l_1l_2\), the torque is \(\tau' = (NIA)B\sin\theta\). This holds for planar coils of any shape in a uniform field.
Moving Coil Galvanometer
A coil of \(N\) turns and area \(A\) is pivoted in a radial magnetic field \(B\), made radial and strong by a soft iron cylindrical core, so that \(\sin\theta = 1\) at all positions and the torque is \(\tau = NIAB\).
A spring provides a restoring torque \(K\phi\), where \(K\) is the torsional constant. At balance, \(K\phi = NIAB\), giving the deflection \(\phi = \left(\dfrac{NAB}{K}\right)I\).
The deflection is thus directly proportional to the current. Modern instruments use digital ammeters and voltmeters instead.
Magnetic Dipole Moment and Potential Energy
A current-carrying coil is described by its magnetic dipole moment \(\vec{\mu}\), of magnitude \(\mu = NIA\), directed along the normal \(\hat{n}\) to the plane of the coil. Its SI unit is A m\(^2\), or J/T.
The torque on such a dipole in a uniform field is \(\vec{\tau} = \vec{\mu} \times \vec{B}\), analogous to the torque \(\vec{\tau} = \vec{p} \times \vec{E}\) on an electric dipole.
The magnetic potential energy of a dipole is \(U = -\vec{\mu} \cdot \vec{B} = -\mu B\cos\theta\).
For \(\theta = 0\), \(U = -\mu B\), the minimum energy, with \(\vec{\mu}\) parallel to \(\vec{B}\). For \(\theta = 180^\circ\), \(U = +\mu B\), the maximum energy, with \(\vec{\mu}\) antiparallel to \(\vec{B}\).
Biot-Savart Law and its Applications
The Biot-Savart law gives the field of a current element: \(dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^{2}}\), and vectorially \(d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l} \times \vec{r}}{r^{3}}\). It is an experimentally deduced inverse square law.
The permeability constant of free space is \(\mu_0 = 4\pi \times 10^{-7}\) T. m/A \(\approx 1.26 \times 10^{-6}\) T. m/A.
Infinitely long straight wire, at perpendicular distance \(R\): \(B = \dfrac{\mu_0 I}{2\pi R}\). For a semi-infinite straight wire, \(B = \dfrac{\mu_0 I}{4\pi R}\). In both cases the field varies inversely as the distance.
Circular arc of radius \(r\) subtending an angle \(\theta\) (in radians) at the centre: \(B = \dfrac{\mu_0}{4\pi}\dfrac{I\theta}{r}\).
Centre of a full circular loop: \(B = \dfrac{\mu_0 I}{2r}\); for a coil of \(N\) turns, \(B = \dfrac{\mu_0 NI}{2R}\).
On the axis of a circular loop of radius \(R\), at a distance \(z\) from the centre: \(B_z = \dfrac{\mu_0 IR^{2}}{2\left(z^{2}+R^{2}\right)^{3/2}}\), which reduces to \(B_0 = \dfrac{\mu_0 I}{2R}\) at the centre.
At large axial distances, \(z \gg R\), the loop behaves as a magnetic dipole: \(B_z = \dfrac{\mu_0 IA}{2\pi z^{3}}\), and with \(\vec{m} = I\vec{A}\), \(\vec{B}_z = \dfrac{\mu_0}{4\pi}\dfrac{2\vec{m}}{z^{3}}\) — analogous to the axial field \(\vec{E} = \dfrac{2\vec{p}}{4\pi\varepsilon_0 z^{3}}\) of an electric dipole. The field lines of a loop resemble those of a bar magnet, one face acting as the North pole and the other as the South pole.
Force between Parallel Currents
The field of wire 1 at wire 2, a distance \(d\) away, is \(B = \dfrac{\mu_0 I_1}{2\pi d}\), and the force on a length \(L\) of wire 2 is \(F_{21} = \dfrac{\mu_0 I_1I_2}{2\pi d}L\), with \(F_{21} = -F_{12}\).
The force per unit length is \(\dfrac{F}{L} = \dfrac{\mu_0 I_1I_2}{2\pi d}\), of the same magnitude in both cases: parallel currents attract, antiparallel currents repel.
The ampere: two infinitely long parallel wires 1 m apart in vacuum, carrying a constant current that produces a force of \(2 \times 10^{-7}\) N per metre of length on each other, carry a current of 1 ampere (A).
Ampere’s Law, Solenoid and Toroid
Ampere’s circuital law: \(\oint \vec{B} \cdot d\vec{s} = \mu_0 I\), where the integral runs over a closed Amperian loop and \(I\) is the net current encircled by it. Signs of the currents are fixed by the curled-palm right hand rule; currents outside the loop contribute nothing to the integral.
