NCERT Solutions Class 12 Physics Chapter 6

Q.1 Predict the direction of induced current in the situations described by the following Figs. 6.18(a) to (f).

Ans.

Lenz’s law gives the direction of the induced current in a closed loop. The given pair of figures indicate the direction of induced current when the North pole of a bar magnet is moved towards and away from a closed loop.

According to Lenz’s law, the direction of induced current in the given situations can be given as follows:
(a) The induced current is directed along qrpq.
(b) The induced current is directed along prqp.
(c) The induced current is directed along yzxy.
(d) The induced current is directed along zyxz.
(e) The induced current is directed along xryx.
(f) No current is induced because the magnetic field lines are lying in the plane of the closed loop.

Q.2 Use Lenz’s law to determine the direction of induced current in the situation described by Fig. 6.19:

(a) A wire of irregular shape turning into a circular shape;

(b) A circular loop being deformed into a narrow straight wire.

Ans.

According to Lenz’s law, the direction of the induced current is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.

(a)When a wire of irregular shape changes to a circular loop, the area of the loop tends to increase. Thus, magnetic flux associated with the loop increases. As per Lenz’s law, the direction of induced current must oppose the magnetic field. Therefore, the induced current is flowing in anticlockwise direction in the loop from adcb.

(b)When the shape of a circular loop is deformed into a narrow straight wire, the magnetic flux tends to decrease. Thus, the induced current flows along a’d’c’b’.

Q.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

Ans.

\begin{array}{l}{\text{Here, number of turns on the solenoid = 15 turns cm}}^{\text{-1}}{\text{= 1500 turns m}}^{\text{-1}}\\ \text{Number of turns per unit length, n = 1500 turns}\\ \text{Area of the small loop of solenoid, A = 2}{\text{.0 cm}}^{\text{2}}{\text{= 2 × 10}}^{\text{-4}}{\text{m}}^{\text{2}}\\ \text{As current in the solenoid changes from 2 A to 4 A,}\\ \therefore \text{Change in current , di = 4 A – 2 A = 2 A}\\ \text{Time}\text{taken}\text{,}\text{dt = 0}\text{.1}\text{s}\\ \text{Rate}\text{of}\text{change}\text{of}\text{current =}\frac{\text{di}}{\text{dt}}\text{=}\frac{\text{2}}{\text{0}\text{.1}}\text{= 20}{\text{As}}^{\text{-1}}\\ \text{According}\text{to Faraday’s law, induced emf in the solenoid is given}\text{as:}\\ \text{e =}\frac{\text{df}}{\text{dt}}\mathrm{\dots .}\text{(i)}\\ \text{Here,}\varphi \text{= Induced flux through the small loop}\\ \varphi \text{= BA}\mathrm{\dots }\left(\text{ii}\right)\\ {\text{B = Magnetic field = μ}}_{\text{0}}\text{ni}\\ {\text{μ}}_{\text{0}}\text{= Permeability of free space = 4}\pi {\text{×10}}^{\text{-7}}{\text{Hm}}^{\text{-1}}\\ \therefore \text{Equation}\left(\text{i}\right)\text{becomes:}\\ \text{e =}\frac{\text{d}}{\text{dt}}\left(\text{BA}\right){\text{= Aμ}}_{\text{0}}\text{n ×}\left(\frac{\text{di}}{\text{dt}}\right)\\ {\text{= 2×10}}^{\text{-4}}{\text{m}}^{\text{2}}{\text{× 4π ×10}}^{\text{-7}}{\text{Hm}}^{-1}\text{× 1500 ×}\frac{\text{2 A}}{\text{0}\text{.1 s}}\\ \text{=7}{\text{.54 × 10}}^{\text{-6}}\text{V}\\ \therefore \text{The induced voltage in the loop =}\text{7}{\text{.54 ×10}}^{\text{-6}}\text{V}\end{array}

Q.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s⁻¹ in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Ans.

