A train is moving from south to north with a velocity of 90 km/h. The vertical component of earth's magnetic induction is 0.4\times { 10 }^{ -4 }\quad Wb/{ m }^{ 2 }. If the distance between the two rails is 1 m, what is the induced e.m.f. in its axle?

By taraday's law,Einduced - \frac{-d\phi }{dt} \phi = \vec{B}.\vec{A} = |\vec{B}| |\vec{A}| cos\theta Here \theta = 0 so, \phi = B.A \dfrac{-d\phi }{dt} = \dfrac{dB.A}{dt} = \dfrac{-BdA}{dt}-\dfrac{AdB}{dt} As B is constant, \dfrac{dB}{dt} = 0 = \dfrac{d\phi }{dt} = \dfrac{-B.dA}{dt} Here, \dfrac{dA}{dt} = \dfrac{dlb}{dt} = \dfrac{ldb}{dt}+\dfrac{bdl}{dt} b(breadth) of the train is notchanging it a length that is goinginto the field is varying with speed90km/h(25m/s),so. \dfrac{dA}{dt} = l(0)+bv = (1)(25) = 25 so, \dfrac{d\phi }{dt} = -(0.4) \times 10^{-4} \times 25 = 10^{-3}v Einduced = 10^{-3}v = 1 mv Option - A is correct.

Given L_1=1 mH, R_1=1\Omega, L_2=2 mH, R_2=2\Omega Neglecting mutual inductance, the time constants (in ms) for circuits (i), (ii), and (iii) are:

For (i): \displaystyle L_{eq}=L_1+L_2=1+2=3 mH and R_{eq}=R_1+R_2=1+2=3 \Omega Time constant \displaystyle =\dfrac{L_{eq}}{R_{eq}}=\dfrac{3}{3}=1 ms For (ii): \displaystyle L_{eq}=\dfrac{L_1L_2}{L_1+L_2}=\dfrac{1(2)}{1+2}=2/3 mH and R_{eq}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{1(2)}{1+2}=2/3 \Omega Time constant \displaystyle =\dfrac{L_{eq}}{R_{eq}}=\dfrac{2/3}{2/3}=1 ms For (iii): \displaystyle L_{eq}={L_1+L_2}={1+2}=3 mH and R_{eq}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{1(2)}{1+2}=2/3 \Omega Time constant \displaystyle =\dfrac{L_{eq}}{R_{eq}}=\dfrac{3}{2/3}=9/2 ms

Electromagnetic Induction - Top 10 Questions

Q: In Figure , a stiff wire bent into a semicircle of radius a = 2.0 cm is rotated at constant angular speed 40 rev/s in a uniform 20 mTmagnetic field. What are the (a) frequency and (b) amplitude of the emf induced in the loop?

(a) The frequency is f=\dfrac{\omega}{2 \pi}=\dfrac{(40\mathrm{rev} / \mathrm{s})(2 \pi \mathrm{rad} / \mathrm{rev})}{2 \pi}=40 \mathrm{Hz} (b) First, we define angle relativeto the plane of Figure such that the semicircular wire is in the \theta=0 position and a quarter of a period (of revolution) later it will be in the \theta=\pi/2 position (where its midpoint will reach a distance of a above theplane of the figure). At the moment it is in the \theta=\pi / 2 position,the area enclosed by the "circuit" will appear to us (as we look downat the figure) to that of a simple rectangle (call this area A_{0} which isthe area it will again appear to enclose when the wire is in the \theta=3 \pi/ 2 position). since the area of the semicircle is \pi a^{2} / 2, thenthe area (as it appears to us) enclosed by the circuit, as a function of ourangle \theta , is A=A_{0}+\dfrac{\pi a^{2}}{2} \cos\theta where(since \theta is increasing at a steady rate) the angle depends linearly ontime, which we can write either as \theta=\omega t or \theta=2 \pi f ifwe take t=0 to be a moment when the arc is in the \theta=0 position.since \vec{B} is uniform (in space) and constant (in time), Faraday's lawleads to \varepsilon=-\dfrac{d \Phi_{B}}{dt}=-B \dfrac{d A}{d t}=-B \dfrac{d\left(A_{0}+\left(\pi a^{2} / 2\right) \cos\theta\right)}{d t}=-B \dfrac{\pi a^{2}}{2} \dfrac{d \cos (2 \pi f t)}{d t} which yields \varepsilon=B \pi^{2}a^{2} f \sin (2 \pi f t) . This (due to the sinusoidal dependence) reinforcesthe conclusion in part (a) and also (due to the factors in front of the sine)provides the voltage amplitude: \varepsilon_{m}=B \pi^{2} a^{2}f=(0.020 \mathrm{T}) \pi^{2}(0.020 \mathrm{m})^{2}(40 / \mathrm{s})=3.2 \times10^{-3} \mathrm{V}