Ampere’s law can be derived from the Biot-Savart law, and plays the same role in magnetostatics that Gauss’ law plays in electrostatics, just as the Biot-Savart law corresponds to Coulomb’s law.
Applied to a long straight wire with a circular Amperian loop, it gives back \(B = \dfrac{\mu_0 I}{2\pi r}\).
Inside a long solenoid with \(n\) turns per unit length carrying a current \(i\): \(B = \mu_0 n i\), uniform and along the axis. Outside an ideal solenoid, \(B = 0\). The result applies well to points in the middle of a real solenoid, but not to points near its ends.
For a toroid of \(N\) total turns carrying a current \(i\), with an Amperian loop of radius \(R\) along the central axis: \(B \cdot 2\pi R = \mu_0 iN\), so \(B = \dfrac{\mu_0 iN}{2\pi R}\). Since \(B \propto 1/R\), the field is not constant over the cross-section of a toroid, unlike that of a solenoid.
10.18 Objective Questions
- Choose the correct option.
- A conductor has 3 segments; two straight and of length \(L\) each and a semicircular with radius \(R\). It carries a current \(I\). What is the magnetic field \(B\) at point P?
Diagram
Fig. – A conductor made of a straight segment I of length \(L\), a semicircular segment II of radius \(R\) centred at P, and a straight segment III of length \(L\), all carrying a current \(I\).
- \(\dfrac{\mu_0}{4\pi}\dfrac{I}{R}\) (B) \(\dfrac{\mu_0}{4\pi}\dfrac{I}{R^2}\) (C) \(\dfrac{\mu_0}{4}\dfrac{I}{R}\) (D) \(\dfrac{\mu_0 I}{4\pi}\)
- Figures a and b show two Amperian loops associated with the conductors carrying current \(I\) in the sense shown. The \(\oint \vec{B}\cdot d\vec{l}\) in the cases a and b will be, respectively,
Diagram
Fig. – (a) A current-carrying conductor threading a closed Amperian loop in one sense; (b) the same conductor lying outside a closed Amperian loop.
- \(-\mu_0 I,\ 0\) (B) \(\mu_0 I,\ 0\) (C) \(0,\ \mu_0 I\) (D) \(0,\ -\mu_0 I\)
- A proton enters a perpendicular uniform magnetic field \(B\) at the origin along the positive \(x\) axis with a velocity v as shown in the figure. Then it will follow the following path. [The magnetic field is directed into the paper.]
Diagram
Fig. – A region of uniform magnetic field directed into the plane of the paper, with a proton entering at the origin O along the positive \(x\) axis with velocity v.
- It will continue to move along the positive \(x\) axis.
- It will move along a curved path, bending towards the positive \(x\) axis.
- It will move along a curved path, bending towards the negative \(y\) axis.
- It will move along a sinusoidal path along the positive \(x\) axis.
- A conducting thick copper rod of length 1 m carries a current of 15 A and is located on the Earth’s equator. There the magnetic flux lines of the Earth’s magnetic field are horizontal, with the field of \(1.3 \times 10^{-4}\) T, south to north. The magnitude and direction of the force on the rod, when it is oriented so that current flows from west to east, are
- \(14 \times 10^{-4}\) N, downward.
- \(20 \times 10^{-4}\) N, downward.
- \(14 \times 10^{-4}\) N, upward.
- \(20 \times 10^{-4}\) N, upward.
- A charged particle is in motion having initial velocity \(\vec{v}\) when it enters into a region of uniform magnetic field perpendicular to \(\vec{v}\). Because of the magnetic force the kinetic energy of the particle will
- remain unchanged. (B) get reduced. (C) increase. (D) be reduced to zero.
10.19 Check Your Concepts
A magnetic field can bend the path of a moving charged particle, but it can never change the particle’s speed. Explain why, using the relation between the magnetic force and the velocity.
A charge is placed at rest in a very strong uniform magnetic field and feels no force at all. The same charge, set moving, immediately experiences a force. What does this tell you about the difference between the electric part and the magnetic part of the Lorentz force?
In the hanging-wire experiment described at the beginning of the chapter, the wires attract or repel as soon as the current starts. Why can this force not be electrostatic in origin, even though the current consists of moving electrons?
A charged particle enters a uniform magnetic field at an angle of \(40^\circ\) to the field direction. Describe the resulting path, and explain which component of the velocity is responsible for each feature of that path.
Why is the frequency of revolution of a charged particle in a cyclotron independent of its speed and of the radius of its path? Explain how this fact makes the design of the machine possible.