\begin{array}{l}\text{Here, length of rectangular wire,l = 8 cm = 0}\text{.08 m}\\ \text{Width of rectangular wire,b = 2 cm = 0}\text{.02 m}\\ \therefore \text{Area of rectangular loop,}\\ \text{A = lb}\\ \text{= 0}\text{.08 m × 0}\text{.02 m}\\ {\text{= 16×10}}^{\text{-4}}{\text{m}}^{\text{2}}\\ \text{Magnetic field strength, B = 0}\text{.3 T}\\ \text{Velocity of}{\text{rectangular loop, v =1 cms}}^{\text{-1}}\\ \text{=0}{\text{.01 ms}}^{\text{-1}}\\ \left(\text{a}\right)\text{When}\text{velocity}\text{is}\text{normal}\text{to}\text{the}\text{longer}\text{side}\\ \text{Emf developed in the loop is given by,}\\ \text{e = Blv}\\ \text{= 0}\text{.3 T × 0}\text{.08 m × 0}{\text{.01 ms}}^{\text{-1}}\\ \text{= 2}{\text{.4 × 10}}^{\text{-4}}\text{V}\\ \text{Time}\text{taken}\text{to}\text{move}\text{along}\text{the}\text{width,}\\ \text{t =}\frac{\text{Distance}\text{travelled}}{\text{Velocity}}\\ \text{=}\frac{{\text{2 × 10}}^{\text{-2}}\text{m}}{{\text{10}}^{\text{-2}}{\text{ms}}^{\text{-1}}}\text{=}\text{2}\text{s}\\ \text{Hence, the induced voltage is 2}{\text{.4 × 10}}^{\text{-4}}\text{V which}\\ \text{lasts for 2 s}\text{.}\\ \left(\text{b}\right)\text{When}\text{velocity}\text{is}\text{normal}\text{to}\text{the}\text{shorter}\text{side,}\\ \text{Emf developed}\text{is}\text{given}\text{as: e = Bbv}\\ \text{= 0}\text{.3 T × 0}\text{.02 m × 0}{\text{.01 ms}}^{\text{-1}}\\ \text{= 0}{\text{.6 × 10}}^{\text{-4}}\text{V}\\ \text{Time}\text{taken}\text{to}\text{move}\text{along}\text{the}\text{length}\text{is}\text{given}\text{as:}\\ \text{t =}\frac{\text{Distance}\text{travelled}}{\text{Velocity}}\\ \text{=}\frac{{\text{8 × 10}}^{\text{-2}}m}{{\text{10}}^{\text{-2}}{\text{ms}}^{\text{-1}}}\text{=}\text{8}\text{s}\\ \therefore \text{The induced voltage is 0}{\text{.6 × 10}}^{\text{-4}}\text{V which lasts for 8 s}\text{.}\end{array}

Q.5 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s⁻¹ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Ans.

\begin{array}{l}\text{Here,}\text{length of the rod, l = 1 m}\\ \text{Angular frequency}\text{of}{\text{rotation,ω = 400 rads}}^{\text{-1}}\\ \text{Magnetic field strength, B = 0}\text{.5 T}\\ \text{Linear velocity}\text{of}\text{one end of the rod has zero ,}\\ \text{and the linear velocity}\text{of}\text{the}\text{other end}\text{is lω}\text{.}\\ \text{Average linear velocity,v =}\frac{\text{0 + lω}}{\text{2}}\text{=}\frac{\text{lω}}{\text{2}}\\ \text{Emf induced between the centre and the ring,}\\ \text{e = Blv}\\ \therefore \text{e = Bl}\left(\frac{\text{lω}}{\text{2}}\right)\\ \text{=}\frac{{\text{Bl}}^{\text{2}}\text{ω}}{\text{2}}\\ \text{=}\frac{\text{0}\text{.5 T ×}{\left(\text{1 m}\right)}^{\text{2}}{\text{× 400 rads}}^{\text{-1}}}{\text{2}}\\ \text{= 100}\text{V}\\ \therefore \text{EMF developed between the centre and the ring is 100 V}\text{.}\end{array}

Q.6 A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s⁻¹ in a uniform horizontal magnetic field of magnitude 3.0 × 10⁻² T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω? Calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?

Ans.

\begin{array}{l}\text{Here,}\text{radius of the circular coil,}\text{r = 0}\text{.08}\text{m}\\ \therefore \text{Area of the circular coil, A =}\pi {\text{r}}^{\text{2}}\\ \text{=}\pi \text{×}{\left(\text{0}\text{.08}\right)}^{\text{2}}{\text{m}}^{\text{2}}\\ \text{Number of turns on the circular coil, N = 20}\\ {\text{Angular speed of circular coil, ω = 50 rads}}^{\text{-1}}\\ {\text{Magnetic field strength, B = 3 × 10}}^{\text{-2}}\text{T}\\ \text{Resistance of the closed loop, R = 10 Ω}\\ \text{Maximum induced emf is given by:}\text{e = NωAB}\\ {\text{= 20 × 50 rads}}^{\text{-1}}\text{×}\pi \text{×}{\left(\text{0}\text{.08 m}\right)}^{\text{2}}{\text{× 3 × 10}}^{\text{-2}}T\\ \text{= 0}\text{.603 V}\\ \text{Thus, maximum emf induced in the coil = 0}\text{.603 V}\text{.}\\ \text{Over a full cycle,}\text{the average emf induced in the coil= 0}\\ \text{Maximum current is given by}\text{the}\text{relation:}\\ \text{I =}\frac{\text{e}}{\text{R}}\\ \text{=}\frac{\text{0}\text{.603 V}}{\text{10}\Omega }\\ \text{= 0}\text{.0603}\text{A}\\ \text{Average power loss due to joule heating}\text{is}\text{given}\text{as:}\\ \text{P =}\frac{\text{eI}}{\text{2}}\\ \text{=}\frac{\text{0}\text{.603 V × 0}\text{.0603 A}}{\text{2}}\\ \text{= 0}\text{.018}\text{W}\\ \text{The induced}\text{current in the coil develops a torque opposing the rotation of the coil}\text{. The}\text{external agent}\\ \text{(rotor) must supply a torque to counter this torque in order to}\text{keep the coil rotating uniformly}\text{.}\\ \therefore \text{The}\text{external}\text{rotor}\text{acts}\text{as}\text{the}\text{source}\text{of}\text{power dissipated as}\text{heat}\text{in}\text{the}\text{coil}\text{.}\end{array}

