Topics

The electric fields and magnetic fields considered up to now have been produced by stationary charges and moving charges (currents), respectively.

It was observed that when a closed current carrying loop can set up a magnetic field. By nature of symmetry an obvious question comes in mind, can a current be generated with the help of magnetic field?

Answer to this care in 1831 when michacl faraday discovered that a change in magnetic filed can produce a current in a loop. The phenomenon is known as electromagnetic induction. Figure illustrates Faraday's experiments.

Faraday's Experiment

Faraday observed that no current is registered in the galvanometer when bar magnet is stationary with respect to the loop. However, a current is induced in the loop when a relative motion exists between the bar magnet and the loop. In particular, the galvanometer deflects in one direction as the magnet approaches the loop, and the opposite direction as it moves away.

Faraday concluded that whenever the flux of magnetic field through the area bounded by a closed conducting loop changes, an emf is produced in the loop.

Lenz's Law (Deciding direction of inducted e.m.f.)

The induced emf resulting from a changing magnetic ftux has a polarity that leads to an induced current whose direction is such that the induced magnetic field opposes the original ftux change. We shall discuss later on how it is in accordance with conservation of energy

Practice Exercise

Q. 1 Determine the direction of the induced current for the following situations.

(a)

(b)

Q. 2 A metal ring is placed near a solenoid, as shown in Figure. Find direction of induced current in the ring
a. at the instant the switch in the circuit containing the solenoid is thrown closed,
b. after the switch has been closed for several seconds, and
c. at the instant the switch is thrown open.

Q. 3 Two circular loops \(A\) and \(B\) have their planes parallel to each other, as shown in Figure.

Loop \(A\) has a current flowing in the counterclockwise direction, viewed from above.
(a) If the current in loop \(A\) decreases with time, what is the direction of the induced current in loop \(B\) ? Will the two loops attract or repel each other?
(b) If the current in loop \(A\) increases with time, what is the direction of the induced current in loop \(B\) ? Will the two loops attract or repel each other?
Q. 4 Fig. illustrates plane figures made of thin conductors which are located in a uniform magnetic field directed away from a reader beyond the plane of the drawing. The magnetic induction starts diminishing. Find how the currents induced in these loops are directed.

(a)

(b)

(c)

(d)

Answers

Q. 1 (a) anticlockwise when seen from the side of magnet (b) clockwise when seen from the side of magnet
Q. 2 (a) clockwise when seen from the side of solenoid
(b) zero
(c) anticlockwise when seen from the side of solenoid
Q. 3 (a) anticlockwise when seen from above, attract (b) clockwise when seen from above , repel
Q. 4 (a) In the round conductor the current flows clockwise there is no current in the connector ; (b) in the outside conductor, clockwise ; (c) in both round conductors, clockwise ; no current in the connector, (d) in the left-hand side of the figure eight, clockwise

First figure (below) shows a bar magnet moving toward a conducting loop. It is the motion of the bar magnet to the right that induces an emf and current in the loop. Lenz's law tells us that this induced emf and current must be in a direction to oppose the motion of the bar magnet. That is, the current induced in the loop produces a magnetic field of its own, and this magnetic field must exert a force to the left on the approaching bar magnet.

Second figure shows the induced magnetic moment of the current loop when the magnet is moving toward it. The loop acts like a small magnet with its north pole to the left and its south pole to the right. Because like poles repel, the induced magnetic moment of the loop repels the bar magnet; that is, it opposes its motion toward the loop. This result means the direction of the induced current in the loop must be as shown in Second figure .
Suppose the induced current in the loop shown in Second figure was opposite to the direction shown. Then there would be a magnetic force toward the right on the approaching bar magnet, causing the bar magnet to gain speed. This gain in speed would cause an increase in the induced current, which in turn would cause the force on the bar magnet to increase, and so on. This result is too good to be true. Any time we nudge a bar magnet toward a conducting loop it would move toward the loop with ever increasing speed and with no significant effort on our part. Were this situation to occur, it would be a violation of energy conservation. The reality, however, is that energy is conserved, and Lenz's law is consistent with this reality.

Faraday's Law

Whenever the flux of magnetic field through the area bounded by a closed conducting loop changes, an emf is produced in the loop. The emf is given by

\[:\ \varepsilon = - \frac{d\Phi_{B}}{dt}\]

Note:

(1) The -ve sign is according to Lenz's law (opposition)
(2) For a coil that consists of \(N\) loops, the induced emf would be:

\[\varepsilon = - N\frac{d\Phi_{B}}{dt}\]

(3) Average emf is is given by

\[\varepsilon = - \frac{\Delta\Phi_{B}}{\Delta t}\]

(4) We can see that an emf may be induced in the following ways:
(i) by varying the magnitude of \(\overrightarrow{B}\) with time (illustrated in Figure)

(ii) by varying the magnitude of \(\overrightarrow{A}\), i.e., the area enclosed by the loop with time (illustrated in Figure)

(iii) varying the angle between \(\overrightarrow{B}\) and the area vector \(\overrightarrow{A}\) with time (illustrated in Figure)

(5) Induced current is given by

\[i = - \frac{1}{R}\frac{d\Phi_{B}}{dt}\]

(6) Charge flown through a coil

\[\begin{matrix} & \Delta q = \int_{}^{}\ idt = \int_{t_{1}}^{t_{2}}\mspace{2mu}\mspace{2mu} - \frac{1}{R}\frac{\text{ }d\Phi}{dt}dt = \frac{1}{R}\int_{\Phi_{1}}^{\Phi_{2}}\mspace{2mu}\mspace{2mu} - d\Phi \\ & \Delta q = \frac{\Phi_{1} - \Phi_{2}}{R} \end{matrix}\]

(7) Heat developed will be

\[H = \int_{}^{}\ i^{2}Rdt\]

(8) You can also evaluate the direction of \(\xi\) from equation \(\varepsilon = - \frac{d\Phi_{B}}{dt}\)

The procedure to decide the direction is as follows:
Put an arrow on the loop to choose the positive sense of current. This choice is arbitrary. Using righthand thumb rule find the positive direction of the normal to the area bounded by the loop. If the fingers curl along the loop in the positive sense, the thumb represents the positive direction of the normal. Calculate the flux through the area bounded by the loop. If the flux increases with time, \(\frac{d\Phi_{B}}{dt}\) is positive and \(\varepsilon\) is negative from equation \(\varepsilon = - \frac{d\Phi_{B}}{dt}\). Correspondingly, the current is negative. It is, therefore, in the direction opposite to the arrow put on the loop. If the flux decreases with time, \(\frac{d\Phi_{B}}{dt}\) is negative, \(\varepsilon\) is positive and the current is along the arrow.

Practice Exercise

Q. 1 A bar magnet falls through a circular loop, as shown in Figure.
(a) Describe qualitatively the change in magnetic flux through the loop when the bar magnet is above and below the loop.
(b) Make a qualitative sketch of the graph of the induced current in the loop as a function of time, choosing \(I\) to be positive when its direction is counterclockwise as

viewed from above.
Q. 2 A small coil is introduced between the poles of an electromagnet so that its axis coincides with the magnetic field direction. The cross-sectional area of the coil is equal to \(S = 3.0{\text{ }mm}^{2}\), the number of turns is \(N = 60\). When the coil turns through \(180^{\circ}\) about its diameter, a ballistic galvanometer connected to the coil indicates a charge \(q = 4.5\mu C\) flowing through it. Find the magnetic induction magnitude between the poles provided the total resistance of the electric circuit equals R=40 \(\Omega\).
Q. 3 A connector AB can slide without friction along a shaped conductor located in a horizontal plane. The connector has a length 1 , mass \(m\), and resistance \(R\). The whole system is located in a uniform magnetic field of induction B directed vertically. At the moment \(t = 0\) a constant horizontal force F starts acting on the connector shifting it translationwise to the right. Find how the velocity of the connector varies with timet.

Q. 4 In the previous question find the velocity of rod as a function of time, if initially it hasbeen given a velocity vo towards right and F is not acting.
Q. 5 A copper connector of mass \(m\) slides down on two smooth copper bars, set at an angle \(\alpha\) to the horizontal, due to gravity (Fig). At the top the bars are interconnected through a resistance R. The separation between the bars is equal to \(l\). The system is located in a uniform magnetic field of induction B , perpendicular to the plane in which the connector slides. The resistances of the bars, the connector and the sliding contacts, as well as the self-inductance of the loop, are assumed to be negligible. Find the steady-state velocity of the connector.

Answers

Q. 1 (a)

(b)

Q. \(2\text{ }B = 1/2R/SN = 0.5\text{ }T\)
Q. \(3v = \frac{FR}{B^{2}I^{2}}\left( 1 - e^{- \frac{B{\overline{2}}^{2}}{mR^{t}}} \right)\)
Q. \(4\ v = v_{0}e^{- \frac{B^{2}I^{2}}{mR}t}\)
Q. \(5v = \frac{mgRsin\alpha}{B^{2}l^{2}}\)

One of the most important applications of Faraday's law of induction is to generators and motors. A generator converts mechanical energy into electric energy, while a motor converts electrical energy into mechanical energy.