A cyclotron designed to accelerate protons is to be used for alpha particles instead, with the same magnetic field. Explain what change must be made to the frequency of the applied alternating voltage, and why.
A closed loop of wire of irregular shape carries a current and is placed in a uniform magnetic field. What is the net force on it? Would your answer change if the same loop were placed near the pole of a strong bar magnet? Give reasons.
Two equal and opposite forces act on the opposite arms of a rectangular current loop in a uniform magnetic field, yet the loop rotates. Explain how this is possible without a net force.
Why is the magnetic field in a moving coil galvanometer made radial by using a soft iron cylindrical core? What would the scale of the instrument look like if an ordinary uniform field were used instead?
A magnetic dipole is placed in a uniform magnetic field, first parallel to the field and then antiparallel to it. In both positions the torque on it is zero. Explain why only one of these positions is stable.
The Biot-Savart law is an inverse square law, yet the field of a long straight wire falls off only as \(1/R\). Explain why there is no contradiction.
State the essential difference between Coulomb’s law and the Biot-Savart law as regards the direction of the field produced. How does this difference show up in the shapes of electric and magnetic field lines?
Two long parallel wires carrying currents in the same direction attract each other, whereas two like electric charges repel each other. Why is there no inconsistency between these two facts?
In applying Ampere’s law, a current lying outside the Amperian loop contributes nothing to \(\oint \vec{B}\cdot d\vec{l}\). Does this mean that such a current produces no magnetic field at points on the loop? Justify your answer.
Ampere’s law is always true, but it can be used to calculate \(\vec{B}\) only in certain situations. State the condition that must be satisfied, and explain why the law cannot be used to find the field of a short, finite straight wire.
The magnetic field inside a long solenoid is uniform, whereas inside a toroid it varies across the cross-section. Account for this difference, referring to the expressions obtained for the two cases.
Explain why a circular current-carrying loop behaves, when viewed from a large distance along its axis, exactly like a small bar magnet. State one fundamental way in which a magnetic dipole nevertheless differs from an electric dipole.
A coaxial cable carries equal and opposite currents in its inner and outer conductors. Show, using Ampere’s law, why the magnetic field outside the cable is zero, and explain why this property makes such cables useful for carrying signals.
Two solenoids are wound with the same total number of turns and carry the same current, but one is twice as long as the other. Which produces the stronger field along its axis, and why?
A current-carrying wire is placed in a magnetic field and experiences no force at all, although both the current and the field are large. What can you conclude about the orientation of the wire? Support your answer with the appropriate expression.
10.20 Practice with Numericals
An alpha particle carrying a charge of \(3.2 \times 10^{-19}\) C moves with a speed of \(4.0 \times 10^{5}\) m s\(^{-1}\) in a uniform magnetic field of 0.60 T. Calculate the magnitude of the magnetic force on it when its velocity makes an angle of (a) \(90^\circ\), (b) \(45^\circ\) and (c) \(180^\circ\) with the field.
A magnetic field is quoted as 2500 gauss. Express it in tesla, and calculate the force experienced by an electron moving at \(8.0 \times 10^{5}\) m s\(^{-1}\) perpendicular to this field. (Charge on electron \(= 1.6 \times 10^{-19}\) C)
A proton moves in a circular path of radius 0.18 m in a uniform magnetic field of 0.45 T. Calculate its linear momentum and its speed. (Mass of proton \(= 1.67 \times 10^{-27}\) kg, charge \(= 1.6 \times 10^{-19}\) C)
A deuteron of mass \(3.34 \times 10^{-27}\) kg and charge \(1.6 \times 10^{-19}\) C moves in a uniform magnetic field of 0.90 T. Calculate its period of revolution and its cyclotron frequency.
In a cyclotron operating with a magnetic field of 1.2 T, protons leave through the exit slit at a radius of 0.45 m. Calculate the maximum kinetic energy of the extracted protons in joule and in MeV. (Mass of proton \(= 1.67 \times 10^{-27}\) kg, charge \(= 1.6 \times 10^{-19}\) C, 1 eV \(= 1.6 \times 10^{-19}\) J)
An electron enters a uniform magnetic field of \(4.0 \times 10^{-3}\) T with a speed of \(6.0 \times 10^{6}\) m s\(^{-1}\), its velocity making an angle of \(30^\circ\) with the field. Calculate the component of its velocity perpendicular to the field and the radius of the circular part of its motion. (Mass of electron \(= 9.1 \times 10^{-31}\) kg, charge \(= 1.6 \times 10^{-19}\) C)
A straight wire of length 0.60 m carrying a current of 12 A is placed in a uniform magnetic field of 0.35 T. Calculate the force on the wire when it is placed (a) perpendicular to the field, (b) at \(45^\circ\) to the field and (c) along the field.