Q: A pair of parallel horizontal conducting rails ot negligible resistance connected to a 2 F capacitor is fixed on a table.The distance between the rails is 1 m. A conducting rod of mass 0.5 kg can slide on the rails frictionlessly. The rod is tied at midpoint to a light string passing over a smooth pulley fixed to the edge of the table. A mass of 0.5 kg tied to the other end of the string hangs vertically as shown in the Figure. A uniform magnetic field of 0.5T can be applied in vertically downward direction. If the system is released from rest, then, point out incorrect statements from the following: [ g being the acceleration due to gravity]

Let v ne the instanteous velocity of the rod when the system is released from rest and a be the acceleration.Charge in C is q=|e|C=Blv.C \therefore i=BlC.\dfrac{dv}{dt}=BlCa \therefore force on the rod due to induced current in magnetic field B=ilB=B^{2}l^{2}Ca F.B.D. of the rod and mass m : T-F_{M}=ma mg-T=ma mg-F_{M}=2ma i.e., \ mg-B^{2}l^{2}Ca=2ma \therefore a=\dfrac{m}{(m+m+B^{2}l^{2}C)}g=\dfrac{m}{2m+B^{2}l^{2}C}g Hence, a=\dfrac{0.5}{(1+0.25\times 1\times 2)}g=\dfrac{g}{3} When B=0 , a'=\dfrac{g}{2} Hence, by putting B on a decrease from \dfrac{g}{2}to\dfrac{g}{3}. In the absence of magnetic field, let mass of the rod (effective) be M. T=Ma,mg-T=ma \therefore a=\dfrac{m}{(n+M)}g By comparison, M=m+B^{2}l^{2}C or mass of the rod should be increased by B^{2}l^{2}C=0.25\times 1\times 2=0.5kg

Q: An e.m.f. of 15 V is applied in a circuit containing 5 henry inductance and 10 ohm resistance. The ratio of currents at time t= \infty and t=1 second is

B. \dfrac{e^{2}}{e^{2}-1} Given, L=5H R= 10\Omega V_0= 12V The current in RL circuit is, I=I_0(1-e^{-t/\tau}) where \tau= \dfrac{L}{R} \tau=\dfrac{1}{2} sec I_0=\dfrac{V_0}{R}A The current in LR circuit at time t=\infty sec , I_1=I_0 The current in LR circuit at time 1 sec I_2=I_0(1-e^{-2}) The ratio of currents at time t=\infty sec and t=1 sec is \dfrac{I_1}{I_2}= \dfrac{I_0}{I_0(1-e^{-2})} \dfrac{I_1}{I_2}=\dfrac{e^2}{e^2-1}

Q: The magnetic induction at the point O . If the wire carries a current i , is

Hint: The resultant magnetic field is the vector sum of {B_x} and {B_y} .Correct Option: C Explanation for Correct Option: Step 1: Find the magnetic field due to the half ring wire, Magnetic field due to the half ring wire, {B_1} = \dfrac{{{\mu _o}i}}{{4R}} , where i is the current through the wire and R is the radius of the wire. Step 2: Magnetic Field induction at point O due to the two straight wires, Magnetic field at O, {B_2} = 2[\dfrac{{{\mu _o}i}}{{4\pi R}}(\cos 0 - \cos 90)] Step 3: Find the resultant magnetic field at point O . Resultant magnetic field, B = \sqrt {{B_1}^2 + {B_2}^2} B = \dfrac{{{\mu _o}i}}{{4\pi R}}{({\pi ^2} + {2^2})^{\dfrac{1}{2}}} So, the magnetic induction at point P is B = \dfrac{{{\mu _o}i}}{{4\pi R}}{({\pi ^2} + {2^2})^{\dfrac{1}{2}}}