Q.7 A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s⁻¹, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10⁻⁴ Wb m⁻².

(a) What is the instantaneous value of the emf induced in the wire?

(b) What is the direction of the emf?

(c) Which end of the wire is at the higher electrical potential?

Ans.

\begin{array}{l}\text{Here},\text{length of wire},\text{l}=\text{1}0\text{m}\\ \text{Speed of the wire},\text{v}=\text{5}.0{\text{ms}}^{-\text{1}}\\ \text{Magnetic field strength},\text{B}=0.\text{3}\times \text{1}{0}^{-\text{4}}{\text{Wbm}}^{-\text{2}}\\ \left(\text{a}\right)\text{Induced}e\text{mf in the wire, e}=\text{Blv}\\ =0.\text{3}\times \text{1}{0}^{-\text{4}}{\text{Wbm}}^{-\text{2}}\times \text{10 m}\times 5.0{\text{ms}}^{-\text{1}}\\ =\text{1}.\text{5}\times \text{1}{0}^{-\text{3}}\text{V}\\ \text{(b) From Fleming’s right hand rule, it can be concluded that the direction of the induced emf is from West to East}\text{.}\\ \\ \text{(c) The western end of the wire must be at higher potential}\text{.}\end{array}

Q.8 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.

Ans.

\begin{array}{l}\text{Here,}{\text{initial current, I}}_{\text{1}}\text{= 5}\text{.0 A}\\ {\text{Final current, I}}_{\text{2}}\text{= 0}\text{.0 A}\\ \text{Change in current,}{\text{dI = I}}_{\text{1}}{\text{– I}}_{\text{2}}\text{= 5}\text{A}\\ \text{Time taken for the change, dt = 0}\text{.1 s}\\ \text{Rate}\text{of}\text{change}\text{of}\text{current =}\frac{\text{dI}}{\text{dt}}\text{=}\frac{\text{5 A}}{\text{0}\text{.1 s}}\text{=}\text{50}{\text{As}}^{\text{-1}}\\ \text{Average induced}\text{emf, e = 200 V}\\ \text{The self-inductance}\left(\text{L}\right)\text{is}\text{related}\text{to}\text{average}\text{induced emf}\text{as:}\\ \text{e = L}\left(\frac{\text{dI}}{\text{dt}}\right)\\ \therefore \text{L =}\frac{\text{e}}{\left(\frac{\text{dI}}{\text{dt}}\right)}\text{=}\frac{\text{200}}{\text{50}}\text{=}\text{4}\text{H}\\ \therefore \text{The self induction of the coil is 4 H}\text{.}\end{array}

Q.9 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

Ans.

\begin{array}{l}\text{Here, mutual}\text{inductance,}\text{μ = 1}\text{.5}\text{H}\\ {\text{Initial current, I}}_{\text{1}}\text{= 0 A}\\ {\text{Final current, I}}_{\text{2}}\text{= 20 A}\\ {\text{Change in current,dI = I}}_{\text{2}}{\text{– I}}_{\text{1}}\\ \text{= 20 A – 0 A = 20}\text{A}\\ \text{Time taken, dt = 0}\text{.1 s}\\ \text{Induced}\text{emf}\text{is}\text{given}\text{as: e =}\frac{\text{d}\varphi }{\text{dt}}\to \text{(i)}\\ \text{Here,}\text{d}\varphi \text{= Change}\text{in}\text{the}\text{flux}\text{linkage}\text{with}\text{coil}\\ \text{Relation}\text{between}\text{emf}\text{and}\text{mutual}\text{inductance}\text{is}\text{given as:}\\ \text{e = μ}\frac{\text{dI}}{\text{dt}}\to \text{(ii)}\\ \text{Comparing}\text{equation}\text{(i)}\text{and}\text{(ii),}\text{we}\text{obtain:}\\ \frac{\text{d}\varphi }{\text{dt}}\text{= μ}\frac{\text{dI}}{\text{dt}}\\ \therefore \text{d}\varphi \text{= μdI}\\ \therefore \text{d}\varphi \text{= 1}\text{.5 H × 20 A = 30}\text{Wb}\\ \therefore \text{The change in the flux linkage is 30 Wb}\text{.}\end{array}

Q.10 A jet plane is travelling towards west at a speed of 1800 kmh^-1. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 ×10⁻⁴ T and the dip angle is 30°.