Figure : (a) A simple generator. (b) The rotating loop as seen from above.
Figure (a) is a simple illustration of a generator. It consists of an \(N\)-turn loop rotating in a magnetic field which is assumed to be uniform. The magnetic flux varies with time, thereby inducing an emf. From Figure (b), we see that the magnetic flux through the loop may be written as

\[\Phi_{B} = \overrightarrow{B} \cdot \overrightarrow{\text{ }A} = BAcos\theta = BAcos\omega t\]

The rate of change of magnetic flux is

\[\frac{d\Phi_{B}}{dt} = - BA\omega sin\omega t\]

Since there are N turns in the loop, the total induced emf across the two ends of the loop is

\[\varepsilon = - N\frac{\text{ }d\Phi_{B}}{dt} = NBA\omega sin\omega t\]

If we connect the generator to a circuit which has a resistance \(R\), then the current generated in the circuit
is given by

\[I = \frac{|\varepsilon|}{R} = \frac{NBA\omega}{R}sin\omega t\]

The current is an alternating current which oscillates in sign and has an amplitude \(I_{0} = NBA\omega/R\). The power delivered to this circuit is \(P = I|\varepsilon| = \frac{(NBA\omega)^{2}}{R}\sin^{2}\omega t\)

On the other hand, the torque exerted on the loop is

\[\tau = \mu Bsin\theta = \mu Bsin\omega t\]

Thus, the mechanical power supplied to rotate the loop is

\[P_{m} = \tau\omega = \mu B\omega sin\omega t\]

Since the dipole moment for the N-turn current loop is

\[\mu = NIA = \frac{N^{2}{\text{ }A}^{2}\text{ }B\omega}{R}sin\omega t\]

the above expression becomes

\[P_{m} = \left( \frac{N^{2}A^{2}B\omega}{R}sin\omega t \right)B\omega sin\omega t = \frac{(NAB\omega)^{2}}{R}\sin^{2}\omega t\]

As expected, the mechanical power in put is equal to the electric power output.

Motional EMF

Consider a conducting bar of length \(l\) moving through a uniform magnetic field which points into the page, as shown in Figure. Particles with charge \(q > 0\) inside experience a magnetic force \({\overrightarrow{F}}_{B} = q\overrightarrow{v} \times \overrightarrow{B}\) which tends to push them upward, leaving negative charges on the lower end.

Figure : A conducting bar moving through a uniform magnetic field

The separation of charge gives rise to an electric field \(\overrightarrow{E}\) inside the bar, which in turn produces a downward electric force \({\overrightarrow{F}}_{c} = q\overrightarrow{E}\). At equilibrium where the two forces cancel,
we have \(qvB = qE\) or \(E = vB\). Between the two ends of the conductor, there exists a potential difference given by

\[V_{ab} = V_{a} - V_{b} = \varepsilon = E\mathcal{l} = B\mathcal{l}v\]

Since \(\varepsilon\) arises from the motion of the conductor, this potential difference is called the motional emf. In general, motional emf around a closed conducting loop can be written as

\[|\varepsilon| = \mid \text{∮}\ \ (\overrightarrow{v} \times \overrightarrow{B}) \cdot d\overrightarrow{\text{ }s}\]

where \(d\overrightarrow{s}\) is a differential length element.

Solution:

The rod is supposed to be the combination of a number of infinitesimal elements. Speed of each element is different. Consider an element at a distance \(\mathcal{l}\) from \(O\).

Speed of this element is \(\omega\mathcal{l}\), and is perpendicular to its length
\[\Rightarrow \varepsilon = \int vBd\mathcal{l}\]

\[\begin{matrix} \text{~}\text{As}\text{~}v = \omega\mathcal{l} \\ \varepsilon = \omega B\int_{0}^{L}\mspace{2mu}\mspace{2mu}\mathcal{l}\text{ }d\mathcal{l} = \frac{1}{2}\omega{BL}^{2} \\ \text{~}\text{'}\text{~}O\text{~}\text{' turns out to be '}\text{~} + \text{~}\text{' and}\text{~}P\text{~}\text{the}\text{~} - \text{~}\text{ve}\text{ terminal}\text{~} \end{matrix}\]

Practice Exercise

Q. 1 For the situation shown, find the emf induced in the loop. Explain your answer by both flux theory and theory of motional emf.

Q. 2 Figure shows a straight, long wire carrying a current \(i\) and a rod of length \(l\) coplanar with the wire and perpendicular to it. The rod moves with a constant velocity v in a direction parallel to the wire. The distance of the wire from the near end of the rod is a. Find the motional emf induced in the rod.

Q. 3 A rectangular loop with a sliding connector of length \(l\) is located in a uniform magnetic field perpendicular to the loop plane (Fig.). The magnetic induction is equal to \(B\). The connector has an electric resistance \(R\), the sides \(AB\) and \(CD\) have resistances \(R_{1}\) and \(R_{2}\) respectively. Neglecting the self-inductance of the loop, find the current flowing in the connector during its motion with a constant velocity v .

Q. 4 A square frame with side \(a\) and a long straight wire carrying a current \(I\) are located in the same plane as shown in Fig. The frame translates to the right with a constant velocity v . Find the emf induced in the frame as a function of distance \(x\).

Q. 5 A spherical conducting shell is placed in a vertical time-varying uniform magnetic field. Is there an induced current along the equator?
Q. 6 A metal rod of mass \(m\) can rotate about a horizontal axis \(O\), sliding along a circular conductor of radius a (Fig.). The arrangement is located in a uniform magnetic field of induction B directed perpendicular to the ring plane. The axis and the ring are connected to an emf source to form a circuit of resistance \(R\). Neglecting the friction, circuit inductance, and ring resistance, find the law according to which the source emf must vary to make the rod rotate with a constant angular velocity \(\omega\).

Answers

Q. 1 Zero
Q. \(2\varepsilon = \frac{\mu_{0}IV}{2\pi}ln\left( 1 + \frac{\mathcal{l}}{a} \right)\)
Q. \(3I = Bvl/\left( R + R_{m} \right)\), where \(R_{m} = R_{1}R_{2}/\left( R_{1} + R_{2} \right)\)
Q. \(4\ \xi_{i} = \frac{\mu_{0}}{4\pi}\frac{2{Ia}^{2}v}{x(x + a)}\)
Q. 5 Yes Q. \(6\ \xi_{i} = \frac{\omega a^{3}{\text{ }B}^{3} + 2mgsin\omega t}{2aB}\)

We have seen that the electric potential difference between two points \(A\) and \(B\) in an electric field \(\overrightarrow{E}\) can be written as

\[\Delta V = V_{B} - V_{A} = - \int_{A}^{B}\mspace{2mu}\overrightarrow{E} \cdot \text{ }d\overrightarrow{\text{ }s}\]

When the electric field is conservative, as is the case of electrostatics, the line integral of it is path-independent, which implies \(\text{∮}\ \overrightarrow{E} \cdot d\overrightarrow{s} = 0\)

Faraday's law shows that as magnetic flux changes with time, an induced current begins to flow. What
causes the charges to move? It is the induced emf which is the work done per unit charge. However, since magnetic field can do no work, as we have shown, the work done on the mobile charges must be electric, and the electric field in this situation cannot be conservative because the line integral of a conservative field must vanish. Therefore, we conclude that there is a non-conservative electric field \({\overrightarrow{E}}_{nc}\) associated with an induced emf : \(\varepsilon = \text{∮}\ {\overrightarrow{E}}_{nc} \cdot d\overrightarrow{s}\)

Combining with Faraday's law then yields :

\[\text{∮}\ \ {\overrightarrow{E}}_{nc} \cdot \text{ }d\overrightarrow{\text{ }s} = - \frac{d\Phi_{B}}{dt}\]

The above expression implies that a changing magnetic flux will induce a non-conservative electric field which can vary with time. It is important to distinguish between the induced, non-conservative electric field and the conservative electric field which arises from electric charges.
As an example, let's consider a uniform magnetic field which points into the page and is confined to a circular region with radius \(R\), as shown in Figure. Suppose the magnitude of \(\overrightarrow{B}\) increases with time, i.e., \(dB/dt > 0\). Let's find the induced electric field everywhere due to the changing magnetic field.

Since the magnetic field is confined to a circular region, from symmetry arguments we choose the integration path to be a circle of radius \(r\). The magnitude of the induced field \({\overrightarrow{E}}_{nc}\) at all points on a circle is the same. According to Lenz's law, the direction \({\overrightarrow{E}}_{nc}\) of must be such that it would drive the induced current to produce a magnetic field opposing the change in magnetic flux. With the area vector \(\overrightarrow{A}\) pointing out of the page, the magnetic flux is negative or inward. With \(dB/dt > 0\), the inward magnetic flux is increasing. Therefore, to counteract this

change the induced current must flow counterclockwise to produce more outward flux. The direction of \({\overrightarrow{E}}_{nc}\) is shown in Figure.

Let's proceed to find the magnitude of \({\overrightarrow{E}}_{nc}\).

In the region \(\mathbf{r} > \mathbf{R}\)

the rate of change of magnetic flux is :

\[\frac{d\Phi_{B}}{dt} = \frac{d}{dt}(\overrightarrow{B} \cdot \overrightarrow{A}) = \frac{d}{dt}( - BA) = - \left( \frac{dB}{dt} \right)\pi r^{2} \]Using equation, we have

\[\text{∮}\ \ {\overrightarrow{E}}_{nc} \cdot \text{ }d\overrightarrow{\text{ }s} = E_{nc}(2\pi r) = - \frac{d\Phi_{B}}{dt} = \left( \frac{dB}{dt} \right)\pi r^{2}\]

which implies

\[E_{nc} = \frac{r}{2}\frac{\text{ }dB}{dt}\]

Similarly, for \(r > R\)

the induced electric field may be obtained as

\[E_{nc}(2\pi r) = - \frac{d\Phi_{B}}{dt} = \left( \frac{dB}{dt} \right)\pi R^{2}\]

or

\[E_{nc} = \frac{R^{2}}{2r}\frac{\text{ }dB}{dt}\]

The pattern of \(E_{nc}\) as a function of \(r\) is shown in figure.

