JEE Physics Electromagnetic Induction MCQs Set 11

Answer: (d) (i), (ii) and (iii) only

Explanation:
1. Flux: \( \Phi=BA\cos\omega t \) and emf \( \varepsilon=BA\omega\sin\omega t \).
2. Flux is maximum when the plane is perpendicular to B, where the emf is zero; flux is zero when the plane is parallel to B, where the emf is maximum.
3. So (i), (ii), (iii) are correct and (iv) is wrong; flux and emf are a quarter cycle (phase \( \pi/2 \)) apart.

Teacher's Note:
1. Flux and emf are like cosine and sine of the same angle.

Answer: (a) all four

Explanation:
1. (i) Emf \( vB\ell \) over resistance r gives \( \frac{vB\ell}{r} \).
2. (ii) \( mg-B\ell i=ma\Rightarrow a=g-\frac{B^2\ell^2v}{mr}=g-\frac{g\,v}{v_m} \).
3. (iii) Solving \( \frac{dv}{dt}=g(1-v/v_m) \) gives the exponential; (iv) integrating it gives the displacement. All four are correct.

Teacher's Note:
1. This is the standard terminal-velocity problem with an exponential approach.

Answer: (c) (i), (ii) and (iii) only

Explanation:
1. The current cannot jump in an inductive circuit, so it is zero at t = 0.
2. Energy \( \frac12Li^2 \) and power \( \varepsilon i \) are therefore zero.
3. The induced emf is not zero: it equals the full battery emf and opposes it.

Teacher's Note:
1. The inductor takes the entire emf at t = 0.

Answer: (d) (i) and (iii) only

Explanation:
1. Rate of storage: \( \frac{dU}{dt}=Li\frac{di}{dt}=i(E-iR)=i(2-i) \).
2. This is maximum when \( i=1 \) A, with value \( 1\times1=1 \) W.
3. So (i) and (iii) are correct.

Teacher's Note:
1. Maximise \( i(E-iR) \) with respect to i: the maximum is at \( i=E/2R \).

Answer: (b) (i), (iii) and (iv) only

Explanation:
1. With the same di/dt, \( V=L\frac{di}{dt} \) gives \( \frac{V_2}{V_1}=\frac{L_2}{L_1}=\frac14 \) (iv).
2. Equal power \( Vi \): \( V_1i_1=V_2i_2\Rightarrow\frac{i_1}{i_2}=\frac{V_2}{V_1}=\frac14 \) (i).
3. Energy \( W=\frac12Li^2 \): \( \frac{W_2}{W_1}=\frac{2\times16}{8\times1}=4 \) (iii).

Teacher's Note:
1. Use proportionalities rather than numbers.

Answer: (c) 0.16 A, clockwise

Explanation:
1. Loop area \( 0.2\times0.4=0.08 \) \( m^2 \). Emf: \( A\frac{dB}{dt}=0.08\times2=0.16 \) V.
2. Resistance 1 ohm, so \( i=0.16 \) A.
3. The outward flux increases, so the induced current produces an inward field: clockwise.

Teacher's Note:
1. Increasing field out of the page gives a clockwise induced current.

Answer: (a) \( I=\frac{E}{R} \)

Explanation:
1. In steady state the current is constant, so there is no induced emf in the inductor, whatever its value.
2. The battery emf then drops only across the resistance: \( I=\frac ER \).

Teacher's Note:
1. The steady current depends only on resistance, never on inductance.

Answer: (b) \( B=5\ \mathrm{T},\ 75\ \mathrm{K}<T_C(B)<100\ \mathrm{K} \)

Explanation:
1. Since \( T_C \) decreases monotonically with B, a field smaller than 7.5 T gives \( T_C \) between 75 K and 100 K.
2. So for 5 T, \( 75<T_C<100 \) K, but its exact value (such as 80 K) cannot be fixed.
3. For 10 T, \( T_C<75 \) K, but not necessarily 70 K.

Teacher's Note:
1. Only inequalities follow from monotonic behaviour.

Answer: (d) (i) and (iii) only

Explanation:
1. The wire's field above the diameter points one way through the loop's plane and below it the opposite way.
2. By symmetry the flux through the two halves is equal and opposite, so the net flux is always zero.
3. Hence the emf is zero whether the current is constant or changing: (i) and (iii).

Teacher's Note:
1. Look for symmetry: opposite fluxes cancel.