Ans.

\begin{array}{l}\text{Here},\text{speed of jet plane},\text{v}=\text{18}00{\text{kmh}}^{-\text{1}}\\ =\text{5}00{\text{ms}}^{-\text{1}}\\ \text{Wing span of the}\text{jet plane},l=\text{25 m}\\ \text{Magnetic field strength of Earth},\text{B}=\text{5}.0\times \text{1}{0}^{-\text{4}}\text{T}\\ \text{Angle of dip},\delta =\text{3}0°\\ \text{Vertical component of Earth}’\text{s magnetic field}\text{is}\text{given}as:\\ {\text{B}}_{\text{V}}=\text{B sin}\delta \\ =\text{5}\times \text{1}{0}^{-\text{4}}\text{sin 3}0°\\ =\text{2}.\text{5}\times \text{1}{0}^{-\text{4}}\text{T}\\ \text{Voltage difference between the ends of the wing is}\\ \text{given}as:\\ \text{e}=\left({\text{B}}_{\text{V}}\right)\times \text{l}\times \text{v}\\ =\text{2}.\text{5}\times \text{1}{0}^{-\text{4}}\text{T}\times \text{25 A}\times \text{5}00{\text{ms}}^{-\text{1}}\\ =\text{3}.\text{125 V}\\ \therefore \text{Voltage difference developed between the ends of}\\ \text{the wings =}\text{3}.\text{125 V}.\end{array}

Q.11 Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s⁻¹. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?

Ans.

\begin{array}{l}\text{Here,}\text{sides of the rectangular loop are 8 cm and 2 cm}\text{.}\\ \therefore \text{Area of the rectangular loop,A = length × width}\\ {\text{= 8 cm × 2 cm = 16 cm}}^{\text{2}}\\ {\text{= 16 × 10}}^{\text{-4}}{\text{m}}^{\text{2}}\\ \text{Initial magnetic field,}{\text{B}}_{\text{1}}\text{= 0}\text{.3}\text{T}\\ \text{Rate of decrease of magnetic field,}\frac{\text{dB}}{\text{dt}}\text{= 0}\text{.02}{\text{Ts}}^{\text{-1}}\\ \text{Induced}\text{emf developed in the rectangular}\text{loop is given as: e =}\frac{\text{d}\varphi }{\text{dt}}\\ \text{Here, d}\varphi \text{= Change in the}\text{flux through the loop}\text{area = AB}\\ \therefore \text{e =}\frac{\text{d}\left(\text{AB}\right)}{\text{dt}}\text{= A}\frac{\text{dB}}{\text{dt}}\\ {\text{= 16 × 10}}^{\text{-4}}{\text{m}}^2\text{× 0}{\text{.02 Ts}}^{\text{-1}}\\ \text{= 0}{\text{.32 × 10}}^{\text{-4}}\text{V}\\ \text{Resistance of loop, R = 1}\text{.6 Ω}\\ \text{Induced}\text{current,}\text{i =}\frac{\text{e}}{\text{R}}\text{=}\frac{\text{0}{\text{.32 ×10}}^{\text{-4}}\text{V}}{\text{1}\text{.6 Ω}}\\ \text{= 0}{\text{.2 × 10}}^{\text{-4}}\text{A}\\ \text{Power dissipated as heat is}\text{given}\text{as:}\\ {\text{P = i}}^{\text{2}}\text{R}\\ \text{=}{\left({\text{2 × 10}}^{\text{-5}}\text{A}\right)}^{\text{2}}\text{×1}\text{.6 Ω}\\ \text{= 6}{\text{.4 × 10}}^{\text{-10}}\text{W}\\ \text{The source of this heat loss is an external agent, which}\\ \text{causes the}\text{changing the}\text{magnetic field with time}\text{.}\end{array}

Q.12 A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s⁻¹ in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10⁻³ T cm⁻¹ along the negative x-direction (that is it increases by 10^{− 3} T cm⁻¹ as one moves in the negative x-direction), and it is decreasing in time at the rate of 10⁻³ T s⁻¹. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

Ans.