A plot of \(E_{nc}\) as a function of r is shown in figure.

Practice Exercise

Q. 1 Figure shows two circular regions of radii \(R_{1}\& R_{2}.\left( R_{2} \right.\ \) being the bigger circle \()\). The magnetic field in both are decreasing at a rate of \(\frac{dB}{dt}\). Calculate \(\text{∮}\ E \cdot d\overrightarrow{l}\) for each of the three paths indicated.

Q. 2 A uniform magnetic field fills a cylindrical volume of radius \(R\). If the magnetic field is decreasing at a rate \(\frac{dB}{dt}\), find induced emf at rod's end.

Answers

Q. \(1\ - \frac{dB}{dt}\pi R_{1}^{2}\) for path 1, \(- \frac{dB}{dt}\pi R_{2}^{2}\) for path 2, \(\frac{dB}{dt}\pi\left( R_{2}^{2} - R_{1}^{2} \right)\) for path 3
Q. \(2\frac{L}{2}\sqrt{R^{2} - \frac{L^{2}}{4}}\frac{dB}{dt}\)

We have seen that when a conducting loop moves through a magnetic field, current is induced as the result of changing magnetic flux. If a solid conductor were used instead of a loop, as shown in Figure, currents can also be induced along any closed loop in the conductor. The induced current are called an eddy current.

The induced eddy currents also generate a magnetic force that opposes the motion, making it more difficult to move the conductor across the magnetic field (Figure).

Since the conductor has non-vanishing resistance \(R\), Joule heating causes a loss of power by an amount \(P = \varepsilon^{2}/R\). Therefore, by increasing the value of R , power loss can be reduced. One way to increase \(R\) is to laminate the conducting slab, or construct the slab by using gluing together thin strips that are insulated from one another (see Figure-a). Another way is to make cuts in the slab, thereby disrupting the conducting path (Figure-b).

Figure (a)

Figure (b)

There are important applications of eddy currents. For example, the currents can be used to suppress unwanted mechanical oscillations. Another application is the magnetic braking systems in high-speed transit cars.

Self-Inductance :

Consider again a coil consisting of \(N\) turns and carrying current \(I\) in the counterclockwise direction, as shown in Figure. If the current is steady, then the magnetic flux through the loop will remain constant. However, suppose the current \(I\) changes with time, then according to Faraday's law, an induced emf will arise to oppose the change.According to lenz law the induced emf will be clockwise if \(\frac{dI}{dt} > 0\), and counterclockwise if \(\frac{dI}{dt} < 0\). The property of the loop in which its own magnetic field opposes any change in current is called "self-inductance," and the emf generated is called the self-induced emf or back emf, which we denote as \(\varepsilon_{L}\). All current-carrying loops exhibit this property. In particular, an inductor is a circuit element (symbol \(\_\_\_\_\) ) which has a large self-inductance.

Mathematically, the self-induced emf can be written as

\[\varepsilon_{L} = - N\frac{\text{ }d\Phi_{B}}{dt} = - N\frac{\text{ }d}{dt}\int_{}^{}\ \overrightarrow{\text{ }B} \cdot \text{ }d\overrightarrow{\text{ }A}\]

and is related to the self-inductance \(L\) by

\[\varepsilon_{L} = - L\frac{dI}{dt}\]

The above two expressions can be combined to yield

\[L = \frac{N\Phi_{B}}{I}\]

Physically, the inductance \(L\) is a measure of an inductor's "opposition" to the change of current; larger the value of \(L\), lower the rate of change of current.

Practice Exercise

Q. 1 Calculate the self-inductance of a toroid which consists of \(N\) turns and has a rectangular cross section, with inner radius \(a\), outer radius \(b\) and height \(h\), as shown in Figure.

Q. 2 How many metres of a thin wire are required to manufacture a solenoid of length \(l_{0} = 100\text{ }cm\) and inductance \(L = 1.0mH\) if the solenoid's cross-sectional diameter is considerably less than its length? Ignore edge effects
Ans. \(l = \sqrt{4\pi l_{0}\text{ }L/\mu_{0}} = 100\text{ }m\)

Practice Exercise

Q. \(1L = \frac{\mu_{0}N^{2}}{\mathcal{l}}\frac{h}{b - a}\)
Q. \(2l = \sqrt{4\pi l_{0}\text{ }L/\mu_{0}} = 100\text{ }m\)

Since an inductor in a circuit serves to oppose any change in the current through it, work must be done by an external source such as a battery in order to establish a current in the inductor. From the workenergy theorem, we conclude that energy can be stored in an inductor.

The power, or rate at which an external emf \(\varepsilon_{\text{ext}\text{~}}\) works to overcome the self-induced emf \(\varepsilon_{L}\) and pass current \(I\) in the inductor is

\[P_{L} = \frac{{dW}_{ext}}{dt} = I\varepsilon_{ext}\]

If only the external emf and the inductor are present, then \(\varepsilon_{\text{ext}\text{~}} = - \varepsilon_{L}\) which implies

\[P_{L} = \frac{{dW}_{ext}}{dt} = - I\varepsilon_{L} = + IL\frac{dI}{dt}\]

If the current is increasing with \(dI/dt > 0\), then \(P > 0\) which means that the external source is doing positive work to transfer energy to the inductor. Thus, the internal energy \(U_{B}\) of the inductor is increased. On the other hand, if the current is decreasing with \(dI/dt < 0\), we then have \(P < 0\). In this case, the external source takes energy away from the inductor, causing its internal energy to go down. The total work done by the external source to increase the current form zero to \(I\) is then :

\[W_{ext} = \int_{}^{}\ {dW}_{ext}\int_{0}^{I}\mspace{2mu}{LI}^{'}{dI}^{'} = \frac{1}{2}{LI}^{2}\]

This is equal to the magnetic energy stored in the inductor:

\[U_{B} = \frac{1}{2}{LI}^{2}\]

We comment that from the energy perspective there is an important distinction between an inductor and a resistor. Whenever a current I goes through a resistor, energy flows into the resistor and dissipates in the form of heat regardless of whether I is steady or time-dependent (recall that power dissipated in a resistor is \(P_{R} = IV_{R} = I^{2}R\) ). On the other hand, energy flows into an ideal inductor only when the current is varying with \(dI/dt > 0\). The energy is not dissipated but stored there; it is released later when the current decreases with \(dI/dt < 0\). If the current that passes through the inductor is steady, then there is no change in energy since \(P_{L} = LI(dI/dt) = 0\). Also note that there is an important distinction between an inductor and a resistor. The potential difference across a resistor depends on I , while the potential difference across an inductor depends on dI / dt. The self-induced emf does not oppose the current itself, but the rate of change of current.

Energy density in a magnetic field

A long solenoid with length \(\mathcal{l}\) and a radius \(R\) consists of \(N\) turns of wire. A current \(I\) passes through the coil.We have to find the energy density in the magnetic field..
Energy of the solenoid is given by

\[U_{B} = \frac{1}{2}{LI}^{2} = \frac{1}{2}\mu_{0}n^{2}I^{2}\pi R^{2}\mathcal{l}\]

The result can be expressed in terms of the magnetic field strength \(B = \mu_{0}nI\) :

\[U_{B} = \frac{1}{2\mu_{0}}\left( \mu_{0}nI \right)^{2}\left( \pi R^{2}\mathcal{l} \right) = \frac{B^{2}}{2\mu_{0}}\left( \pi R^{2}\mathcal{l} \right)\]

Since \(\pi R^{2}\mathcal{l}\) is the volume within the solenoid, and the magnetic field inside is uniform, the term
Hence magnetic energy density ( i.e.the energy per unit volume) of the magnetic field will be given by

\[u_{B} = \frac{U_{B}}{\pi r^{2}\mathcal{l}} = \frac{B^{2}}{2\mu_{0}}\]

may be identified as the magnetic energy density, or the energy per unit volume of the magnetic field. The above expression holds true even when the magnetic field is non-uniform. The result can be compared with the energy density associated with an electric field:

\[u_{E} = \frac{\varepsilon_{0}E^{2}}{2}\]

Growth and decay of current in L-R circuit

Growth of Current :

Consider the \(RL\) circuit shown in Figure. At \(t = 0\) the switch is closed. We find that the current does not rise immediately to its maximum value \(\varepsilon/R\). This is due to the presence of the self-induced emf in the inductor.