Answer: (a) \( \frac{BR}{2} \)

Explanation:
1. Induced emf around the orbit: \( \varepsilon=\pi R^2\frac{dB}{dt}=\pi R^2B \) (since B rises by B in 1 s).
2. Electric field: \( E=\frac{\varepsilon}{2\pi R}=\frac{BR}{2} \).

Teacher's Note:
1. The induced field is tangential and uniform in magnitude around the orbit.

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

Explanation:
1. Both inductors have the same voltage at all times, so \( L_1i_1=L_2i_2 \) and \( \frac{i_1}{i_2}=\frac{L_2}{L_1} \) always: (iv) true.
2. Finally the total current is \( V/R \), divided as \( i_1=\frac VR\frac{L_2}{L_1+L_2} \) and \( i_2=\frac VR\frac{L_1}{L_1+L_2} \): (i) and (iii) true.
3. At t = 0 the inductors carry no current, so the current through R is zero: (ii) false.

Teacher's Note:
1. Parallel inductors share current in the inverse ratio of their inductances, at every instant.

Answer: (d) \( 0.63\times10^{-3} \) A

Explanation:
1. Motional emf: \( B\ell v=1\times0.1\times0.01=10^{-3} \) V.
2. Time constant \( \tau=\frac LR=1 \) ms and final current \( \frac{10^{-3}}{1}=10^{-3} \) A.
3. After one time constant: \( i=10^{-3}(1-e^{-1})=0.63\times10^{-3} \) A.

Teacher's Note:
1. The moving wire is a constant-emf source driving an LR growth circuit.

Answer: (b) 0.15 mV

Explanation:
1. Velocity (east), field (north) and aerial (vertical) are mutually perpendicular.
2. Emf: \( B\ell v=5.0\times10^{-5}\times2\times1.5=1.5\times10^{-4} \) V = 0.15 mV.

Teacher's Note:
1. Use the horizontal field component since the aerial is vertical and the boat moves horizontally.

Answer: (c) 3 mV

Explanation:
1. Emf: \( B\ell v=0.30\times10^{-4}\times20\times5.0=3\times10^{-3} \) V = 3 mV.

Teacher's Note:
1. The wire, its velocity (down) and the horizontal field (north-south) are mutually perpendicular.

Answer: (a) \( \frac{5B\omega\ell^2}{2} \)

Explanation:
1. The rod's ends are at distances \( 2\ell \) and \( 3\ell \) from the fixed pivot.
2. Emf: \( \frac12B\omega\left[(3\ell)^2-(2\ell)^2\right]=\frac12B\omega\cdot5\ell^2 \).
3. So \( \varepsilon=\frac{5B\omega\ell^2}{2} \).

Teacher's Note:
1. Subtract the squares of the two end distances.

Answer: (b) 1 s

Explanation:
1. The flux through the outer coil is the solenoid flux \( \propto I(t) \), so the induced current is proportional to \( -\frac{dI}{dt} \).
2. Compute \( \frac{dI}{dt}=k(1-t)e^{-t} \), which changes sign at t = 1 s.
3. So the current reverses direction at t = 1 s.

Teacher's Note:
1. The induced current follows the sign of dI/dt, which is zero where I is maximum.

Answer: (d) 175

Explanation:
1. The loop is made of the square ABCD in the xy-plane (area vector \( 25\hat k \)) and the square ADEF in the plane x = 0 (area vector \( 25\hat i \)).
2. The vector area of the whole loop is \( \vec A=25\hat i+25\hat k \).
3. Flux: \( \vec B\cdot\vec A=3\times25+4\times25=175 \) Wb.

Teacher's Note:
1. For a bent loop, add the vector areas of its planar parts.

Answer: (a) reverses direction at t = 0.5 s, where the induced emf is zero

Explanation:
1. Induced emf in the ring is proportional to \( \frac{dI}{dt}=I_0(1-2t) \).
2. This is zero at t = 0.5 s and changes sign there.
3. So the current reverses and the emf is zero at t = 0.5 s.

Teacher's Note:
1. The ring's current follows the sign of dI/dt.

Answer: (c) \( \frac{EL}{eR^2} \)

Explanation:
1. Current: \( i=\frac ER(1-e^{-t/\tau}) \) with \( \tau=\frac LR \).
2. Charge: \( q=\frac ER\int_0^\tau(1-e^{-t/\tau})dt=\frac ER\left[\tau-\tau(1-e^{-1})\right]=\frac ER\cdot\frac\tau e \).
3. With \( \tau=\frac LR \): \( q=\frac{EL}{eR^2} \).