A horizontal wire of mass 15 g and length 0.80 m is to be held in mid-air by the magnetic force on it, in a horizontal magnetic field of 0.50 T perpendicular to the wire. Calculate the current that must be passed through it. (Take \(g = 9.8\) m s\(^{-2}\))
A flat circular coil of 75 turns, each of radius 6.0 cm, carries a current of 1.8 A and is placed in a uniform magnetic field of 0.40 T. Calculate the torque on the coil when the normal to its plane makes an angle of \(60^\circ\) with the field, and the maximum possible torque on it.
A moving coil galvanometer has a coil of 45 turns and area \(6.0 \times 10^{-4}\) m\(^2\), suspended in a radial magnetic field of 0.06 Wb/m\(^2\). The torsional constant of the spring is \(2.5 \times 10^{-9}\) N m per degree. Calculate the deflection produced by a current of \(4.0 \times 10^{-5}\) A.
A circular coil of 150 turns and radius 7.0 cm carries a current of 0.80 A. Calculate its magnetic dipole moment, and the potential energy of the coil when its moment is antiparallel to a uniform magnetic field of 0.20 T.
A magnetic dipole of moment \(6.0 \times 10^{-2}\) A m\(^2\) is placed in a uniform magnetic field of 0.35 T. Calculate the work that must be done to turn it from the position of minimum potential energy to a position in which its moment makes an angle of \(60^\circ\) with the field.
A long straight wire carries a current of 18 A. Calculate the magnetic field at perpendicular distances of 3.0 cm and 9.0 cm from the wire. (\(\mu_0 = 4\pi \times 10^{-7}\) T m/A)
Two long parallel wires 5.0 cm apart carry currents of 8.0 A and 12 A in opposite directions. Calculate (a) the force per unit length between them, stating whether it is attractive or repulsive, and (b) the magnitude of the resultant magnetic field at a point midway between the wires.
A wire is bent into a circular arc of radius 8.0 cm subtending an angle of \(120^\circ\) at its centre and carries a current of 6.0 A. Calculate the magnetic field produced at the centre of the arc. (\(\mu_0 = 4\pi \times 10^{-7}\) T m/A)
A circular coil of 40 turns and radius 10 cm carries a current of 3.0 A. Calculate the magnetic field (a) at the centre of the coil and (b) at a point on its axis 10 cm from the centre.
A solenoid 50 cm long is wound with 1500 turns of wire and carries a current of 4.0 A. Calculate the magnetic field at a point well inside the solenoid, and state what the field would be just outside an ideal solenoid.
A toroid of mean radius 12 cm is wound with 900 turns of wire. What current must be passed through the wire to produce a magnetic field of \(3.0 \times 10^{-2}\) T along its central axis? (\(\mu_0 = 4\pi \times 10^{-7}\) T m/A)
An Amperian loop encloses four long straight wires carrying currents of 12 A and 7.0 A out of the plane of the paper, and 9.0 A and 4.0 A into the plane of the paper. A fifth wire carrying 15 A lies outside the loop. Calculate the value of \(\oint \vec{B}\cdot d\vec{l}\) around the loop.
A circular coil of radius 5.5 cm is to produce a magnetic field of \(2.0 \times 10^{-4}\) T at its centre with a current of 1.2 A. Calculate the number of turns required in the coil. (\(\mu_0 = 4\pi \times 10^{-7}\) T m/A)
A piece of straight wire has mass 20 g and length 1 m. It is to be levitated using a current of 1 A flowing through it and a perpendicular magnetic field \(B\) in a horizontal direction. What must be the magnitude of \(B\)?
