JEE Physics Electromagnetic Induction Magnetism and Capacitors MCQs Set 01

Answer: (d) \(3 \times 10^{-3}\) Wb

Explanation:
1. The field exists only inside the solenoid, so only the cross-section of the solenoid contributes to the flux.
2. \(\Phi = B \times A_{\text{solenoid}} = 2 \times 15 \times 10^{-4} = 3 \times 10^{-3}\) Wb.
3. The remaining area of the coil has no field.

Teacher's Note:
1. Flux depends on the area that actually contains the field, not on the whole area of the coil.
2. The field outside a long solenoid is negligible.

Answer: (d) 2.0 V

Explanation:
1. \(e = -\dfrac{d\phi}{dt} = -(10t - 4) = 4 - 10t\).
2. At \(t = 0.2\) s: \(e = 4 - 2 = 2.0\) V.

Teacher's Note:
1. The resistance is not needed for the emf; it would be used for the current.
2. The current here would be 0.2 A.

Answer: (b) Anticlockwise in A and clockwise in B

Explanation:
1. With current upwards, the field on the left of the wire points out of the page and on the right points into the page.
2. As the current decreases, the outward flux through A decreases, so by Lenz's law A develops an outward field, which needs an anticlockwise current.
3. The inward flux through B decreases, so B develops an inward field, which needs a clockwise current.

Teacher's Note:
1. Lenz's law: the induced current opposes the change of flux.
2. The field of a straight wire has opposite directions on its two sides.

Answer: (a) Along z

Explanation:
1. The motional emf is \(\vec{v} \times \vec{B} \cdot \vec{\ell}\).
2. Here \(\vec{\ell}\) is along y and \(\vec{B}\) is along x, so \(\vec{v} \times \vec{B}\) must have a component along y.
3. Only a velocity along z gives \(\hat{z} \times \hat{x} = \hat{y}\).
4. Motion along x (parallel to B) or along y (along the rod) gives no emf.

Teacher's Note:
1. Emf appears only when the conductor cuts the field lines.
2. Velocity, field and length must be mutually perpendicular for the maximum emf.

Answer: (a) \(\dfrac{5}{7}\) A

Explanation:
1. Just after the key is closed, the inductor opposes any change of current, so it carries zero current and acts like an open circuit.
2. The right branch then carries no current (\(I_3 = 0\)).
3. The circuit is the 6 ohm resistor in series with the middle 8 ohm resistor.
4. \(I_1 = \dfrac{10}{6 + 8} = \dfrac{5}{7}\) A.

Teacher's Note:
1. At t = 0 an inductor acts as an open circuit, and long afterwards it acts as a plain wire.
2. Replace it accordingly before applying Ohm's law.

Answer: (c) \(\pi\ \mu\text{V}\)

Explanation:
1. \(\phi = B\pi r^2\), so \(\left|\dfrac{d\phi}{dt}\right| = 2\pi r B\left|\dfrac{dr}{dt}\right|\).
2. \(= 2\pi \times 0.02 \times 0.025 \times 10^{-3} = \pi \times 10^{-6}\) V.
3. The emf is \(\pi\ \mu\text{V}\).

Teacher's Note:
1. Differentiate the area with respect to time using the chain rule.
2. Keep all units in SI (metre, tesla) before multiplying.

Answer: (a) \(8 \times 10^{-4}\) V

Explanation:
1. Emf induced in Q is \(M\dfrac{dI_P}{dt}\): \(2 \times 10^{-3} = M \times 5\), so \(M = 4 \times 10^{-4}\) H.
2. Mutual inductance is the same in both directions.
3. Emf in P \(= M\dfrac{dI_Q}{dt} = 4 \times 10^{-4} \times 2 = 8 \times 10^{-4}\) V.

Teacher's Note:
1. Mutual inductance is symmetric: \(M_{PQ} = M_{QP}\).
2. Find M from the first case and reuse it in the second.

Answer: (a) Clockwise in A and anticlockwise in B

Explanation:
1. For an upward current the field is out of the page on the left of the wire and into the page on the right.
2. As the current increases, the outward flux through A increases, so A produces an inward field, which means a clockwise current.
3. The inward flux through B increases, so B produces an outward field, which means an anticlockwise current.

Teacher's Note:
1. Use the right-hand grip rule for the wire, then Lenz's law for each loop.
2. Increasing flux is opposed, decreasing flux is supported by the induced current.