\(RL\) circuit is described by the following differential equation:

\[\varepsilon - IR - L\frac{dI}{dt} = 0\]

The above equation can be rewritten as

\[\frac{dI}{I - \varepsilon/R} = - \frac{dt}{\text{ }L/R}\ \Rightarrow \ \frac{dI}{I - \varepsilon/R} = - \frac{dt}{\tau}\]

where

\[\tau = \frac{L}{R}\]

is the time constant of the RL circuit. The qualitative behaviour of the current as a function of time is depicted in figure.
Integrating over both sides with prper limits we get

\[\begin{matrix} & \ \int_{0}^{I}\mspace{2mu}\mspace{2mu}\frac{dI}{I - \varepsilon/R} = - \int_{0}^{t}\mspace{2mu}\mspace{2mu}\frac{dt}{\tau} \\ & I = \frac{\varepsilon}{R}\left( 1 - e^{- t/\tau} \right) \end{matrix}\]

Note that after a sufficiently long time, the current reaches its equilibrium value \(\varepsilon/R\). The time constant \(\tau\) is a measure of how fast the equilibrium state is attained; the larger the value of \(L\), the longer it takes to build up the current. A comparison of the behavior of current in a circuit with or without an inductor is shown in Figure.
Similarly, the magnitude of the self-induced emf can be obtained as

\[\left| \varepsilon_{L} \right| = \left| - L\frac{dI}{dt} \right| = \varepsilon e^{- t/\tau}\]

which is at a maximum when \(t = 0\) and vanishes as \(t\) approaches infinity. This implies that a sufficiently long time after the switch is closed, self-induction disappears and the inductor simply acts as a conducting wire connecting two parts of the circuit.
To see that energy is conserved in the circuit, we multiply Eq. by \(I\) and obtain

\[P = \ I\varepsilon = I^{2}R + LI\frac{dI}{dt}\]

The left-hand side represents the rate at which the battery delivers energy to the circuit. On the other hand, the first term on the right-hand side is the power dissipated in the resistor in the form of heat, and the second term is the rate at which energy is stored in the inductor. While the energy dissipated through the resistor is irrecoverable, the magnetic energy stored in the inductor can be released later.

Decay of Current :

Next we consider the \(RL\) circuit shown in Figure. Suppose the switch \(S_{1}\) has been closed for a long time so that the current is at its equilibrium value \(\varepsilon/R\). What happens to the current when at \(t = 0\) switches \(S_{1}\) is opened and \(S_{2}\) closed?

Writhing loop equation

\[- L\frac{dI}{dt} - IR = 0\]

which can be rewritten as

\[\frac{dI}{I} = - \frac{dt}{\text{ }L/R}\ \Rightarrow \ \frac{dI}{I} = - \frac{dt}{\tau}\]

Integrating over both sides with prper limits we get

\[\int_{0}^{I}\mspace{2mu}\frac{dI}{I} = - \int_{0}^{t}\mspace{2mu}\frac{dt}{\tau}\]

We get

\[I = \frac{\varepsilon}{R}e^{- t/\tau}\]

where \(\tau = L/R\) is the same time constant as in the case of rising current. A plot of the current as a function of time is shown in Figure.

Practice Exercise

Q. 1 Consider the circuit shown in figure below :

Determine the current through each resistor
(a) immediately after the switch is closed.
(b) a long time after the switch is closed.

Suppose the switch is reopened a long time after it's been closed. What is each current
(c) immediately after it is opened?
(d) after a long time?

Figure \(RL\) circuit

Q. 2 A closed circuit consists of a source of constant emt \(\xi\) and a choke coil of inductance L connected in series. The active resistance of the whole circuit is equal to \(R\). At the moment \(t = 0\) the choke coil inductance was decreased abruptly \(\eta\) times. Find the current in the circuit as a function of time \(t\). Instruction. During a stepwise change of inductance the total magnetic flux (flux linkage) remains constant.
Q. 3 Find the time dependence of the current flowing through the inductance \(L\) of the circuit shown in Fig. after the switch Sw is shorted at the moment \(t = 0\).

Q. 4 A coil of inductance \(L = 2.0\mu H\) and resistance \(R = 1.0\Omega\) is connected to a source of constant emf \(\xi = 3.0\text{ }V\) (Fig.). A resistance \(R_{0} = 2.0\Omega\) is connected in parallel with the coil. Find the amount of heat generated in the coil after the switch \(Sw\) is disconnected. The internal resistance of the source is negligible.

Practice Exercise

Q. 1 (a) \(I_{1} = I_{2} = \frac{\varepsilon}{R_{1} + R_{2}}\)
(b) \(\ I_{1} = \frac{\left( R_{2} + R_{3} \right)\varepsilon}{R_{1}R_{2} + R_{1}R_{3} + R_{2}R_{3}},I_{2} = \frac{R_{3}\varepsilon}{R_{1}R_{2} + R_{1}R_{3} + R_{2}R_{3}},I_{3} = \frac{R_{2}\varepsilon}{R_{1}R_{2} + R_{1}R_{3} + R_{2}R_{3}}\)
(c) \(I_{3} = - I_{2} = \frac{R_{2}\varepsilon}{R_{1}R_{2} + R_{1}R_{3} + R_{2}R_{3}}\ \) (d) \(\ I_{1} = I_{2} = I_{3} = 0\).
Q. \(2\ I = \frac{\xi}{R}\left\lbrack 1 + (\eta - 1)e^{- t\eta R/L}\ Q.3\ I = \frac{\xi}{R}\left( 1 - e^{- tR/2\text{ }L} \right)\ Q.4\ Q = \frac{L\xi^{2}}{2R^{2}\left( 1 + R_{0}/R \right)} = 3\mu\text{ }J \right.\ \)

Suppose two coils are placed near each other, as shown in Figure.

The first coil has \(N_{l}\) turns and carries a current \(I_{l}\) which gives rise to a magnetic field \({\overrightarrow{B}}_{1}\). Since the two coils are close to each other, some of the magnetic field lines through coil 1 will also pass through coil 2 . Let \(\Phi_{21}\) denote the magnetic flux through one turn of coil 2 due to \(I_{l}\). Now, by varying \(I_{l}\) with time, there will be an induced emf associated with the changing magnetic flux in the second coil :

\[e_{21} = - N_{2}\frac{\text{ }d\Phi_{21}}{dt} = - \frac{N_{2}\text{ }d}{dt}\int_{\text{coil}\text{~}2}^{}\mspace{2mu}{\overrightarrow{\text{ }B}}_{1} \cdot \text{ }d{\overrightarrow{\text{ }A}}_{2}\]

The time rate of change of magnetic flux \(\Phi_{21}\) in coil 2 is proportional to the time rate of change of the current in coil 1 :

\[N_{2}\frac{\text{ }d\Phi_{21}}{dt} = M_{21}\frac{{dI}_{1}}{dt}\]

where the proportionality constant \(M_{21}\) is called the mutual inductance. It can also be written as

\[M_{21} = \frac{N_{2}\Phi_{21}}{I_{1}}\]

The SI unit for inductance is the henry \((H)\) :

\[1\text{~}\text{henry}\text{~} = 1H = 1\text{ }T.m^{2}/A\]

We shall see that the mutual inductance \(M_{2I}\) depends only on the geometrical properties of the two coils such as the number of turns and the radii of the two coils.

In a similar manner, suppose instead there is a current \(I_{2}\) in the second coil and it is varying with time (Figure). Then the induced emf in coil 1 becomes

\[\varepsilon_{12} = - N_{1}\frac{\text{ }d\Phi_{12}}{dt} = - \frac{N_{1}\text{ }d}{dt}\int_{\text{coil}\text{~}1}^{}\mspace{2mu}{\overrightarrow{\text{ }B}}_{2} \cdot \text{ }d{\overrightarrow{\text{ }A}}_{1}\]

This changing flux in coil 1 is proportional to the changing current in coil 2 ,

\[N_{1}\frac{\text{ }d\Phi_{12}}{dt} = M_{12}\frac{{dI}_{2}}{dt}\]

where the proportionality constant \(M_{12}\) is another mutual inductance and can be written as

\[M_{12} = \frac{N_{1}\Phi_{12}}{I_{2}}\]

According to reciprocity theorem

\[M_{12} = M_{21} \equiv M\]

Practice Exercise

Q. 1 Calculate the mutual inductance of a long straight wire and a rectangular frame with sides a and b . The frame and the wire lie in the same plane, with the side \(b\) being closest to the wire, separated by a distance \(l\) from it and oriented parallel to it.
Q. 2 Two thin concentric wires shaped as circles with radii a and b lie in the same plane. Allowing for \(a \ll b\), find:
(a) their mutual inductance;
(b) the magnetic flux through the surface enclosed by the outside wire, when the inside wire carries a current I.
Q. 3 There are two stationary loops with mutual inductance \(L_{12}\).. The current in one of the loops starts to be vang as \(I_{1} = \alpha t\), where \(\alpha\) is a constant, \(t\) is time. Find the time dependence \(I_{2}(t)\) of the current in the other loop whose inductance is \(L_{2}\) and resistance R .

Answers

Q. \(1\ {\text{ }L}_{12} = \frac{\mu_{0}\text{ }b}{2\pi}ln\left( 1 + \frac{a}{l} \right)\)
Q. 2 (a) \(L_{12} \approx \mu_{0}\pi a^{2}/2\text{ }b;\Phi_{21} = \mu_{0}\pi a^{2}I/2\text{ }b\)
Q. \(3\ I_{2} = \frac{\alpha L_{12}}{R}\left( 1 - e^{- t/\tau} \right)\) where \(\tau = L_{2}/R\)

Consider an \(LC\) circuit in which a capacitor is connected to an inductor, as shown in Figure.

Suppose the capacitor initially has charge \(Q_{0}\). When the switch is closed, the capacitor begins to discharge and the electric energy is decreased. On the other hand, the current created from the discharging process generates magnetic energy which then gets stored in the inductor. In the absence of resistance, the total energy is transformed back and forth between the electric energy in the capacitor and the magnetic energy in the inductor. This phenomenon is called electromagnetic oscillation.

