JEE Physics Electromagnetic Induction MCQs Set 21

Answer: (c) \( \dfrac{Br^2\omega}{2R} \)

Explanation:
1. Only the radial arm OA (or OD) is moving through the field; the flux changes at the rate \( \dfrac{d\Phi}{dt}=B\dfrac{r^2}{2}\omega \).
2. \( i=\dfrac{\varepsilon}{R}=\dfrac{Br^2\omega}{2R} \).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (d) 0.02 m/s

Explanation:
1. The balanced bridge (four 3 \( \Omega \) arms and a 3 \( \Omega \) diagonal carrying no current) has resistance \( \dfrac{6\times6}{6+6}=3\ \Omega \). Total resistance \( =1+3=4\ \Omega \).
2. \( \varepsilon=IR_{tot}=10^{-3}\times4=4\times10^{-3} \) V \( =Blv_0=2\times0.1\times v_0\Rightarrow v_0=0.02 \) m/s.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (b) \( 6.67\times10^{-4}\ \text{Wb} \)

Explanation:
1. Mutual inductance: \( M=\dfrac{\Phi_{Q}}{I_P}=\dfrac{10^{-3}}{3} \) H.
2. By reciprocity \( \Phi_P=MI_Q=\dfrac{10^{-3}}{3}\times2=6.67\times10^{-4} \) Wb.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (c) \( 1/L \)

Explanation:
1. \( L_{ind}=\mu_0n^2AL=\mu_0\dfrac{N^2}{L}A \) with \( n=N/L \).
2. For fixed N and A, \( L_{ind}\propto\dfrac1L \).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (a) \( \dfrac{n_2}{n_1} \)

Explanation:
1. Mutual inductance: \( M=\mu_0n_1n_2\pi r_1^2l \) (flux of the outer solenoid through the inner coil). Self-inductance of the inner coil: \( L_1=\mu_0n_1^2\pi r_1^2l \).
2. \( \dfrac M{L_1}=\dfrac{n_2}{n_1} \).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (b) \( l^2/L \)

Explanation:
1. Field at the centre due to the large loop: \( B\propto\dfrac{\mu_0I}{L} \), nearly uniform over the small loop.
2. Flux \( =Bl^2\propto\dfrac{\mu_0Il^2}{L} \), so \( M\propto\dfrac{l^2}{L} \).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (c) (i), (iii) and (iv)

Explanation:
1. Same \( dI/dt \): \( V=L\dfrac{dI}{dt}\Rightarrow\dfrac{V_1}{V_2}=\dfrac{L_1}{L_2}=4 \) (iv).
2. Same power \( Vi \): \( \dfrac{i_1}{i_2}=\dfrac{V_2}{V_1}=\dfrac14 \) (i). Energy \( \tfrac12Li^2 \): \( \dfrac{W_1}{W_2}=\dfrac{8\times1}{2\times16}=\dfrac14 \) (iii).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (a) \( \text{decreasing as }1/r \)

Explanation:
1. Apply \( \oint\vec E\cdot d\vec l=-\dfrac{d\Phi}{dt} \) to a circle of radius r: outside the field region the flux is \( B\pi a^2 \).
2. \( E\cdot2\pi r=\pi a^2\dfrac{dB}{dt}\Rightarrow E=\dfrac{a^2}{2r}\dfrac{dB}{dt}\propto\dfrac1r \).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (b) \( \dfrac{EL}{2.7R^2} \)

Explanation:
1. \( i=\dfrac ER\left(1-e^{-Rt/L}\right) \), so \( q=\int_0^{L/R}i\,dt=\dfrac ER\left[t-\dfrac LR\left(1-e^{-Rt/L}\right)\right]_0^{L/R} \).
2. \( q=\dfrac ER\cdot\dfrac LR\,e^{-1}=\dfrac{EL}{eR^2}\approx\dfrac{EL}{2.7R^2} \).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (d) \( 2\ln2 \)

Explanation:
1. \( \tau=\dfrac LR=2 \) s and \( i=\dfrac ER(1-e^{-t/\tau}) \).
2. Heat rate \( =i^2R \), magnetic energy rate \( =Li\dfrac{di}{dt} \). Equal when \( iR=L\dfrac{di}{dt} \), i.e. \( E(1-e^{-t/\tau})=Ee^{-t/\tau}\Rightarrow e^{-t/\tau}=\dfrac12 \), so \( t=\tau\ln2=2\ln2 \).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (a) 6 A

Explanation:
1. After a long time the inductor behaves as a plain wire (no back emf).
2. The branches are then 5 \( \Omega \) and 5 \( \Omega \) in parallel: \( 2.5\ \Omega \), so \( I=\dfrac{15}{2.5}=6 \) A.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (c) -1

Explanation:
1. With no source in the loop, Kirchhoff's law gives \( V_R+V_L=0 \) at every instant.
2. Therefore \( V_R/V_L=-1 \), at \( t=L/R \) as at any other time.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (d) 1 ms

Explanation:
1. \( I=\dfrac VR\left(1-e^{-t/\tau}\right)=2(1-e^{-t/\tau}) \) with \( \tau=\dfrac{8.4\times10^{-3}}{6}=1.4 \) ms.
2. \( I=1 \) A gives \( e^{-t/\tau}=\dfrac12\Rightarrow t=\tau\ln2=0.97\approx1 \) ms.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (b) After a time \( 2\tau \) the charge on the capacitor is \( CV(1-e^{-2}) \)

Explanation:
1. \( q=CV(1-e^{-t/\tau}) \): at \( t=\tau \), \( q=0.63\,CV \); at \( t=2\tau \), \( q=CV(1-e^{-2}) \).
2. The source does work \( CV^2 \) while the capacitor stores \( \tfrac12CV^2 \) and the resistor dissipates \( \tfrac12CV^2 \): the work done is twice the heat, not half.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.