\begin{array}{l}\text{Here,}\text{side of square loop, is = 12 cm = 0}\text{.12 m}\\ \therefore \text{Area of square loop, A = 0}\text{.12 m × 0}\text{.12 m = 0}{\text{.0144 m}}^{\text{2}}\\ \text{Velocity of square}{\text{loop, v = 8 cms}}^{\text{-1}}\text{= 0}{\text{.08 ms}}^{\text{-1}}\\ \text{Gradient of magnetic field along negative x-direction:}\\ \frac{\text{dB}}{\text{dx}}{\text{= 10}}^{\text{-3}}{\text{Tcm}}^{\text{-1}}{\text{=10}}^{\text{-1}}{\text{Tm}}^{\text{-1}}\\ \text{Rate of decrease of the magnetic field:}\frac{\text{dB}}{\text{dt}}{\text{= 10}}^{\text{-3}}{\text{Ts}}^{\text{-1}}\\ \text{Resistance of loop,}\text{R = 4}\text{.5}\text{mΩ = 4}{\text{.5 ×10}}^{\text{-3}}\text{Ω}\\ \text{Rate of change of magnetic flux because}\text{of motion of}\\ \text{loop in a non-uniform magnetic field:}\frac{\text{d}\varphi }{\text{dt}}\text{= A ×}\frac{\text{dB}}{\text{dx}}\text{× v}\\ \text{= 0}{\text{.0144 m}}^{\text{2}}{\text{× 10}}^{\text{-1}}{\text{Tm}}^{\text{-1}}\text{× 0}{\text{.08 ms}}^{\text{-1}}\\ \text{=11}{\text{.52 × 10}}^{\text{-5}}{\text{Tm}}^{\text{2}}{\text{s}}^{\text{-1}}\\ \text{Rate of change of magnetic}\text{flux due to explicit time}\\ \text{change}\text{in}\varphi \text{=A}\left(\frac{\text{dB}}{\text{dt}}\right)\\ \text{= 0}{\text{.0144 × 10}}^{\text{-3}}{\text{Wbs}}^{\text{-1}}\text{= 1}{\text{.44 ×10}}^{\text{-5}}{\text{Wbs}}^{\text{-1}}\\ \text{Since,}\text{rate of change of magnetic flux = induced emf}\\ \therefore \text{Total induced emf,}\text{e = 1}{\text{.44 × 10}}^{\text{-5}}{\text{Wbs}}^{\text{-1}}\text{+ 11}{\text{.52 × 10}}^{\text{-5}}{\text{Tm}}^{\text{2}}{\text{s}}^{\text{-1}}\text{= 12}{\text{.96 ×10}}^{\text{-5}}\text{V}\\ \therefore \text{Induced current,}\text{i =}\frac{\text{e}}{\text{R}}\text{=}\frac{\text{12}{\text{.96 × 10}}^{\text{-5}}V}{\text{4}{\text{.5 ×10}}^{\text{-3}}\Omega }\\ \text{i = 2}{\text{.88 × 10}}^{\text{-2}}\text{A}\\ \therefore \text{The direction of induced current is such that there is}\\ \text{an increase in the magnetic}\text{flux}\text{through the loop along}\\ \text{positive z-direction}\text{.}\end{array}

Q.13 It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm² with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.

Ans.

\begin{array}{l}\text{Here,}{\text{area of the small flat search coil, A = 2 cm}}^{\text{2}}{\text{= 2 × 10}}^{\text{-4}}{\text{m}}^{\text{2}}\\ \text{Number of turns}\text{on}\text{the}\text{coil, N = 25}\\ \text{Total charge flowing through the coil, Q = 7}\text{.5 mC = 7}{\text{.5 × 10}}^{\text{-3}}\text{C}\\ \text{Total resistance of the coil and galvanometer,}\text{R = 0}\text{.50 Ω}\\ \text{Induced current through the coil}\text{is}\text{given}\text{as:}\text{I =}\frac{\text{e}}{\text{R}}\to \text{(i)}\\ \text{Induced emf,}\text{e}\text{=}\text{-N}\frac{\text{d}\varphi }{\text{dt}}\to \text{(ii)}\\ \text{Here,}\text{d}\varphi \text{= Change in flux}\\ \text{Comparing equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{,}\text{we}\text{obtain:}\\ \text{I = –}\frac{\text{N}\frac{\text{d}\varphi }{\text{dt}}}{\text{R}}\\ \therefore \text{Idt = –}\frac{\text{N}}{\text{R}}\text{d}\varphi \to \text{(iii)}\\ \text{Initial flux through coil}\text{is}\text{given}\text{as:}{\varphi }_{\text{i}}\text{= BA}\\ \text{Here,}\text{B = Strength}\text{of}\text{magnetic field}\\ \text{Final flux through coil,}{\varphi }_{\text{f}}\text{= 0}\\ \text{Integrating equation (iii)}\text{on both sides, we obtain:}\\ {\displaystyle \int \text{Idt}}\text{=-}\frac{\text{N}}{\text{R}}{\displaystyle \underset{{\varphi }_{\text{i}}}{\overset{{\varphi }_{\text{f}}}{\int }}\text{d}\varphi }\\ \text{Total charge,}\text{Q =}{\displaystyle \int \text{Idt}}\\ \therefore \text{Q = –}\frac{\text{N}}{\text{R}}\left({\varphi }_{\text{f}}\text{–}{\varphi }_{\text{i}}\right)\text{= –}\frac{\text{N}}{\text{R}}\left(\text{0 –}{\varphi }_{\text{i}}\right)\text{=}\frac{\text{N}{\varphi }_{\text{i}}}{\text{R}}\\ \text{Q =}\frac{\text{NBA}}{\text{R}}\\ \therefore \text{B =}\frac{\text{QR}}{\text{NA}}\\ \text{=}\frac{\text{7}{\text{.5 ×10}}^{\text{-3}}\text{C × 0}\text{.5 Ω}}{{\text{25 × 2 ×10}}^{\text{-4}}{\text{m}}^{\text{2}}}\\ \text{= 0}\text{.75}\text{T}\\ \therefore \text{The magnetic}\text{field strength of the magnet is 0}\text{.75 T}\text{.}\end{array}