Q: A circular loop of radius 0.3 cm lies parallel to a much bigger circular loop of radius 20 cm . The centre of the small loop is on the axis of the bigger loop. The distance between their centres is 15 cm . If a current of 2.0 A flows through the smaller loop, then the flux linked with bigger loop is

As field due to current loop 1 at an axial point B_1 = \dfrac{\mu_0 I_1 R^2}{2 (d^2 + R^2)^{3/2}} Flux linked with smaller loop 2 due to B_1 is \phi_2 = B_1 A_2 = \dfrac{\mu_0 I_1 R^2}{2 (d^2 + R^2)^{3/2}} \pi r^2 The coefficient of mutual inductance between the loops is M = \dfrac{\phi_2 }{I_1} = \dfrac{\mu_0 R^2 \pi r^2 }{2 (d^2 + R^2)^{3/2}} Flux linked with bigger loop 1 is \phi = MI_2 = \dfrac{\mu_0 R^2 \pi r^2 I_2}{2 (d^2 + R^2)^{3/2}} Substituting the given values, we get \phi_1 = \dfrac{4 \pi\times 10^{-7} \times (20 \times 10^{-2})^2 \times \pi \times (0.3 \times 10^{-2})^2 \times 2}{2 [(15 \times 10^{-2})^2 + (20 \times 10^{-2})^2]^{3/2} } \phi_1 = 9.1 \times 10^{-11} weber Hence, the correct answer is (D)

Q: This circuit is equivalent to a single inductor having inductance L_{eq} =18 H . Determine the value of the inductance L .

The equivalent of 20H and 60H inductors connected in paralle is \dfrac{20\times 60}{20+60}H=15H This is connected in series with 25H inductor giving an equivalent of 40H .Now this inductor is connected in parallel with 20H inductor and L , which are in turn connected to 10H inductor.Hence, L_{eq}=18H=10H+\dfrac{L(20)+L(40)+(20)(40)}{20.40.L} \implies L=20H

Q: An ac generator has emf \xi=\xi _{m}\sin \omega_{d}t , with \xi _{m}=25.0V and \omega _{d}=377rad/s . It is connected to a 12.7 H inductor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is -12.5 V and increasing in magnitude, what is the current?

(a) The circuit consists of one generator across one inductor; therefore, \varepsilon _{m}=V_{L} . The current amplitude is I_m=\dfrac{\varepsilon _{m}}{X_{L}}=\dfrac{\varepsilon _{m}}{\omega _{d}L}=\dfrac{25.0V}{\left ( 377rad/s \right )\left ( 12.7H \right )}=5.22 \times 10^{-3}A .(b) I = I_m sin(\omega _d t - \phi) --(i)Current will be maximum when sin(\omega _d t - \phi) = 1 or, \omega _d t - \phi = \dfrac{\pi}{2} At t = 0 potential difference equals zero as sin(\omega _d t) = 0 \phi = -\dfrac{\pi}{2} This is the phase angle between current and potential difference.Now, I = I_m sin(\omega_d t + \dfrac{\pi}{2}) or, I = I_m(cos \omega _d t) So, maximum current occurs when cos(\omega _d t) = 1 That gives t = 0 At this instant potential difference is zero: \varepsilon = 0 (c) Consider equation \xi =\xi _{m}\sin \omega _{d}t with \varepsilon =-\dfrac{\varepsilon _{m}}{2} . In order to satisfy this equation, we require \sin \left ( \omega _{d}t \right )=-1/2 . Now we note that the problem states that ε is increasing in magnitude, which (since it is already negative) means that it is becoming more negative. Thus, differentiating equation with respect to time (and demanding the result be negative) we must also require \cos \left ( \omega _{d}t \right )< 0 . These conditions imply that \omega t must equal \left ( 2n\pi -5\pi /6 \right ) [ n = integer]. Consequently, equation i=I\sin \left ( \omega _{d}t+\phi\right) yields (for all values of n ) i=I\sin \left ( 2n\pi -\frac{5\pi}{6}-\frac{\pi}{2} \right )=\left ( 5.22 \times 10^{-3}A \right )\left ( \frac{\sqrt{3}}{2} \right )=4.51 \times 10^{-3}A .

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Electromagnetic Induction

- Running short of time? Practice top 10 questions of the chapter and boost your confidence.