The total energy in the \(LC\) circuit at some instant after closing the switch is

\[U = U_{C} + U_{L} = \frac{1}{2}\frac{Q^{2}}{C} + \frac{1}{2}{LI}^{2}\]

The fact that U remains constant implies that

\[\begin{matrix} & \frac{dU}{dt} = \frac{d}{dt}\left( \frac{1}{2}\frac{Q^{2}}{C} + \frac{1}{2}{LI}^{2} \right) = \frac{Q}{C}\frac{dQ}{dt} + LI\frac{dI}{dt} = 0 \\ & \frac{Q}{C} + L\frac{{\text{ }d}^{2}Q}{{dt}^{2}} = 0 \end{matrix}\]

where

\[I = - \frac{dQ}{dt}\]

and

\[\frac{dI}{dt} = - \frac{d^{2}Q}{{dt}^{2}}\]

Notice the sign convention we have adopted here. The negative sign implies that the current \(I\) is equal to the rate of decrease of charge in the capacitor plate immediately after the switch has been closed.

The general solution to equation is

\[Q = Q_{0}cos\left( \omega_{0}t + \phi \right)\]

where \(Q_{0}\) is the amplitude of the charge and \(\phi\) is the phase. The angular frequency \(\omega_{0}\) is given by

\[\omega_{0} = \frac{1}{\sqrt{LC}}\]

The corresponding current in the inductor is

\[I = - \frac{dQ}{dt} = \omega_{0}Q_{0}sin\left( \omega_{0}t + \phi \right) = I_{0}sin\left( \omega_{0}t + \phi \right)\]

where \(\ I_{0} = \omega_{0}Q_{0}\).
From the initial conditions \(Q(\) at \(t = 0) = Q_{0}\) and \(I(\) at \(t = 0) = 0\), the phase \(\phi\) can be determined to \(\phi =\) 0 . Thus, the solutions for the charge and the current in our LC circuit are
and \(\ I(t) = I_{0}sin\omega_{0}t\)
The time dependence of \(Q(t)\) and \(I(t)\) are depicted in figure.

Using Eqs., we see that at any instant of time, the electric energy and the magnetic energies are given by

\[U_{E} = \frac{Q^{2}(t)}{2C} = \left( \frac{Q_{0}^{2}}{2C} \right)\cos^{2}\omega_{0}t\]

and

\[U_{B} = \frac{1}{2}LI^{2}(t) = \frac{LI_{0}^{2}}{2}\sin^{2}\omega t = \frac{L\left( - \omega_{0}Q_{0} \right)^{2}}{2}\sin^{2}\omega_{0}t = \left( \frac{Q_{0}^{2}}{2C} \right)\sin^{2}\omega_{0}t\]

The electric and magnetic energy oscillation is illustrated in figure.

Mechanical analogy

The mechanical analog of the \(LC\) oscillations is the mass-spring system, shown in Figure.

If the mass is moving with a speed \(v\) and the spring having a spring constant \(k\) is displaced from its
equilibrium by \(x\), then the total energy of this mechanical system is

\[U = K + U_{sp} = \frac{1}{2}{mv}^{2} + \frac{1}{2}{kx}^{2}\]

where \(K\) and \(U_{\text{sp}\text{~}}\) are the kinetic energy of the mass and the potential energy of the spring, respectively. In the absence of friction, \(U\) is conserved and we obtain

\[\frac{dU}{dt} = \frac{d}{dt}\left( \frac{1}{2}{mv}^{2} + \frac{1}{2}{kx}^{2} \right) = mv\frac{dv}{dt} + kx\frac{dx}{dt} = 0\]

Using \(\ v = \frac{dx}{dt}\ \) and \(\frac{dv}{dt} = \frac{d^{2}x}{dt^{2}}\),
the above equation may be rewritten as

\[m\frac{{\text{ }d}^{2}x}{{dt}^{2}} + kx = 0\]

The general solution for the displacement is

\[x = x_{0}cos\left( \omega_{0}t + \phi \right)\]

where

\[\omega_{0} = \sqrt{\frac{k}{\text{ }m}}\]

is the angular frequency and \(x_{0}\) is the amplitude of the oscillations. Thus, at any instant in time, the energy of the system may be written as

\[\begin{matrix} U & \ = \frac{1}{2}mx_{0}^{2}\omega_{0}^{2}\sin^{2}\left( \omega_{0}t + \phi \right) + \frac{1}{2}kx_{0}^{2}\cos^{2}\left( \omega_{0}t + \phi \right) \\ & \ = \frac{1}{2}kx_{0}^{2}\left\lbrack \sin^{2}\left( \omega_{0}t + \phi \right) + \cos^{2}\left( \omega_{0}t + \phi \right) = \frac{1}{2}kx_{0}^{2} \right.\ \end{matrix}\]

In figure we illustrate the energy oscillations in the LC circuit and the mass spring system (harmonic oscillator).

LC Circuit	Mass-spring System	Energy
\[L\]		\[\square\]
		\[U_{E}\]	\[U_{B}\]
		\[U_{SP}\]	\[K\]
	\[x = 0\]		\[\square\]
		\[U_{E}\] \[U_{SP}\]	\[U_{B}\] \[K\]
	\[x = 0\]	I
		\[U_{E}\]	\[U_{B}\]
		\[U_{SP}\]	\[K\]
			\[U_{B}\] \[K\]
		\[U_{E}\] \[U_{SP}\]
		I
			\[\begin{matrix} U_{E} & U_{B} \\ U_{SP} & K \end{matrix}\]

Practice Exercise

Q. 1 Consider the circuit shown in Figure. Suppose the switch which has been connected to point \(a\) for a long time is suddenly thrown to \(b\) at \(t = 0\).

Find the following quantities :
(a) the frequency of oscillation of the \(LC\) circuit.
(b) the maximum charge that appears on the capacitor.
(c) the maximum current in the inductor.
(d) the total energy the circuit possesses at any time \(t\).

Answer

Q. 1
(a) \(f = \frac{1}{2\pi\sqrt{LC}}\)
(b) \(Q = C\varepsilon\)
(c) \(I_{0} = \varepsilon\sqrt{\frac{C}{L}}\)
(d) \(\frac{1}{2}C\varepsilon^{2}\)

Q. 1 A variable magnetic field creates a constant emf E in a conductor \(ABCDA\). The resistances of portion \(ABC,CDA\) and \(AMC\) are \(R_{1},R_{2}\) and \(R_{3}\) respectively. What current will be shown by meter \(M\) ? The magnetic field is concentrated near the axis of the circular conductor.

Sol. Let \(E_{1}\) and \(E_{2}\) be the emfs developed in ABC and CDA , respectively. Then \(E_{1} + E_{2} = E\). There is no net emf in the loop AMCBA as it does not enclose the magnetic field. If \(E_{3}\) is the emf in AMC then \(E_{1} - E_{3} = 0\). The equivalent circuit and distribution of current is shown in figure.

By the loop rule

\[\begin{matrix} & \text{~}\text{and}\text{~}\ R_{1}(x - y) + R_{2}x = E_{1} + E_{2} = E \\ & R_{3}y - R_{1}(x - y) = E_{3} - E_{1} = 0 \\ & \text{~}\text{Solving for}\text{~}y,y = \frac{ER_{1}}{R_{1}R_{2} + R_{2}R_{3} + R_{3}R_{1}}. \end{matrix}\]

Q. 2 A rectangular frame ABCD made of a uniform metal wire has a straight connection between E & F made of the same wire as shown in the figure. AEFD is a square of side 1 m & \(EB = FC = 0.5\text{ }m\). The entire circuit is placed in a steadily increasing uniform magnetic

field directed into the place of the paper & normal to it. The rate of change of the magnetic field is \(1\text{ }T/s\), the resistance per unit length of the wire is \(1\Omega/m\). Find the current in segments \(AE,BE\) & EF.
Sol. Induced e.m.f.
\[= - \frac{d\Phi}{dt} = - \frac{d}{dt}(BA) = - A\frac{dB}{dt} \therefore\ \] Induced e.m.f. in \(AEFD = 1 \times 1 = 1\text{ }V\ \left( \right.\ \) area \(\left. \ = 1{\text{ }m}^{2} \right)\)
Induced e.m.f. in \(EBCF = \frac{1}{2} \times 1 = 0.5\text{ }V\ \left( \right.\ \) area \(\left. \ = \frac{1}{2}{\text{ }m}^{2} \right)\)
Total induced e.m.f \(= 1 + 0.5 = 1.5\text{ }V\)
Given that resistance per unit length of the wire is \(1\Omega/m\). Hence the equivalent circuit take the form as shown in figure. The resistances \(3\Omega\) and \(2\Omega\) are in parallel. Hence their equivalent resistance would

\[\frac{(3 \times 2)}{(3 + 2)} = \left( \frac{6}{5} \right)\]

\(\therefore\ \) Current from \(EF = \frac{1.5}{(6/5)} = \frac{5}{4}\) amp.

Current in \(AE = \frac{5}{4} \times \frac{2}{5} = \frac{1}{2}amp\).

Current in \(BE = \frac{5}{4} \times \frac{3}{5} = \frac{3}{4}\) amp.
Q. 3 A very small circular loop of area \(5 \times 10^{- 4}{\text{ }m}^{2}\), resistance 2 ohm and negligible inductance is initially coplanar and concentric with a much larger fixed circular loop of radius 0.1 m . A constant current of 1 Amp . is passed in the bigger loop and the smaller loop is rotated with angular velocity \(\omega rad/s\) about a diameter. Calculate (a) the flux linked with the smaller loop (b) induced

emf and induced current in the smaller loop as a function of time.

