JEE Physics Electromagnetic Induction MCQs Set 15

Answer: (a) \( 1.05\times10^{-2} \) Wb/m²

Explanation:
1. \( B=\mu_0nI \) so \( \dfrac{B_2}{B_1}=\dfrac{n_2I_2}{n_1I_1}=\dfrac{100}{200}\times\dfrac13=\dfrac16 \).
2. \( B_2=\dfrac{6.28\times10^{-2}}{6}=1.05\times10^{-2} \) Wb/m².

Teacher's Note:
1. Field inside a solenoid is proportional to turns per unit length times current.
2. Use the ratio instead of calculating \( \mu_0 \).

Answer: (c) -10 V

Explanation:
1. \( \varepsilon=-\dfrac{d\phi}{dt}=-(20t-50) \).
2. At t = 3 s: \( \varepsilon=-(60-50)=-10 \) V.

Teacher's Note:
1. Differentiate the flux with respect to time.
2. The negative sign comes from Lenz's law.

Answer: (b) \( 2.4\pi\times10^{-4} \) H

Explanation:
1. \( M=\dfrac{\mu_0N_1N_2A}{\ell}=\dfrac{4\pi\times10^{-7}\times300\times400\times10\times10^{-4}}{0.2} \).
2. Evaluating, \( M=2.4\pi\times10^{-4} \) H.

Teacher's Note:
1. Mutual inductance of coaxial solenoids is \( \mu_0N_1N_2A/\ell \).
2. Convert area and length to SI units.

Answer: (a) 3 mV

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

Teacher's Note:
1. Motional emf is \( B\ell v \) when all three are mutually perpendicular.
2. Wire, velocity and horizontal field are perpendicular here.

Answer: (b) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

Explanation:
1. Infrared waves give heat treatment for muscular strain, radio waves are used for broadcasting, X-rays show bone fractures, and ultraviolet radiation is absorbed by the ozone layer.
2. So (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv).

Teacher's Note:
1. Link each region of the spectrum with its best-known use.
2. X-rays penetrate soft tissue but are absorbed by bone.

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

Explanation:
1. Mutual inductance of a small loop (radius a) on the axis of a large loop (radius b) at distance x: \( M=\dfrac{\mu_0\pi a^2b^2}{2(b^2+x^2)^{3/2}} \).
2. Here \( b^2+x^2=0.0625 \) and so \( M=\dfrac{4\pi\times10^{-7}\times\pi\times(3\times10^{-3})^2\times0.04}{2\times0.015625}=4.5\times10^{-11} \) H.
3. Flux \( \phi=MI=4.5\times10^{-11}\times2\approx9.1\times10^{-11} \) Wb.

Teacher's Note:
1. Treat the small loop as a dipole in the large loop's axial field, or use reciprocity.
2. The flux equals mutual inductance times current.

Answer: (b) electric energy density is equal to the magnetic energy density

Explanation:
1. The electric energy density is \( \tfrac12\varepsilon E^2 \) and the magnetic energy density is \( \dfrac{B^2}{2\mu} \).
2. Since \( E=vB \) and \( v=1/\sqrt{\mu\varepsilon} \), the two are equal.

Teacher's Note:
1. The energy of an electromagnetic wave is shared equally between the electric and magnetic fields.
2. Use \( E=vB \).

Answer: (b) (4), (2), (1), (3)

Explanation:
1. Energy per quantum is \( E=h\nu \), increasing with frequency.
2. Radio waves have the lowest frequency, then yellow light, then blue light, and X-rays have the highest.

Teacher's Note:
1. Higher frequency means higher energy per photon.
2. Order of the spectrum: radio, visible (red to violet), X-ray.

Answer: (a) 250 Wb

Explanation:
1. Induced current \( i=\dfrac{\varepsilon}{R}=-\dfrac{1}{R}\dfrac{d\phi}{dt} \), so \( |\Delta\phi|=R\int i\,dt \).
2. Area under the current-time graph \( =\tfrac12\times10\times0.5=2.5 \) A s.
3. So \( |\Delta\phi|=100\times2.5=250 \) Wb.

Teacher's Note:
1. The area under an i-t graph is the charge that flowed.
2. Flux change equals resistance times charge flowed.

Answer: (d) zero

Explanation:
1. The field is constant in time and parallel to the rod; no flux change takes place through any loop in the rod, and the charges moving (if any) have velocity along B, so no magnetic force.
2. Hence no emf and no current is induced.

Teacher's Note:
1. Induction needs a changing flux or a velocity component perpendicular to B.
2. A static field parallel to the rod does nothing.

Answer: (c) (1) and (3) only

Explanation:
1. Electrostatic and gravitational fields of static sources are conservative: the line integral around a closed loop is zero, so their field lines cannot be closed circles.
2. Magnetic field lines and induced electric field lines (from changing flux) can form closed loops.

Teacher's Note:
1. Closed field lines imply a non-zero circulation, i.e. a non-conservative field.
2. Induced electric fields are non-conservative.

Answer: (d) No friction, hence less power consumption

Explanation:
1. The train floats above the track, so there is no contact friction between the train and the track.
2. This reduces power consumption and allows very high speeds.

Teacher's Note:
1. Levitation removes rolling friction.
2. Electric power is still needed for propulsion.

Answer: (a) By Lenz's law the train experiences a drag

Explanation:
1. The induced currents oppose the cause (the motion), so they produce a backward force on the moving train.
2. Hence the main disadvantage is the electromagnetic drag by Lenz's law.

Teacher's Note:
1. Lenz's law means opposition to the change that produces the current.
2. Drag here is electromagnetic and not due to friction.