Answer: (b) The north pole faces the ring and the magnet moves towards it

Explanation:
1. An anticlockwise current seen from the magnet side makes the near face of the ring a north pole.
2. By Lenz's law the ring opposes the approach of the magnet by repelling it, so a like pole must face the magnet.
3. That happens when the north pole of the magnet is facing the ring and the magnet moves towards it.

Teacher's Note:
1. Anticlockwise looks like N, clockwise looks like S, when viewed from that side.
2. The induced pole always opposes the motion of the magnet.

Answer: (c) transverse

Explanation:
1. In an electromagnetic wave the electric and magnetic fields oscillate perpendicular to each other and to the direction of travel.
2. Therefore the wave is transverse and it travels, so it is not stationary.

Teacher's Note:
1. Transverse means the oscillation is perpendicular to propagation.
2. Light can be polarised only because it is transverse.

Answer: (d) \(\vec{E} = E_0\cos\left(\omega t - \dfrac{2\pi}{\lambda}y\right)\hat{z}\)

Explanation:
1. A wave travelling in the +y direction has the phase \(\omega t - ky\). This rules out options (b) and (c).
2. The wave direction is along \(\vec{E} \times \vec{B}\). With \(\vec{B}\) along \(\hat{x}\), \(\vec{E}\) must be along \(\hat{z}\) because \(\hat{z} \times \hat{x} = \hat{y}\).
3. So \(\vec{E} = E_0\cos(\omega t - ky)\,\hat{z}\).

Teacher's Note:
1. E, B and the direction of propagation form a right-handed set.
2. The electric field cannot be along the magnetic field.

Answer: (a) \(2.83 \times 10^{-8}\) T

Explanation:
1. \(E_0 = \sqrt{2}\,E_{rms} = 6\sqrt{2} = 8.49\) V/m.
2. \(B_0 = \dfrac{E_0}{c} = \dfrac{8.49}{3 \times 10^{8}} = 2.83 \times 10^{-8}\) T.

Teacher's Note:
1. Convert rms to peak by multiplying by \(\sqrt{2}\).
2. Only then divide by c to get the peak magnetic field.

Answer: (c) \(\sqrt{\dfrac{P}{2\pi c\varepsilon_0 r^2}}\)

Explanation:
1. Intensity at distance r is \(I = \dfrac{P}{4\pi r^2}\).
2. Also \(I = \dfrac{1}{2}\varepsilon_0 c E_0^2\).
3. Equating: \(E_0^2 = \dfrac{P}{2\pi\varepsilon_0 c\, r^2}\), so \(E_0 = \sqrt{\dfrac{P}{2\pi c\varepsilon_0 r^2}}\).

Teacher's Note:
1. Intensity falls off as the inverse square of distance.
2. The peak field therefore falls as 1/r.

Answer: (d) not a conduction current but a current caused by a time varying electric field

Explanation:
1. Displacement current is defined as \(I_d = \varepsilon_0\dfrac{d\Phi_E}{dt}\).
2. It arises from a changing electric field, for example between the plates of a charging capacitor, and involves no flow of charge.

Teacher's Note:
1. Maxwell added displacement current to Ampere's law to make it consistent for charging capacitors.
2. It produces a magnetic field just like a conduction current.

Answer: (d) Both ferromagnetic and paramagnetic

Explanation:
1. Diamagnetic atoms have zero net magnetic moment because their electron moments cancel.
2. Paramagnetic and ferromagnetic atoms have unpaired electrons, so each atom has a permanent magnetic moment.
3. Therefore both (b) and (c) are correct.

Teacher's Note:
1. Diamagnetic materials have only induced moments.
2. Ferromagnets also show domains.

Answer: (a) 6 V, plane of the coil normal to the magnetic meridian

Explanation:
1. The coil's field at its centre is \(B = \dfrac{\mu_0 N I}{2r}\) with \(r = 0.05\) m.
2. \(12\pi \times 10^{-6} = \dfrac{4\pi \times 10^{-7} \times 5 \times I}{0.1}\), so \(I = 0.6\) A.
3. \(V = IR = 0.6 \times 10 = 6\) V.
4. The coil's field must point along the magnetic meridian to cancel the horizontal earth field, so its axis lies in the meridian and its plane is normal to the meridian.

Teacher's Note:
1. The axis of a current loop is perpendicular to its plane.
2. The current direction must be chosen so that the loop field is opposite to the earth's field.

Answer: (a) Maximum

Explanation:
1. Torque is \(\tau = NIAB\sin\theta\), where \(\theta\) is the angle between the field and the normal to the plane.
2. When the field is parallel to the plane, \(\theta = 90^\circ\) and \(\sin\theta = 1\), so the torque is maximum.