Sol. (a) The situation is shown in figure. The field at the center of larger loop,

\[B_{1} = \frac{\mu_{0}}{4\pi}\frac{2\pi I}{R} = {10}^{- 7}\frac{2\pi \times 1}{0.1} = 2\pi \times 10^{- 6}\frac{\text{ }Wb}{{\text{ }m}^{2}}\]

is initially along the normal to the area of smaller loop. Now as the smaller loop (and hence normal to its plane) is rotating at angular velocity \(\omega\), so in time t it will turn by an angle \(\theta = \omega\) t w.r. to \(\overrightarrow{B}\) and hence the flux linked with the smaller loop at time t ,

\[\begin{matrix} & \phi_{2} = B_{1}{\text{ }S}_{2}cos\theta = \left( 2\pi \times 10^{- 6} \right)\left( 5 \times 10^{- 4} \right)cos\omega t \\ & \text{~}\text{i.e.,}\text{~}\phi_{2} = \pi \times 10^{- 9}cos\omega tWb \end{matrix}\]

(b) The induced emf in the smaller loop,

\[e_{2} = - \frac{d\phi_{2}}{dt} = - \frac{d}{dt}\left( \pi \times 10^{- 9}cos\omega t \right)\]

\[\text{~}\text{i.e.,}\text{~}e_{2} = \pi \times 10^{- 9}\omega sin\omega t\text{~}\text{volt}\text{~}\]

(c) The induced current in the smaller loop,

\[I_{2} = \frac{e_{2}}{R} = \frac{1}{2}\pi\omega \times 10^{- 9}sin\omega t\]

Q. 4 A wire frame of area A and resistance R is suspended freely from a 0.392 m long thread. There is a uniform magnetic field of \(B\) tesla and the plane of wire-frame is perpendicular to the magnetic field. The frame is forcely made to oscillate using external force with small angular amplitude \(\theta_{0}\) along the direction of magnetic field according to the law \(\theta = \theta_{0}sin\omega\). The plane of the frame is always along the direction of thread and does not rotate about it. What is the inducted e.m.f. in wire-frame as a function of time? Also find the maximum current is the frame.
Sol. The situation is shown in figure. The instantaneous flux through the frame when displaced through an angle \(\theta\) is given by

Equation of S.H.M.

\[\theta = \theta_{0}sin\omega t\]

Now

\[\Phi = BAcos\theta\]

Instantaneous induced e.m.f.

\[\begin{matrix} & e = - \frac{d\Phi}{dt} = BAsin\theta\frac{\text{ }d\theta}{dt} = BA\theta\frac{\text{ }d\theta}{dt}\ (\because sin\theta = \theta) \\ & e = BA\left( \theta_{0}sin\omega t \right)\frac{d}{dt}\left( \theta_{0}sin\omega t \right) \\ & \ = BA\theta_{0}sin\omega t\ \theta_{0}\omega cos\omega t \\ & \ = \frac{1}{2}BA\omega\theta_{0}^{2}sin2\omega t \end{matrix}\]

and

\[i_{\max} = BA\omega\theta_{0}^{2}sin2\omega t/2R\]

Q. 5 A coil A C D of radius R and number of turns n carries a current i amp. and is placed in the plane of paper. A small conducting ring \(P\) of radius \(r\) is placed at a distance \(y\) from the centre and above the coil ACD . Calculate the induced e.m.f. produced in the ring when the ring is allowed to fall freely. Express induced e.m.f. in terms of speed of the ring.

Sol. We know that magnetic induction at some point on the axis of a current carrying coil at a distance \(y\) from its centre is given by

\[B = \frac{\mu_{0}}{4\pi} \times \frac{2\pi{niR}^{2}}{\left( R^{2} + y^{2} \right)^{3/2}}\]

Now, the magnetic flux linked with ring P is

\[\Phi = BA = \frac{\mu_{0}}{4\pi} \times \frac{2\pi n{nr}^{2}R^{2}}{\left( R^{2} + y^{2} \right)^{3/2}} \times \pi r^{2} = \frac{\mu_{0}}{2} \times \frac{\pi nR^{2}r^{2}}{\left( R^{2} + y^{2} \right)^{3/2}}\]

Let the ring \(P\) falls with velocity \(v\). Now \(y\) varies with time as

\[y = y_{0} - vt\]

The induced e.m.f. in the ring

\[\begin{matrix} & e = \frac{d\Phi}{dt} = \frac{\mu_{0}}{2} \times \pi niR^{2}r^{2}\frac{d}{dt}\left( R^{2} + y^{2} \right)^{- 3/2} = - \frac{3}{2}\mu_{0}\pi niR^{2}r^{2}\left( R^{2} + y^{2} \right)^{- 5/2} \cdot 2y\left( \frac{dy}{dt} \right) \\ & \ = - \frac{3}{2}\frac{\mu_{0}\pi niR^{2}r^{2}}{\left( R^{2} + y^{2} \right)^{5/2}}y( - v) = \frac{3}{2}\frac{\mu_{0}\pi niR^{2}r^{2}yv}{\left( R^{2} + y^{2} \right)^{5/2}}. \end{matrix}\]

Q. 6 An infinitesimally small bar magnet of dipole moment \(\overline{M}\) is pointing and moving with the speed v in the x -direction. A small closed circular conducting loop of radius a and of negligible self-inductance lies in the \(y\)-z plane with its center at \(x = 0\), and its axis coinciding with the \(x\)-axis. Find the force opposing the motion of the magnet, if the resistance of the loop is R. Assume that the distance \(x\) of the magnet from the center of the loop is much greater than a.
Sol. Field due to the bar magnet at distance x (near the loop)

\[B = \frac{\mu_{0}}{4\pi}\frac{2M}{x^{2}}\]

\(\Rightarrow \ \) Flux lined with the loop : \(\phi = BA = \pi a^{2} \cdot \frac{\mu_{0}}{4\pi}\frac{2M}{x^{3}}\)
emf induced in the loop: \(e = - \frac{d\phi}{dt} = \frac{\mu_{0}}{4\pi}\frac{6\pi{Ma}^{2}}{x^{4}}\frac{dx}{dt}\)

\[= \frac{\mu_{0}}{4\pi}\frac{6\pi{Ma}^{2}}{x^{4}}v\]

\(\Rightarrow\) Induced current: \(i = \frac{e}{R} = \frac{\mu_{0}}{2\pi}\frac{3\pi{Ma}^{2}}{{Rx}^{4}}.V. = \frac{3\mu_{0}{Ma}^{2}}{{Rx}^{4}}\)
Let \(F =\) force opposing the motion of the magnet
Power due to the opposing force \(=\) Heat dissipated in the coil per second.

\[\begin{matrix} \Rightarrow \ Fv & \ = i^{2}R \Rightarrow \ \text{ }F = \frac{i^{2}R}{v} = \left( \frac{\mu_{0}}{4\pi} \right)^{2} \times \left( \frac{6\pi{Ma}^{2}}{{Rx}^{4}} \right)^{2} \times v^{2} \times \frac{R}{v} \\ \text{ }F & \ = \frac{9}{4}\frac{\mu_{0}^{2}M^{2}a^{4}v}{{Rx}^{4}} \end{matrix}\]

Q. 7 A square loop of side ' a ' with a capacitor of capacitance C is located between two current carrying long parallel wires as shown. The value of I in the wire is given as \(I = I_{0}sin\omega\).
(a) calculate maximum current in the square loop.
(b) Draw a graph between charge on the lower plate of the capacitor v/s time.

[JEE 2003]

Sol. (a) \(\ I = I_{0}sin\omega t\)
Flux linked with square loop
\[\int d\phi = \int_{a}^{2a}\mspace{2mu}\frac{\mu_{0}\lbrack}{2\pi}\left\lbrack \frac{1}{x} + \frac{1}{3a - x} \right\rbrack adx \]

\[\begin{matrix} & \ = \frac{\mu_{0}Ia}{2\pi}\left\lbrack lnx - \left. \ ln(3a - x) \right|_{a}^{2a} \right.\ \\ & \ = \frac{\mu_{0}Ia}{2\pi}\lbrack ln2a - lna - lna + ln2a\rbrack \\ & \ = \frac{\mu_{0}Ia}{2\pi}2ln2 = \frac{\mu_{0}Ia}{\pi}ln2 \end{matrix}\]

Charge an capacitor \(= \left| C\frac{d\phi}{dt} \right| = C\frac{\mu_{0}a}{\pi}ln2\frac{dI}{dt}\)

\[\begin{array}{r} \ = \frac{\mu_{0}Ca}{\pi}ln2\left( I_{0}\omega \right)cos\omega t\#(i) \\ \ = \frac{\mu_{0}C\left( I_{0}\omega \right)a}{\pi}ln2cos\omega t\#(ii) \end{array}\]

Maximum current \(I = \left| \frac{dq}{dt} \right|_{\max} = \frac{\mu_{0}{CI}_{0}\omega^{2}a}{\pi}ln2\)
(b) The graph for charge and time can be drawn from equation (i) as shown in figure.

\[\begin{matrix} & \frac{\mu_{0}{CI}_{0}\omega a}{\pi}ln2 \\ & \ - \frac{\mu_{0}{CI}_{0}\omega a}{\pi}ln2. \end{matrix}\]

Q. 8 Two parallel vertical metallic rails AB & CD are separated by 1 m . They are connected at the two ends by resistance \(R_{1}\& R_{2}\) as shown in the figure. A horizontally metallic bar \(L\) of mass 0.2 kg slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of 0.6 T perpendicular to the plane of the rails, it is observed that when the terminal velocity is attained, the power dissipated in \(R_{1}\& R_{2}\) are 0.76 W

& 1.2 W respectively. Find the terminal velocity of bar \(L\) & value \(R_{1}\& R_{2}\).
Sol. Atterminal velocity
\(mg = I\mathcal{l}B\ \Rightarrow \ 0.2 \times 9.8 = I \times 1 \times 0.6\ \Rightarrow \ I = \frac{1.96}{0.6}Amp\).

\[\begin{array}{r} Ie = (0.76 + 1.2) = 1.96\#(1) \end{array}\]

From (1) and (2)
\(\frac{1.96}{0.6}e = 1.96\ \Rightarrow \ e = 0.6\) volt
\[{P_{1} = \frac{e^{2}}{R_{1}} = 0.76\ \Rightarrow \ R_{1} = \frac{e^{2}}{0.76} = \frac{0.36}{0.76} = 0.47\Omega }{P_{2} = \frac{e^{2}}{R_{2}} = 1.2\ \Rightarrow \ R_{2} = \frac{e^{2}}{1.2} = 0.3\Omega }{P = Fv\ \Rightarrow \ 1.56 = mgv\ \Rightarrow \ 1.96 = 1.96v\ \Rightarrow \ v = 1\text{ }m/s }\]Q. 9 A rod of length 2 a is free to rotate in a vertical plane, about a horizontal axis O passing through its midpoint. A long straight, horizontal wire is in the same plane and is carrying a constant current \(i\) as shown in figure. At initial moment of time the rod is horizontal and starts to rotate with constant angular velocity \(\omega\). Calculate e.m.f. induced in rod as a function of time.