Q.14 Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod= 9.0 mΩ. Assume the field to be uniform.

(a) Suppose K is open and the rod is moved with a speed of 12 cm s⁻¹ in the direction shown. Give the polarity and magnitude of the induced emf.

(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?

(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.

(d) What is the retarding force on the rod when K is closed?

(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s⁻¹) when K is closed? How much power is required when K is open?

(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?

(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Ans.

\begin{array}{l}\text{Here,}\text{length of rod, l = 15 cm= 0}\text{.15 m}\\ \text{Strength}\text{of}\text{magnetic field , B = 0}\text{.50 T}\\ \text{Resistance of the}\text{closed loop}\text{containing}{\text{rod, R= 9 mΩ = 9 × 10}}^{\text{-3}}\text{Ω}\\ \left(\text{a}\right)\text{Here,}{\text{speed of rod, v = 12 cms}}^{\text{-1}}\text{= 0}{\text{.12 ms}}^{\text{-1}}\\ \text{Magnitude}\text{of}\text{induced emf is given as:e = Bvl}\\ \text{= 0}\text{.5 × 0}\text{.12 × 0}\text{.15= 9 mV}\\ \text{The polarity of the induced emf is such that P acquires}\\ \text{positive polarity}\text{while Q acquires negative polarity}\text{.}\\ \left(\text{b}\right)\text{Yes; when key K is open, excess positive}\text{charge}\\ \text{develops}\text{at P}\text{and}\text{an}\text{equal}\text{negative}\text{excess}\text{charge}\\ \text{develops}\text{at}\text{Q}\text{. When key K is closed, continuous flow of}\\ \text{induced}\text{current}\text{maintains}\text{the}\text{excess charge}\text{.}\\ \left(\text{c}\right)\text{When}\text{key}\text{K}\text{is}\text{open}\text{and}\text{the}\text{rod}\text{is}\text{moving}\text{uniformly,}\\ \text{there}\text{is}\text{no}\text{net}\text{force}\text{on}\text{the}\text{electron}\text{in}\text{the}\text{rod}\text{PQ}\text{.}\text{This}\\ \text{is}\text{because,}\text{the}\text{effect}\text{of}\text{Lorentz}\text{magnetic force is cancelled}\\ \text{by the electric force set-up due to the excess charge of}\\ \text{opposite nature at both ends of the rod}\text{.}\\ \left(\text{d}\right)\text{Here,}\text{induced}\text{current flowing through the rod, I =}\frac{\text{e}}{\text{R}}\\ \text{=}\frac{{\text{9 × 10}}^{\text{-3}}}{{\text{9 × 10}}^{\text{-3}}}\text{= 1}\text{A}\\ \text{Retarding force on the rod, F = BIl}\\ \text{F = 0}\text{.50 T ×1 A × 0}\text{.15 m = 7}{\text{.5 × 10}}^{\text{-2}}\text{N}\\ \left(\text{e}\right)\text{Here,}{\text{speed of rod, v = 12 cms}}^{\text{-1}}\text{= 0}{\text{.12 ms}}^{\text{-1}}\\ \text{Retarding force on the rod, F = 7}{\text{.5 ×10}}^{\text{-2}}\text{N}\\ \therefore \text{Power = F×v}\\ \text{P = 7}{\text{.5 × 10}}^{\text{-2}}\text{N× 0}{\text{.12 ms}}^{\text{-1}}{\text{= 9 × 10}}^{\text{-3}}\text{W}\text{.}\\ \left(\text{f}\right)\text{Power dissipated as heat:}{\text{P = I}}^{\text{2}}\text{R=}{\left(\text{1 A}\right)}^{\text{2}}{\text{× 9 × 10}}^{\text{-3}}\text{Ω}\\ {\text{= 9 × 10}}^{\text{-3}}\text{W = 9 mW}\\ \text{Source of power is external agent in this case}\text{.}\\ \left(\text{g}\right)\text{As}\text{magnetic}\text{field}\text{is}\text{parallel}\text{to}\text{the}\text{rails,}{\text{θ = 0}}^{\text{o}}\\ \therefore \text{Induced}\text{emf}\text{=}\text{Bvlsinθ}\\ {\text{= Bvlsin0}}^{\text{o}}\text{= 0}\\ \text{In this case, motion of the rod does not cross the field}\\ \text{lines}\text{so}\text{that}\text{flux}\text{change = 0}\text{and}\text{therefore,}\text{e = 0}\text{.}\end{array}

