JEE Physics Electromagnetic Induction MCQs Set 07

Answer: (a) 10 mV

Explanation:
1. Area \( A=\pi r^2=\pi\times\frac{100}{\pi}\times10^{-4}=10^{-2}\ \text{m}^2 \).
2. Rate of change of field \( \frac{dB}{dt}=\frac{0.5}{0.5}=1\ \text{T/s} \) (steady, so the emf is the same at 0.25 s).
3. \( \varepsilon=A\frac{dB}{dt}=10^{-2}\ \text{V}=10\ \text{mV} \).

Teacher's Note:
1. A steady rate of change gives a constant emf.
2. Convert 10/√π cm to metres before squaring.

Answer: (c) \( \frac12B L^2\omega \)

Explanation:
1. Speed of an element at distance x from the axis: \( v=\omega x \).
2. \( \varepsilon=\int_0^LB\omega x\,dx=\frac12B\omega L^2 \).

Teacher's Note:
1. The emf is the sum of the motional emfs B v dx of all small elements.
2. The average speed of the rod is ωL/2.

Answer: (a) 3 A

Explanation:
1. Long after closing the switch, the current is steady, so the ideal inductors have no voltage across them and behave like plain wires.
2. Each branch then has a resistance of 12 ohm, and three 12 ohm branches in parallel give \( \frac{12}{3}=4\ \Omega \).
3. \( I=\frac{12}{4}=3\ \text{A} \).

Teacher's Note:
1. In steady state di/dt = 0, so the induced emf of an ideal inductor is zero.
2. The inductance affects only the transient period.

Answer: (a) 10 mV

Explanation:
1. \( \varepsilon=\left|\frac{d\Phi}{dt}\right|=B\frac{d(\pi r^2)}{dt}=2\pi rB\left|\frac{dr}{dt}\right| \).
2. \( \varepsilon=2\pi(0.1)(0.8)(0.02)\approx1.0\times10^{-2}\ \text{V}=10\ \text{mV} \).

Teacher's Note:
1. Differentiate the flux B π r² with respect to time.
2. Convert cm to m: r = 0.1 m and dr/dt = 0.02 m/s.

Answer: (d) 16 V

Explanation:
1. \( \varepsilon=Blv=2\times1\times8=16\ \text{V} \).

Teacher's Note:
1. Motional emf = B l v when B, l and v are mutually perpendicular.
2. Units: T m (m/s) = V.

Answer: (b) 0

Explanation:
1. Each wire is straight and ends at a corner, so use \( B=\frac{\mu_0I}{4\pi d}(\sin\theta_2-\sin\theta_1) \) with perpendicular distance \( d=\frac a2 \) for every wire.
2. Wires AB and ED each subtend angles from 45° to 90° or -90° to -45° (factor 0.29), and wires BC and TE each subtend factors 1.71, in units of \( \frac{\mu_0I}{2\pi a} \).
3. By the right-hand rule AB and TE give fields into the page: \( 0.29+1.71=2 \). BC and ED give fields out of the page: \( 1.71+0.29=2 \).
4. The net field at O is zero.

Teacher's Note:
1. Find the direction of each field with the right-hand rule before adding magnitudes.
2. The arrangement is symmetric about the point O, which makes the contributions cancel in pairs.

Answer: (c) \( M=\frac{2\sqrt2\mu_0R^2}{L} \)

Explanation:
1. Field at the centre due to one side of the square (distance \( \frac L2 \), angles 45° each way): \( B_1=\frac{\mu_0I}{4\pi(L/2)}(2\sin45^\circ)=\frac{\sqrt2\mu_0I}{2\pi L} \).
2. Four sides: \( B=\frac{2\sqrt2\mu_0I}{\pi L} \).
3. The small loop sees an almost uniform field: \( \Phi=B\pi R^2=\frac{2\sqrt2\mu_0R^2}{L}I \), so \( M=\frac\Phi I=\frac{2\sqrt2\mu_0R^2}{L} \).

Teacher's Note:
1. Put the current in the large loop and find the flux through the small loop.
2. Because L is much larger than R, the field over the small loop is nearly constant.

Answer: (a) 2 T

Explanation:
1. The wire length is fixed: \( 4\times2\pi r=2\pi r'\Rightarrow r'=4r \).
2. \( B=\frac{\mu_0NI}{2r}\propto\frac Nr \): the original has \( \frac4r \), the new coil has \( \frac1{4r} \).
3. \( B'=32\times\frac{1/4r}{4/r}=32\times\frac1{16}=2\ \text{T} \).

Teacher's Note:
1. Unwinding keeps the total wire length constant, so the radius becomes four times larger.
2. N decreases by 4 and r increases by 4, so B falls by 16.

Answer: (a) 44 V

Explanation:
1. \( \omega=\frac{2\pi\times500}{60}=\frac{2\times22\times500}{7\times60}=52.4\ \text{rad/s} \).
2. The plane at 60° to the field means the normal is at 30° to the field: \( \varepsilon=NBA\omega\sin30^\circ \).
3. \( \varepsilon=600\times0.4\times70\times10^{-4}\times52.4\times0.5=44\ \text{V} \).

Teacher's Note:
1. Flux = NBA cos(angle between the normal and B), so the emf is NBAω sin(angle).
2. If the plane makes 60° with the field, the normal makes 30°.