Sol. The rotated position of the rod after a time t is shown in figure. Consider a small element of length dx of the rod at a distance \(x\) from the centre.
The velocity of the element \(v = x\omega\) and its distance from the wire is \(r = (d - xsin\omega t)\). Magnetic induction at this position
\[B = \frac{\mu_{0}i}{2\pi r} = \frac{\mu_{0}i}{2\pi(d - xsin\omega t)} \]

The induced e.m.f in this element

\[de = Bvdx = \frac{\mu_{0}i(x\omega)dx}{2\pi(d - xsin\omega t)}\]

In order to obtain the resultant e.m.f., we integrate this expression from (-a) to + a. Hence

\[\begin{matrix} & e = \frac{\mu_{0}i\omega}{2\pi}\int_{- a}^{a}\mspace{2mu}\mspace{2mu}\frac{xdx}{(\text{ }d - xsin\omega t)} \\ & \ = \frac{\mu_{0}i\omega}{2\pi}\left( \frac{- 1}{\sin^{2}\omega t} \right)\left\lbrack 2asin\omega t + dlog\left( \frac{\text{ }d - asin\omega t}{\text{ }d + asin\omega t} \right) \right\rbrack \\ & \ = \frac{\mu_{0}i\omega}{2\pi\sin^{2}\omega t}\left\lbrack - 2asin\omega t + dlog\left( \frac{\text{ }d + asin\omega t}{\text{ }d - asin\omega t} \right) \right\rbrack. \end{matrix}\]

Q. 10 P and Q are two infinite conducting plates kept parallel to each other and separated by a distance \(2r\). A conducting ring of radius \(r\) falls vertically between the plates such that planes are always tangent to the ring. Both the planes are connected by a resistance R . There exists a uniform magnetic field of strength \(B\) perpendicular to the plane of ring. The arrangement is shown in figure. Plane \(Q\) is smooth and friction between the plane P and the ring is enough to prevent slipping. At \(t = 0\), the planes and the ring. Find

(a) the current through R as a function of time
(b) terminal velocity of the ring (assume \(g\) to be constant).

Sol. Let v be the velocity of C.M. of the ring at any time t . e.m.f. across the diameter of ring, \(e = 2\text{ }B\) vr.

\[\begin{array}{r} \therefore\ \text{~}\text{Current through}\text{~}R\text{~}\text{is}\text{~}I = (2Bvr/R)\#(i) \end{array}\]

The different force on the ring are shown in figure.
For rotational motion

\[\begin{array}{r} \begin{matrix} fr & = I\alpha & \\ f & = \frac{I\alpha}{r} = \frac{Ia}{r^{2}} & (\because\alpha = a/r) \end{matrix}\#(ii) \end{array}\]

where \(a =\) acceleration of C.M.

Now \(F_{m} = BiL = \frac{4B^{2}r^{2}}{R}v\ (\because L = 2r)\)

For translational motion

\[\begin{array}{r} mg - f - F_{m} = ma\#(iv) \end{array}\]

Substituting the value of \(f\) and \(F_{m}\) from eqs. (ii) and (iii) in equation (iv), we get

\[mg - \frac{Ia}{r^{2}} - \frac{4{\text{ }B}^{2}r^{2}}{R}v = ma\]

\[mg - \frac{4B^{2}r^{2}}{R}v = ma + \frac{Ia}{r^{2}}\]

or \(\ mg - \frac{4B^{2}r^{2}}{R}v = a\left\lbrack m + \frac{I}{r^{2}} \right\rbrack\)
or \(\ \left( mg - \frac{4B^{2}r^{2}}{B}v \right) = \frac{dv}{dt}\left\lbrack m + \frac{I}{r^{2}} \right\rbrack\)
or \(\ \frac{dv}{\left\lbrack mg - \frac{4B^{2}r^{2}}{R}v \right\rbrack} \times \left( m + \frac{I}{r^{2}} \right) = dt\)
for a ring \(I = {mr}^{2}\)

\[\begin{array}{r} \therefore\ \frac{dv}{\left\lbrack mg - \frac{4{\text{ }B}^{2}r^{2}}{R}v \right\rbrack}(2\text{ }m) = dt\#(v) \end{array}\]

Integrating equation (iv) with proper limits, we get

\[- \frac{mR}{2{\text{ }B}^{2}r^{2}}\log_{e}\left\lbrack mg - \frac{4{\text{ }B}^{2}r^{2}}{R}v \right\rbrack_{0}^{v} = t\]

or \(\ \frac{- mR}{2B^{2}r^{2}}\log_{e}\left\lbrack \frac{mg - \frac{4B^{2}r^{2}}{R}v}{mg} \right\rbrack = t\)

\[\begin{array}{r} \therefore\ v = \frac{mgR}{4{\text{ }B}^{2}r^{2}}\left\lbrack 1 - e^{- \frac{2{\text{ }B}^{2}r^{2}}{mR} - t} \right\rbrack\#(vi) \end{array}\]

For equation (i) and (vi), we get

\[i = \frac{2Bvr}{R} = \frac{mg}{2Br}\left\lbrack 1 - e^{- \frac{2{Br}^{2}r^{2}}{mR},t} \right\rbrack.\]

Q. 11 A rectangular conducting loop in the vertical \(x - z\) plane has length L , width W , mass M and resistance R . It is dropped lengthwise from rest. At \(t = 0\) the bottom of the loop is at a height h above the horizontal x axis. There is a uniform magnetic field \(B\) perpendicular to the \(x - z\) plane, below the \(x\)-axis. The bottom and top of the loop cross this axis at \(t = t_{1}\), and \(t_{2}\) respectively. Obtain the expression for the velocity of the loop for the time \(t_{1} \leq t \leq t_{2}\)

Sol. For time \(t_{1}\), the loop is freely falling under gravity, so velocity attained by loop at

\[\begin{matrix} & t = t_{1} \\ & v_{1} = {gt}_{1} = \sqrt{2gh} \end{matrix}\]

During the time \(t_{1} \leq t \leq t_{2}\), flux linked with the loop is changing, so induced emf

\[e = - \frac{d\phi}{dt} = - Bv\text{ }W\]

and Induced current

\[I = - \frac{B \cup \text{ }W}{R}\text{~}\text{clockwise}\text{~}\]

Magnetic Force

\[F = WIB = - \frac{B^{2}v{\text{ }W}^{2}}{R}\]

So, \(\ m\frac{dv}{dt} = mg - \frac{B^{2}vW^{2}}{R}\)

\[dt = \frac{mdv}{\left\lbrack \left( mg - \frac{B^{2}{\text{ }W}^{2}v}{R} \right) \right\rbrack}\]

Integrating,

\[t = - \frac{mR}{{\text{ }B}^{2}{\text{ }W}^{2}}\log_{e}\left\lbrack mg - \frac{B^{2}{vW}^{2}}{R} \right\rbrack + A\]

At \(t = t_{1},v = v_{1} = {gt}_{1}\)
\[\therefore\ A = t_{1} + \frac{mR}{B^{2}{\text{ }W}^{2}}\log_{e}\left\lbrack mg - \frac{B^{2}v_{1}{\text{ }W}^{2}}{R} \right\rbrack \]Substituting for A,

\[e^{- \frac{B^{2}{\text{ }W}^{2}}{mR}\left( t - t_{1} \right)} = \log_{e}\left\lbrack \frac{mg - \frac{B^{2}vW^{2}}{R}}{mg - \frac{B^{2}v_{1}W^{2}}{R}} \right\rbrack\]

Gives the velocity \(v\) of the loop in the interval \(t_{1} \leq t \leq t_{2}\).
Q. 12 A coil of inductance \(L = 50 \times 10^{- 6}\) henry and resistance \(= 0.5\Omega\) is connected to a battery of emf \(= 5.0\) V. A resistance of \(10\Omega\) is connected parallel to the coil. Now at some instant the connection of the battery is switched off. Find the amount of heat generated in the coil after switching off the battery.
Sol Total energy stored in the inductor \(= \frac{1}{2}{Li}_{0}^{2}\)

\[E_{L} = \frac{1}{2}L\left( \frac{V}{r} \right)^{2}\]