Calculate the induced emf in a closed loop.

In Figure , a stiff wire bent into a semicircle of radius a = 2.0 cm is rotated at constant angular speed 40 rev/s in a uniform 20 mTmagnetic field. What are the (a) frequency and (b) amplitude of the emf induced in the loop?

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Problems on gauss law and faraday law Lenz's law.

A train is moving from south to north with a velocity of

90

km/h. The vertical component of earth's magnetic induction is

0.4 \times 10^{- 4} W b / m^{2} .

If the distance between the two rails is

1 m,

what is the induced e.m.f. in its axle?

1 m V

0.1 m V

10 m V

100 m V

View solution

Solve problem on moving conductor with a constant velocity in a circuit with capacitor.

A pair of parallel horizontal conducting rails ot negligible resistance connected to a

2 F

capacitor is fixed on a table.The distance between the rails is

1 m .

A conducting rod of mass

0.5 k g

can slide on the rails frictionlessly. The rod is tied at midpoint to a light string passing over a smooth pulley fixed to the edge of the table. A mass of

0.5 k g

tied to the other end of the string hangs vertically as shown in the Figure. A uniform magnetic field of

0.5 T

can be applied in vertically downward direction. If the system is released from rest, then, point out incorrect statements from the following:

[

g

being the acceleration due to gravity]

This question has multiple correct options

When the magnetic field is put on, the acceleration of the system increases from

\frac{g}{3}

\frac{g}{2}

Due to induction, the change in the acceleration of the system is same as that obtained by increasing the mass of the rod by

0.5 k g

in the absence of the magnetic field.

Due to induction, the change in the acceleration of the system is same as that obtained by decreasing the mass of the rod by

0.25 k g

in the absence of magnetic field.

When the magnetic field is put on, the acceleration of the system decreases from

\frac{g}{2}

\frac{g}{3}

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Problem on Mutual Inductance

An e.m.f. of 15 V is applied in a circuit containing 5 henry inductance and 10 ohm resistance. The ratio of currents at time $t = \infty$ and $t = 1$ second is

\frac{e ^{1 / 2}}{e ^{1 / 2} - 1}

\frac{e ^{2}}{e ^{2} - 1}

1 - \frac{1}{e}

\frac{1}{e}

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The magnetic induction at the point

O

. If the wire carries a current

i

, is

\frac{μ _{0} i}{2 R}

\frac{μ _{0} i}{2 π R}

\frac{μ _{0} i ( π ^{2} + 4 ) ^{1 / 2}}{4 π R}

\frac{μ _{0} i ( π ^{2} + 4 )}{4 π R}

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Calculate mutual inductance of two inductors of different shapes.

A circular loop of radius

0.3 c m

lies parallel to a much bigger circular loop of radius

20 c m

. The centre of the small loop is on the axis of the bigger loop. The distance between their centres is

15 c m

. If a current of

2.0 A

flows through the smaller loop, then the flux linked with bigger loop is

6 \times 1 0^{- 11} W b

3.3 \times 1 0^{- 11} W b

6.6 \times 1 0^{- 9} W b

9.1 \times 1 0^{- 11} W b

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Calculate the effective inductance of inductors in series and parallel.

This circuit is equivalent to a single inductor having inductance

L_{e q}

= 18 H

. Determine the value of the inductance

L

20 H

40 H

60 H

80 H

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Given

L_{1} = 1 m H, R_{1} = 1 Ω, L_{2} = 2 m H, R_{2} = 2 Ω

Neglecting mutual inductance, the time constants (in ms) for circuits (i), (ii), and (iii) are:

1, 1, \frac{9}{2}

\frac{9}{4}, 1, 1

1, 1, 1

1, \frac{9}{4}, 1

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Solve problems on LR circuits involving multiple resistors and inductors.

The current through the inductor after a long time will be

zero

infinite

\frac{6}{1 3} A

none of these

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Calculate the emf induced in an AC generator.

An ac generator has emf

ξ = ξ_{m} sin ω_{d} t

, with

ξ_{m} = 25.0 V

and

ω_{d} = 377 r a d / s

. It is connected to a

12.7 H

inductor. (a) What is the maximum value of the current? (b) When the current is a maximum, what is the emf of the generator? (c) When the emf of the generator is

- 12.5 V

and increasing in magnitude, what is the current?

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