Teacher's Note:
1. Integrate the growth current over one time constant.

Answer: (d) \( 9.1\times10^{-11} \) Wb

Explanation:
1. By reciprocity the mutual inductance is the same whichever loop carries the current. Put the 2.0 A in the big loop and find the flux through the small one.
2. Axial field of the big loop: \( B=\frac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}=\frac{4\pi\times10^{-7}\times2\times0.04}{2(0.0625)^{3/2}}=3.22\times10^{-6} \) T.
3. Flux through the small loop: \( B\pi r^2=3.22\times10^{-6}\times\pi(3\times10^{-3})^2=9.1\times10^{-11} \) Wb.

Teacher's Note:
1. The small loop is so small that the field over it is taken as uniform.

Answer: (a) \( \frac{\pi}{8}\times10^{-4} \) A

Explanation:
1. The frame's area in the field is half the circle: \( \frac12\pi(0.01)^2=\frac\pi2\times10^{-4} \) \( m^2 \).
2. Emf: \( \frac{dB}{dt}\times\text{area}=1\times\frac\pi2\times10^{-4} \) V.
3. Current: \( \frac{\pi\times10^{-4}/2}{4}=\frac\pi8\times10^{-4} \) A, about \( 3.9\times10^{-5} \) A.

Teacher's Note:
1. Only the part of the frame that overlaps the field region contributes to the flux.

Answer: (b) 2 W

Explanation:
1. Power from the cell: \( 5\times1=5 \) W. Heat in the resistor: \( i^2R=1\times3=3 \) W.
2. The rest goes to the magnetic field: \( 5-3=2 \) W.
3. (Voltage across the inductor is \( 5-3=2 \) V, so \( 2\times1=2 \) W.)

Teacher's Note:
1. Energy conservation: cell power = heat power + rate of increase of magnetic energy.

Answer: (c) 4 C

Explanation:
1. The magnetic force is \( B\ell i \), so the impulse is \( m\,dv=-B\ell\,i\,dt \).
2. Integrating: \( m(u-v_f)=B\ell q \).
3. So \( q=\frac{1\times(5-1)}{1\times1}=4 \) C.

Teacher's Note:
1. Momentum change equals \( B\ell \) times the charge; the details of L and R do not matter.

Answer: (d) \( \frac{2mg}{B\ell} \)

Explanation:
1. Since there is no resistance, \( B\ell v=L\frac{di}{dt} \), so \( i=\frac{B\ell x}{L} \) where x is the distance dropped.
2. Equation of motion: \( m\ddot x=mg-B\ell i=mg-\frac{B^2\ell^2}{L}x \), simple harmonic about \( x_0=\frac{mgL}{B^2\ell^2} \) with amplitude \( x_0 \).
3. The maximum extension is \( 2x_0 \), so \( i_{max}=\frac{B\ell\cdot2x_0}{L}=\frac{2mg}{B\ell} \).

Teacher's Note:
1. The rod performs SHM, so the extreme position is twice the equilibrium displacement.

Answer: (a) \( \frac{g\sqrt{mL}}{B\ell} \)

Explanation:
1. The motion is SHM with \( \omega^2=\frac{B^2\ell^2}{mL} \) and amplitude \( x_0=\frac{mgL}{B^2\ell^2} \).
2. Maximum speed: \( v_{max}=\omega x_0=\frac{B\ell}{\sqrt{mL}}\cdot\frac{mgL}{B^2\ell^2} \).
3. So \( v_{max}=\frac{g\sqrt{mL}}{B\ell} \).

Teacher's Note:
1. Maximum speed is at the equilibrium position of the SHM.

Answer: (b) 22 mA

Explanation:
1. The flux is \( \Phi=B(x,t)\,a^2 \) and the loop moves in x, so \( \frac{d\Phi}{dt}=a^2\left(\frac{\partial B}{\partial t}+v\frac{\partial B}{\partial x}\right) \).
2. Gradient: \( -10^{-3}\ \mathrm{T/cm}=-0.1 \) T/m, so \( v\frac{\partial B}{\partial x}=0.08\times(-0.1)=-0.008 \) T/s.
3. Emf: \( (0.12)^2(7-0.008)=0.1007 \) V; current \( \frac{0.1007}{4.5}=0.0224 \) A.

Teacher's Note:
1. Both the time variation and the motion through the gradient change the flux.