\(\therefore\ \) Fraction of energy lost across inductor

\[\begin{matrix} & \ = E_{L} \cdot \frac{r}{(R + r)} \\ & \ = \frac{LV^{2}}{2r(R + r)} = \frac{50 \times 10^{- 6} \times 5^{2}}{2 \times 0.5(10 + 0.5)} = 1.19 \times 10^{- 4}\text{ }J \end{matrix}\]

Q. 13 A metal rod OA of mass \(m\) & length \(r\) is kept rotating with a constant angular speed \(\omega\) in a vertical plane about a horizontal axis at the end O . The free end A is arranged to slide without friction along a fixed conducting circular ring in the same plane as that of rotation. A uniform & constant magnetic induction \(\overrightarrow{B}\) is applied perpendicular & into the plane of rotation as shown in figure. An inductor \(L\) and an external resistance \(R\) are

connected through a switch S between the point O & a point C on the ring to form an electrical circuit. Neglect the resistance of the ring and the rod. Initially, the switch is open.
(a) What is the induced emf across the terminals of the switch?
(b) (i) Obtain an expression for the current as a function of time after switch S is closed.
(ii) Obtain the time dependence of the torque required to maintain the constant angular speed, given that the \(rodOA\) was along the positive X -axis at \(t = 0\).
Sol. Let v be the velocity of C.M. of the ring at any time t .
e.m.f. across the diameter of ring, \(e = 2\text{ }B \vee r\).
\(\therefore\ \) Current through R is \(I = (2\text{ }Bvr/R)\)

The different force on the ring are shown in figure.
For rotational motion

\[\begin{matrix} & & fr = I\alpha \\ & f = \frac{I\alpha}{r} = \frac{Ia}{r^{2}}\ (\because\alpha = a/r) & \text{(ii)} \end{matrix}\]

where \(a =\) acceleration of C.M.

Now \(F_{m} = BiL = \frac{4{\text{ }B}^{2}r^{2}}{R}v\ (\because L = 2r)\)

For translational motion

\[\begin{array}{r} mg - f - F_{m} = ma\#(iv) \end{array}\]

Substituting the value of \(f\) and \(F_{m}\) from eqs. (ii) and (iii) in equation (iv), we get

\[\begin{matrix} & mg - \frac{Ia}{r^{2}} - \frac{4B^{2}r^{2}}{R}v = ma \\ & mg - \frac{4B^{2}r^{2}}{R}v = ma + \frac{Ia}{r^{2}} \end{matrix}\]

or \(\ mg - \frac{4{\text{ }B}^{2}r^{2}}{R}v = a\left\lbrack m + \frac{I}{r^{2}} \right\rbrack\)
or \(\ \left( mg - \frac{4B^{2}r^{2}}{B}v \right) = \frac{dv}{dt}\left\lbrack m + \frac{I}{r^{2}} \right\rbrack\)
or \(\ \frac{dv}{\left\lbrack mg - \frac{4B^{2}r^{2}}{R}v \right\rbrack} \times \left( m + \frac{I}{r^{2}} \right) = dt\)
for a ring \(I = {mr}^{2}\)

\[\begin{array}{r} \therefore\ \frac{dv}{\left\lbrack mg - \frac{4{\text{ }B}^{2}r^{2}}{R}v \right\rbrack}(2\text{ }m) = dt\#(v) \end{array}\]

Integrating equation (iv) with proper limits, we get

\[- \frac{mR}{2{\text{ }B}^{2}r^{2}}\log_{e}\left\lbrack mg - \frac{4{\text{ }B}^{2}r^{2}}{R}v \right\rbrack_{0}^{v} = t\]

or \(\ \frac{- mR}{2B^{2}r^{2}}\log_{e}\left\lbrack \frac{mg - \frac{4B^{2}r^{2}}{R}v}{mg} \right\rbrack = t\)

\[\begin{array}{r} \therefore\ v = \frac{mgR}{4{\text{ }B}^{2}r^{2}}\left\lbrack 1 - e^{- \frac{2{\text{ }B}^{2}r^{2}}{mR^{2}} +} \right\rbrack\#(vi) \end{array}\]

For equation (i) and (vi), we get

\[i = \frac{2Bvr}{R} = \frac{mg}{2Br}\left\lbrack 1 - e^{- \frac{2{\text{ }B}^{2}r^{2}}{mR} \cdot t} \right\rbrack.\]

Q. 14 A square loop of side a and a straight, infinite conductor are placed in the same plane with two sides of the square parallel to the conductor. The inductance and resistance are equal to \(L\) and \(R\) respectively. The frame is turned through \(180^{\circ}\) about the axis \({OO}^{'}\). Find the electric charge that flows in the square loop.

Sol. By circuit equation \(i = \left( \varepsilon - L\frac{di}{dt} \right)/R\) where \(\varepsilon =\) induced emf and \(L\frac{di}{dt} =\) self-induced emf

\[\begin{matrix} & \ \Rightarrow \ Ri = \varepsilon - L\frac{di}{dt} \Rightarrow \ \int_{}^{}\ Ridt = \int_{}^{}\ \varepsilon dt - \int_{}^{}\ L\frac{di}{dt}dt \\ & \ \Rightarrow \ Rq = \int_{}^{}\ - \frac{d\varphi}{dt}dt - L\lbrack i\rbrack_{i}^{f} = \varphi_{i} - \varphi_{f}\ \left( \because i_{\text{initial}\text{~}} = 0,i_{\text{final}\text{~}} = 0 \right) \\ & \ \Rightarrow \ q = \left( \varphi_{i} - \varphi_{f} \right)/R \end{matrix}\]

Consider a strip at a distance \(x\) in the initial position. Then \(B = \left( \mu_{0}/4\pi \right)(2I/x)\) along the inward normal to the plane.

\[\begin{matrix} \therefore & d\varphi_{1} = \left( \mu_{0}I/2\pi x \right)adxcos0 = \frac{\mu_{0}Ia}{2\pi}\frac{dx}{x} \\ \Rightarrow & \varphi_{i} = \frac{\mu_{0}Ia}{2\pi}\int_{b}^{a + b}\mspace{2mu}\mspace{2mu}\frac{dx}{x} = \frac{\mu_{0}Ia}{2\pi}ln\frac{a + b}{\text{ }b} \end{matrix}\]

Similarly \(\varphi_{f} = \frac{\mu_{0}Ia}{2\pi}ln\frac{2a + b}{a + b}\)

\[\begin{matrix} \therefore & \left| \varphi_{i} - \varphi_{r} \right| = \frac{\mu_{0}Ia}{2\pi}ln\frac{2a + b}{\text{ }b} \\ \therefore & |q| = \frac{\mu_{0}Ia}{2\pi R}ln\frac{2a + b}{\text{ }b}. \end{matrix}\]

Q. 15 A thin wire ring of radius \(a\) and resistance \(r\) is located inside a long

solenoid so that their axes coincide. The length of the solenoid is equal to \(l\) its cross-sectional radius to b. At a certain moment, the solenoid was connected to a source of constant voltage \(V\). The total resistance of the circuit is equal to R. Assuming the inductance of the ring to be negligible, find the maximum value of the radial force acting per unit length of the ring.
Sol. The inductance \(L\) of the solenoid

\[L = \mu_{0}n^{2}\pi{\text{ }b}^{2}l\]

The current through the solenoid varies as

\[I(t) = \frac{V}{R}\left( 1 - e^{- tR/L} \right)\]

\(B(t) =\) magnetic induction

\[\begin{matrix} & \ = \mu_{0}nI(t) \\ & e = \frac{d\phi}{dt}\&\overrightarrow{\text{ }B} \cdot \overrightarrow{\text{ }A} \end{matrix}\]

flux through ring \(= \mu_{0}n.\pi a^{2}\)

\[e = \frac{d\phi}{dt},\mu_{0}n\pi a^{2}\frac{dI}{dt}\]

where

\[i(t)\text{~}\text{is}\text{~}\frac{dI(t)}{dt}\]

Force unit per length on the ring \(= \frac{F}{\mathcal{l}} = BI(t)\)

\[\begin{matrix} \therefore\ & F(t) = \mu_{0}nI(t) \cdot \mu_{0}n\overset{˙}{I}(t)\pi a^{2}/r \\ & = \frac{\mu_{0}^{2} \cdot \pi a^{2}}{r} \cdot n^{2}I(t)\overset{˙}{I}(t) \\ & F(t) = \frac{\mu_{0}^{2}\pi a^{2}}{r} \cdot n^{2}\left\lbrack \frac{V}{R}\left( 1 - e^{- tR/L} \right) \right\rbrack \cdot \left\lbrack \frac{V}{R} \cdot \frac{{Re}^{- tR/L}}{L} \right\rbrack \\ & = \frac{\mu_{0}^{2}\pi a^{2}V^{2}}{r} \cdot e^{- t/RL}\left( 1 - e^{- t/RL} \right)\left( \frac{n^{2}}{RL} \right) \end{matrix}\]

Initial conditions are \(F(t) = 0\) as \(t = 0\)

\[F(t) = 0\text{~}\text{at}\text{~}t = \infty\]

For finding maximum value of force, we differentiate \(F(t)\) w.r.t. t and put to zero. This gives the time at which this occurs.

\[\begin{matrix} e^{- t/RL} = \frac{1}{2} \\ \therefore\ {\text{ }F}_{\max} = \frac{\mu_{0}^{2}\pi a^{2}{\text{ }V}^{2}}{4rR}\frac{n^{2}}{\text{ }L} = \frac{\mu_{0}a^{2}v^{2}}{4\pi R/b^{2}} \end{matrix}\]