Calculate the value of the magnetic field at a distance of 2 cm from a very long straight wire carrying a current of 5 A. (Given: \(\mu_0 = 4\pi \times 10^{-7}\) Wb/Am)
An electron is moving with a speed of \(3.2 \times 10^{7}\) m/s in a magnetic field of \(6.00 \times 10^{-4}\) T perpendicular to its path. What will be the radius of the path? What will be the frequency and the kinetic energy in keV? (Given: mass of electron \(= 9.1 \times 10^{-31}\) kg, charge \(e = 1.6 \times 10^{-19}\) C, 1 eV \(= 1.6 \times 10^{-19}\) J)
An alpha particle (the nucleus of a helium atom, with charge \(+2e\)) is accelerated and moves in a vacuum tube with kinetic energy \(= 10.00\) MeV. On applying a transverse uniform magnetic field of 1.851 T, it follows a circular trajectory of radius 24.60 cm. Obtain the mass of the alpha particle. (Charge of electron \(= 1.62 \times 10^{-19}\) C)
Two wires shown in the figure are connected in a series circuit and the same amount of current of 10 A passes through both, but in opposite directions. The separation between the two wires is 8 mm. The length AB is \(S = 22\) cm. Obtain the direction and magnitude of the magnetic field due to the current in wire 2 on the section AB of wire 1. Also obtain the magnitude and direction of the force on wire 1. (\(\mu_0 = 4\pi \times 10^{-7}\) T.m/A)
Diagram
Fig. – Two parallel wires 8 mm apart carrying equal currents of 10 A in opposite directions, with the section AB of length \(S = 22\) cm marked on wire 1.
A very long straight wire carries a current of 5.2 A. What is the magnitude of the magnetic field at a distance of 3.1 cm from the wire? (\(\mu_0 = 4\pi \times 10^{-7}\) T.m/A)
Current of equal magnitude flows through two long parallel wires having a separation of 1.35 cm. If the force per unit length on each of the wires is \(4.76 \times 10^{-2}\) N, what must be \(I\)?
The magnetic field at a distance 2.4 cm from a long straight wire is 16 \(\mu\)T. What must be the current through the wire?
The magnetic field at the centre of a circular current-carrying loop of radius 12.3 cm is \(6.4 \times 10^{-6}\) T. What will be the magnetic moment of the loop?
A circular loop of radius 9.7 cm carries a current of 2.3 A. Obtain the magnitude of the magnetic field (a) at the centre of the loop and (b) at a distance of 9.7 cm from the centre of the loop but on the axis.
A circular coil of wire is made up of 100 turns, each of radius 8.0 cm. If a current of 0.40 A passes through it, what will be the magnetic field at the centre of the coil?
For proton acceleration, a cyclotron is used in which a magnetic field of 1.4 Wb/m\(^2\) is applied. Find the time period for reversing the electric field between the two Ds.
A moving coil galvanometer has been fitted with a rectangular coil having 50 turns and dimensions 5 cm \(\times\) 3 cm. The radial magnetic field in which the coil is suspended is of 0.05 Wb/m\(^2\). The torsional constant of the spring is \(1.5 \times 10^{-9}\) Nm/degree. Obtain the current required to be passed through the galvanometer so as to produce a deflection of \(30^\circ\).
A solenoid of length \(\pi\) m and 5 cm in diameter has a winding of 1000 turns and carries a current of 5 A. Calculate the magnetic field at its centre along the axis.
A toroid of narrow radius of 10 cm has 1000 turns of wire. For a magnetic field of \(5 \times 10^{-2}\) T along its axis, how much current is required to be passed through the wire?
In a cyclotron, protons are to be accelerated. The radius of its D is 60 cm and its oscillator frequency is 10 MHz. What will be the kinetic energy of the proton thus accelerated? (Proton mass \(= 1.67 \times 10^{-27}\) kg, \(e = 1.6 \times 10^{-19}\) C, 1 eV \(= 1.6 \times 10^{-19}\) J)
A wire loop of the form shown in the figure carries a current \(I\). Obtain the magnitude and direction of the magnetic field at P. (Use the expression given in the original figure for the contribution of segment II.)
Diagram
Fig. – A wire loop made up of straight radial segments and circular arcs of radius \(R\), carrying a current \(I\), with the point P at the common centre of the arcs.
Two long parallel wires going into the plane of the paper are separated by a distance \(R\), and carry a current \(I\) each in the same direction. Show that the magnitude of the magnetic field at a point P equidistant from the wires and subtending an angle \(\theta\) from the plane containing the wires, is \(B = \dfrac{\mu_0 I}{\pi R}\sin 2\theta\). What is the direction of the magnetic field?
The figure shows a section of a very long cylindrical wire of diameter \(a\), carrying a current \(I\). The current density, which is in the direction of the central axis of the wire, varies linearly with radial distance \(r\) from the axis according to the relation \(J = J_0 r/a\). Obtain the magnetic field \(B\) inside the wire at a distance \(r\) from its centre.
Diagram
Fig. – Cross-section of a long cylindrical wire of diameter \(a\) carrying a current \(I\) along its axis, with a circular Amperian loop of radius \(r\) drawn inside the wire.
- In the above problem, what will be the magnetic field \(B\) inside the wire at a distance \(r\) from its axis, if the current density \(J\) is uniform across the cross-section of the wire?