Teacher's Note:
1. Torque is zero when the field is along the normal to the plane.
2. It is maximum when the field lies in the plane.

Answer: (c) an electron

Explanation:
1. The frequency of revolution is \(f = \dfrac{qB}{2\pi m}\), so it depends on the ratio \(q/m\).
2. The electron has the largest \(q/m\) (about \(1.76\times10^{11}\) C/kg, against \(9.6\times10^{7}\) for a proton and half of that for an alpha particle).
3. A neutron has no charge and does not rotate.

Teacher's Note:
1. The cyclotron frequency does not depend on speed or radius.
2. Lighter particles with the same charge spin faster.

Answer: (c) 1

Explanation:
1. Inside the wire, by Ampere's law, \(B = \dfrac{\mu_0 i r}{2\pi a^2}\). At \(r = a/2\): \(B_1 = \dfrac{\mu_0 i}{4\pi a}\).
2. Outside the wire, \(B = \dfrac{\mu_0 i}{2\pi r}\). At \(r = 2a\): \(B_2 = \dfrac{\mu_0 i}{4\pi a}\).
3. The ratio \(B_1 : B_2 = 1\).

Teacher's Note:
1. Inside a uniformly carrying wire B rises linearly with r; outside it falls as 1/r.
2. The maximum is at the surface of the wire.

Answer: (a) increases

Explanation:
1. With the battery disconnected the charge \(Q\) stays constant.
2. Increasing the separation \(d\) decreases \(C = \dfrac{\varepsilon_0 A}{d}\).
3. Energy \(U = \dfrac{Q^2}{2C}\) therefore increases.
4. The extra energy comes from the work done in pulling the plates apart.

Teacher's Note:
1. Isolated capacitor means constant charge; connected to a battery means constant potential.
2. Choose the right formula for each case.

Answer: (c) \(\dfrac{3}{2}CV^2\)

Explanation:
1. In parallel the capacitances add: \(C_{eq} = C + \dfrac{C}{2} = \dfrac{3C}{2}\).
2. Charge supplied by the battery \(Q = C_{eq}V = \dfrac{3}{2}CV\).
3. Work done by the battery \(= QV = \dfrac{3}{2}CV^2\). (Half of this is stored as energy, the rest is lost as heat.)

Teacher's Note:
1. Work done by a battery is QV, while the energy stored is only half of it.
2. Do not confuse the two.

Answer: (d) \(\dfrac{2V}{3}\)

Explanation:
1. Charges are \(CV\) and \(5CV\). With unlike polarities joined, the net charge is \(5CV - CV = 4CV\).
2. The combined capacitance is \(C + 5C = 6C\).
3. Common potential \(= \dfrac{4CV}{6C} = \dfrac{2V}{3}\).

Teacher's Note:
1. Charge is conserved, so add charges with signs.
2. Opposite polarities partly cancel each other.

Answer: (c) \(1.33\ \mu\text{C}\)

Explanation:
1. In the steady state no current flows in the capacitor branch, so \(R_1\) carries no current and has no voltage drop.
2. The current flows through \(R_2\) and the internal resistance: \(I = \dfrac{2}{1 + 0.5} = \dfrac{4}{3}\) A.
3. The capacitor voltage equals the voltage across \(R_2\): \(V = IR_2 = \dfrac{4}{3}\) V.
4. \(Q = CV = 1\ \mu\text{F} \times \dfrac{4}{3}\ \text{V} = 1.33\ \mu\text{C}\).

Teacher's Note:
1. A capacitor is an open circuit in steady state.
2. The resistor in series with it carries no current and drops no voltage.

Answer: (a) 1.76 mm

Explanation:
1. Time in the field: \(t = \dfrac{0.10}{4 \times 10^{7}} = 2.5 \times 10^{-9}\) s.
2. Acceleration: \(a = \dfrac{eE}{m} = \dfrac{1.6\times10^{-19} \times 3200}{9.1\times10^{-31}} = 5.63 \times 10^{14}\) m/s\(^2\).
3. Deviation \(y = \tfrac{1}{2}at^2 = \tfrac{1}{2} \times 5.63\times10^{14} \times (2.5\times10^{-9})^2 = 1.76 \times 10^{-3}\) m = 1.76 mm.

Teacher's Note:
1. Motion is like a horizontal projectile: uniform along the plates, accelerated across them.
2. Time is fixed by the horizontal distance.