Answer: (b) Magnetic force

Explanation:
1. The induced currents in the track coils interact with the magnetic field of the train, producing an upward force on the current-carrying conductors.
2. This is a magnetic force.

Teacher's Note:
1. Forces on currents in magnetic fields are magnetic forces.
2. Induced electric fields only drive the currents.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. The alternating current produces a changing flux through the ring, inducing a current in the ring (Assertion is true: the ring floats).
2. The field of the coil has a horizontal component near the ring edges. The induced current interacts with this component and gives an average upward force, which is exactly why the ring floats. So the Reason is true and explains the Assertion.

Teacher's Note:
1. This is Elihu Thomson's jumping ring.
2. Induced currents repel the source of the changing flux (Lenz's law).

Answer: (a) 6

Explanation:
1. Field inside: \( B=\mu_0\dfrac{I}{L}=\dfrac{\mu_0I_0}{10}\cos300t \).
2. Emf \( =-\pi r^2\dfrac{dB}{dt}=\pi(0.1)^2\cdot\dfrac{\mu_0I_0}{10}\cdot300\sin300t=0.3\pi\mu_0I_0\sin300t \).
3. Current in the loop \( i=\dfrac{0.3\pi\mu_0I_0}{0.005}\sin300t=60\pi\mu_0I_0\sin300t \) and moment \( m=i\pi r^2=60\pi\mu_0I_0\times0.01\pi\sin300t=0.6\pi^2\mu_0I_0\sin300t\approx5.9\mu_0I_0\sin300t \), so N is about 6.

Teacher's Note:
1. Treat the tube as a solenoid with one turn per tube length.
2. Moment of a loop is current times area.

Answer: (d) (1) and (3) only

Explanation:
1. The wire lies along a diameter. The field above the wire is directed into the plane of the loop on one side and out of it on the other side, and the two halves of the loop are symmetric.
2. The total flux through the loop is zero at all times, whatever the current, so the induced emf is always zero.
3. Hence (1) and (3) are correct and (2), (4) are wrong.

Teacher's Note:
1. Symmetry makes the net flux zero.
2. If flux is always zero, its rate of change is also zero.

Answer: (c) 7

Explanation:
1. Field on the axis of the big loop at \( z=\sqrt3R \) for current I: \( B=\dfrac{\mu_0IR^2}{2(R^2+3R^2)^{3/2}}=\dfrac{\mu_0I}{16R} \).
2. Flux through the two-turn square (normal at 45 degrees to the axis): \( \phi=2a^2B\cos45^\circ=\dfrac{2a^2\mu_0I}{16R\sqrt2}=\dfrac{\mu_0a^2I}{8\sqrt2R} \).
3. So \( M=\dfrac{\mu_0a^2}{2^3\cdot2^{1/2}R}=\dfrac{\mu_0a^2}{2^{7/2}R} \) and \( p/2=7/2 \), giving p = 7.

Teacher's Note:
1. Use the axial field of a circular loop and the small area of the square.
2. Flux involves the cosine of the angle between B and the normal.

Answer: (c) \( \dfrac{BR}{2} \)

Explanation:
1. Induced emf \( E(2\pi R)=\pi R^2\dfrac{dB}{dt} \) with \( \dfrac{dB}{dt}=B \) per second.
2. Hence \( E=\dfrac{BR}{2} \).

Teacher's Note:
1. Faraday's law for the circular path gives the induced electric field.
2. The rate of change of field is B per second.

Answer: (b) \( -\gamma\dfrac{BQR^2}{2} \)

Explanation:
1. Induced field \( E=\dfrac{BR}{2} \) is clockwise (to oppose the increasing flux along +z), so the torque on the charge about the centre is \( \tau=QER=\dfrac{QBR^2}{2} \) and it is directed along -z.
2. Change in angular momentum over 1 s: \( \Delta L=-\dfrac{QBR^2}{2} \).
3. So \( \Delta\mu=\gamma\Delta L=-\gamma\dfrac{BQR^2}{2} \).

Teacher's Note:
1. The induced electric field exerts a tangential force on the orbiting charge.
2. Torque is the rate of change of angular momentum.

Answer: (d) (2) and (3) only

Explanation:
1. The two lobes are traversed in opposite senses, so the net flux is \( (2A-A)B\cos\omega t=AB\cos\omega t \) and the net emf is \( AB\omega\sin\omega t \), proportional to \( \sin\omega t \) and to the difference of the areas. So (1) and (4) are false.
2. The rate of change of flux (the emf) is maximum when the loop plane is parallel to B, i.e. perpendicular to the paper: (2) is true.
3. The net amplitude \( AB\omega \) equals the amplitude of the smaller loop alone: (3) is true.

Teacher's Note:
1. Twisting the wire reverses the sense of circulation in one lobe.
2. Emf is largest when the flux is changing fastest, i.e. when the flux is zero.

Answer: (a) (1) and (2) only

Explanation:
1. Radius \( r=\dfrac{p}{QB} \). The particle re-enters region 1 if \( r<\dfrac{3R}{2} \), i.e. \( B>\dfrac{2p}{3QR} \): (1) true.
2. For \( B=\dfrac{8p}{13QR} \), \( r=\dfrac{13R}{8}>\dfrac{3R}{2} \). The circle has centre at \( (0,\,5R/8) \) and meets \( x=3R/2 \) at \( y=0 \) (on the x-axis): (2) true.
3. The distance between P1 and the re-entry point is \( 2r\propto m \), directly (not inversely) proportional to the mass: (3) false. At the farthest point the momentum turns from \( p\hat i \) to \( p\hat j \), a change of magnitude \( \sqrt2p \), not \( p/\sqrt2 \): (4) false.

Teacher's Note:
1. Compare the radius of the path with the width of the region.
2. Draw the circle: its centre is above the entry point.