Q.15 An air-cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10⁻³ s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Ans.

\begin{array}{l}\text{Here,}\text{length of solenoid, l = 30 cm}\\ \text{= 0}\text{.3 m}\\ \text{Cross-sectional}{\text{area, A = 25 cm}}^{\text{2}}\\ {\text{= 25 × 10}}^{\text{-4}}{\text{m}}^{\text{2}}\\ \text{Number of turns, N = 500}\\ \text{Current flowing}\text{through the solenoid, I = 2}\text{.5 A}\\ \text{Time}\text{for}\text{which}{\text{current flows, t = 10}}^{\text{-3}}\text{s}\\ \text{Average back emf}\text{is}\text{given}\text{as:}\text{e =}\frac{\text{d}\varphi }{\text{dt}}\to \text{(i)}\\ \text{Here,}\text{d}\varphi \text{= Change in flux = NAB}\to \left(\text{2}\right)\\ \text{Here,}{\text{B = Magnetic field strength = μ}}_{\text{o}}\frac{\text{NI}}{\text{l}}\to \text{(iii)}\\ \text{Here,}{\text{μ}}_{\text{o}}\text{= Permeability of free space}\\ \text{= 4}\pi {\text{× 10}}^{\text{-7}}{\text{TmA}}^{\text{-1}}\\ \text{Applying equations}\left(\text{ii}\right)\text{and}\left(\text{iii}\right)\text{in equation}\left(\text{i}\right)\text{,}\text{we obtain:}\\ \text{e =}\frac{{\text{μ}}_{\text{o}}{\text{N}}^{\text{2}}\text{IA}}{\text{lt}}\\ \text{=}\frac{\text{4}\pi {\text{× 10}}^{\text{-7}}{\text{TmA}}^{\text{-1}}\text{×}{\left(\text{500}\right)}^{\text{2}}\text{× 2}{\text{.5 A × 25 ×10}}^{\text{-4}}{\text{m}}^{\text{2}}}{\text{0}{\text{.3 m ×10}}^{\text{-3}}\text{s}}\\ \text{= 6}\text{.5}\text{V}\\ \therefore \text{Average back emf induced in the solenoid is 6}\text{.5 V}\text{.}\end{array}

Q.16 (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s. Calculate the induced emf in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Take a small strip\hspace{0.33em}of\hspace{0.33em}width dy in the loop at a}\\ \text{distance y\hspace{0.33em}from the wire carrying\hspace{0.33em}current.}\\ \text{Here,\hspace{0.33em}length\hspace{0.33em}of\hspace{0.33em}small\hspace{0.33em}strip = a}\\ \text{Area of small\hspace{0.33em}strip,\hspace{0.33em}dA = a × dy}\\ \text{Let\hspace{0.33em}current flowing\hspace{0.33em}through the wire = I}\\ \text{Magnetic field\hspace{0.33em}due\hspace{0.33em}to\hspace{0.33em}current\hspace{0.33em}carrying\hspace{0.33em}wire\hspace{0.33em}at\hspace{0.33em}a\hspace{0.33em}}\\ \text{distance\hspace{0.33em}y\hspace{0.33em}from\hspace{0.33em}the\hspace{0.33em}wire\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:\hspace{0.33em}B =}\frac{{\text{μ}}_{\text{o}}\text{I}}{\text{2}\mathrm{\pi }\text{y}}\\ \text{Magnetic flux associated with the\hspace{0.33em}small\hspace{0.33em}strip,\hspace{0.33em}d}\mathrm{\varphi }\text{= BdA}\\ \therefore \text{d}\mathrm{\varphi }\text{\hspace{0.33em}=\hspace{0.33em}}\frac{{\text{μ}}_{\text{o}}\text{I}}{\text{2}\mathrm{\pi }\text{y}}\left(\text{a × dy}\right)\\ \text{=\hspace{0.33em}}\frac{{\text{μ}}_{\text{o}}\text{Ia}}{\text{2}\mathrm{\pi }}\frac{\text{dy}}{\text{y}}\\ {\text{Here,\hspace{0.33em}μ}}_{\text{o}}\text{= Permeability of free space}\\ \text{= 4}\mathrm{\pi }{\text{× 10}}^{\text{-7}}{\text{T m A}}^{\text{-1}}\\ \text{Magnetic flux associated with the\hspace{0.33em}square\hspace{0.33em}loop\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:}\\ \mathrm{\varphi }\text{=}\frac{{\text{μ}}_{\text{o}}\text{Ia}}{\text{2}\mathrm{\pi }}\underset{\text{y = x}}{\overset{\text{y = a+x}}{\int }}\frac{\text{dy}}{\text{y}}\\ \mathrm{\varphi }\text{=}\frac{{\text{μ}}_{\text{o}}\text{Ia}}{\text{2}\mathrm{\pi }}{\left[{\text{log}}_{\text{e}}\text{y}\right]}_{\text{x}}^{\text{a+x}}\\ \mathrm{\varphi }\text{=}\frac{{\text{μ}}_{\text{o}}\text{Ia}}{\text{2}\mathrm{\pi }}{\text{log}}_{\text{e}}\left(\frac{\text{a+x}}{\text{x}}\right)\\ \text{For\hspace{0.33em}mutual\hspace{0.33em}inductance\hspace{0.33em}M,\hspace{0.33em}the\hspace{0.33em}magnetic\hspace{0.33em}flux\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:}\\ \mathrm{\varphi }\text{= MI}\\ \therefore \text{MI =}\frac{{\text{μ}}_{\text{o}}\text{Ia}}{\text{2}\mathrm{\pi }}{\text{log}}_{\text{e}}\left(\frac{\text{a+x}}{\text{x}}\right)\\ \text{M =}\frac{{\text{μ}}_{\text{o}}\text{a}}{\text{2}\mathrm{\pi }}{\text{log}}_{\text{e}}\left(\frac{\text{a}}{\text{x}}\text{+1}\right)\\ \left(\text{b}\right)\text{\hspace{0.33em}Given, I = 50 A}\\ \text{x = 0.2 m}\\ \text{a = 0.1 m}\\ {\text{v = 10 ms}}^{\text{-1}}\\ \text{Induced\hspace{0.33em}emf in the loop\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:}\\ \text{e = Bav}\\ \therefore \text{e =}\left(\frac{{\text{μ}}_{\text{o}}\text{I}}{\text{2}\mathrm{\pi }\text{x}}\right)\text{av}\\ \text{e =}\frac{\text{4}\mathrm{\pi }{\text{× 10}}^{\text{-7}}{\text{T m A}}^{\text{-1}}{\text{× 50 A × 0.1 m × 10 ms}}^{\text{-1}}}{\text{2}\mathrm{\pi }\text{× 0.2 m}}\\ {\text{e = 5 ×10}}^{\text{-5}}\text{\hspace{0.33em}V}\end{array}$

Q.17 A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

B = – B0 k (r ≤ a; a < R)
= 0 (otherwise)

What is the angular velocity of the wheel after the field is suddenly switched off?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.33em}line charge per unit length = λ}\\ \text{Mass of wheel = M}\\ \text{Radius of wheel = R}\\ \text{Line charge per unit length\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:\hspace{0.33em}λ =}\frac{\text{Net\hspace{0.33em}charge}}{\text{Length}}\text{=}\frac{\text{Q}}{\text{2}\mathrm{\pi }\text{r}}\\ \text{Here,\hspace{0.33em}r = Distance of the point within the wheel}\\ \text{Magnetic field\hspace{0.33em}strength\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:\hspace{0.33em}}\overrightarrow{\text{B}}{\text{= -B}}_{\text{o}}\widehat{\text{k}}\\ \text{At distance r,\hspace{0.33em}the\hspace{0.33em}centripetal\hspace{0.33em}force\hspace{0.33em}balances\hspace{0.33em}the\hspace{0.33em}}\\ \text{magnetic force i.e.,}\\ \text{BQv =}\frac{{\text{Mv}}^{\text{2}}}{\text{r}}\\ \text{Here,\hspace{0.33em}v = linear\hspace{0.33em}velocity\hspace{0.33em}of\hspace{0.33em}wheel}\\ \therefore \text{B2}\mathrm{\pi }\text{rλ =}\frac{\text{Mv}}{\text{r}}\\ \text{v =}\frac{\text{B2}\mathrm{\pi }{\text{λr}}^{\text{2}}}{\text{M}}\\ \therefore \text{Angular velocity\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}as:}\\ \text{ω =}\frac{\text{v}}{\text{R}}\text{=}\frac{\text{B2}\mathrm{\pi }{\text{λr}}^{\text{2}}}{\text{MR}}\\ \text{In\hspace{0.33em}case\hspace{0.33em}of\hspace{0.33em}r}\le \text{a\hspace{0.33em}and\hspace{0.33em}a < R,\hspace{0.33em}we\hspace{0.33em}have:}\\ \text{ω = –}\frac{\text{2}\mathrm{\pi }{\text{B}}_{\text{o}}{\text{a}}^{\text{2}}\text{λ}}{\text{MR}}\widehat{\text{k}}\end{array}$