Practice 4: Electromagnetic Induction

10 MCQ + 5 FRQ · Covers Faraday's Law, Lenz's Law, Motional EMF, Inductance, and RL/LC Circuits

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Multiple Choice Questions

Q1. A circular loop of wire lies in the plane of the page. A magnetic field directed into the page is decreasing in magnitude. The induced current in the loop:

Correct! B into page is decreasing. By Lenz's Law, induced current creates a field to oppose the decrease — producing B into page. RHR: clockwise current produces B into page.
Incorrect. The correct answer is (B) Clockwise. Lenz's Law: oppose the decrease of B into page by creating B into page. RHR gives clockwise current.

Q2. A conducting rod of length \(L\) moves with velocity \(v\) perpendicular to a uniform magnetic field \(B\). The motional EMF induced across the rod is:

Correct! For a rod moving perpendicular to B: \(\mathcal{E} = \oint (\vec{v} \times \vec{B}) \cdot d\vec{\ell} = vBL\). Magnetic force on charges: \(F_B = qvB\), creating separation until \(qE = qvB\), so \(\mathcal{E} = EL = BLv\).
Incorrect. The correct answer is (D) \(BLv\). Motional EMF: \(\mathcal{E} = BLv\) when \(\vec{v} \perp \vec{B} \perp \vec{L}\).

Q3. The magnetic flux through a coil changes from 0.50 Wb to 0.10 Wb in 0.20 s. The coil has 100 turns. The average induced EMF is:

Correct! \(\mathcal{E} = -N\frac{\Delta\Phi}{\Delta t} = -100 \times \frac{0.10 - 0.50}{0.20} = -100 \times (-2.0) = 200\ \text{V}\).
Incorrect. The correct answer is (C) \(200\ \text{V}\). \(\mathcal{E} = N|\Delta\Phi/\Delta t| = 100 \times 0.40/0.20 = 200\ \text{V}\).

Q4. A bar magnet is pushed toward a conducting loop (N pole first). The induced current in the loop, as viewed from the magnet side, is:

Correct! N pole approaching increases flux to the right through the loop. By Lenz's Law, induced current creates flux opposing (to the left). From the magnet side, a N pole facing the magnet repels it — requiring counterclockwise current by RHR.
Incorrect. The correct answer is (A) Counterclockwise. Lenz's Law: oppose the approaching N pole by creating a N pole facing it → counterclockwise current.

Q5. In an RL circuit with battery \(\mathcal{E}\), resistor \(R\), and inductor \(L\), the time constant is:

Correct! RL time constant: \(\tau = L/R\). Current rises as \(I(t) = (\mathcal{E}/R)(1 - e^{-t/\tau})\). Analogous to \(\tau = RC\) in RC circuits.
Incorrect. The correct answer is (E) \(L/R\). RL time constant \(\tau = L/R\); current: \(I(t) = (\mathcal{E}/R)(1 - e^{-Rt/L})\).

Q6. A circular loop of radius \(r\) rotates with angular speed \(\omega\) in a uniform magnetic field \(B\). The axis of rotation is perpendicular to \(B\). The maximum induced EMF is:

Correct! Flux: \(\Phi = BA\cos(\omega t)\). \(\mathcal{E} = -d\Phi/dt = BA\omega\sin(\omega t)\). Maximum: \(\mathcal{E}_{\text{max}} = BA\omega = B(\pi r^2)\omega = \pi r^2 B\omega\).
Incorrect. The correct answer is (C) \(\pi r^2 B\omega\). \(\Phi = BA\cos(\omega t)\), \(\mathcal{E}_{\text{max}} = BA\omega = \pi r^2 B\omega\).

Q7. An LC circuit has \(L = 10\ \text{mH}\) and \(C = 100\ \mu\text{F}\). The angular frequency of oscillation is:

Correct! \(\omega = 1/\sqrt{LC} = 1/\sqrt{(10\times10^{-3})(100\times10^{-6})} = 1/\sqrt{10^{-6}} = 1000\ \text{rad/s}\).
Incorrect. The correct answer is (B) \(1000\ \text{rad/s}\). \(\omega = 1/\sqrt{LC} = 1/\sqrt{10^{-6}} = 1000\ \text{rad/s}\).

Q8. The energy stored in an inductor carrying current \(I\) is:

Correct! \(U_L = \frac{1}{2}LI^2\), analogous to \(U_C = \frac{1}{2}CV^2\). The energy is stored in the magnetic field of the inductor.
Incorrect. The correct answer is (D) \(\frac{1}{2}LI^2\). Energy stored in an inductor's magnetic field.

Q9. A conducting loop is pulled out of a uniform magnetic field region at constant speed. The induced current while the loop is partially in the field is:

Correct! Flux change rate \(d\Phi/dt = B \cdot dA/dt = B(Lv)\) is constant while the loop crosses the boundary at constant speed. \(\mathcal{E} = -d\Phi/dt\) is constant, so \(I\) is constant.
Incorrect. The correct answer is (A) Constant. \(d\Phi/dt = BLv\) is constant at constant speed, so \(\mathcal{E}\) and \(I\) are constant.

Q10. In an LC circuit with no resistance, the total energy:

Correct! \(U_{\text{total}} = \frac{1}{2}LI^2 + \frac{1}{2}CV^2 = \text{constant}\). Energy oscillates between magnetic (inductor) and electric (capacitor) forms, analogous to kinetic/potential energy in a spring-mass system.
Incorrect. The correct answer is (E). No resistance means no dissipation: total energy is constant, oscillating between \(U_L = \frac{1}{2}LI^2\) and \(U_C = \frac{1}{2}CV^2\).

Free Response Questions

FRQ

F1: Sliding Rod on Rails

A conducting rod of mass \(m = 0.050\ \text{kg}\) and resistance \(R = 2.0\ \Omega\) slides without friction on two parallel conducting rails separated by \(L = 0.20\ \text{m}\). A uniform magnetic field \(B = 0.80\ \text{T}\) is perpendicular to the plane. The rod is given an initial velocity \(v_0 = 10\ \text{m/s}\) to the right.

  1. Find the induced EMF as a function of velocity \(v\).
  2. Write the differential equation for \(v(t)\) and solve for \(v(t)\).
  3. How far does the rod travel before stopping?
  4. Show that the initial kinetic energy equals the total energy dissipated in the resistor.
Show Solution
(a): \(\mathcal{E} = BLv\). Induced current: \(I = \mathcal{E}/R = BLv/R\). Magnetic force opposes motion (Lenz): \(F = ILB = B^2 L^2 v/R\) (to the left).
(b): \(m\,dv/dt = -B^2 L^2 v/R\). This is \(dv/dt = -(B^2 L^2/mR)v\). Solution: \(v(t) = v_0 e^{-t/\tau}\) where \(\tau = mR/(B^2 L^2)\). With numbers: \(\tau = (0.050)(2.0)/[(0.80)^2(0.20)^2] = 0.10/0.0256 = 3.91\ \text{s}\).
(c): \(dx/dt = v = v_0 e^{-t/\tau}\). Integrate: \(x_{\text{total}} = v_0\tau = 10 \times 3.91 = 39.1\ \text{m}\).
(d): \(P = I^2 R = (BLv/R)^2 R = B^2 L^2 v^2/R\). Total dissipated: \(\int_0^\infty P\,dt = (B^2 L^2 /R) \int_0^\infty v_0^2 e^{-2t/\tau}\,dt = (B^2 L^2 v_0^2/R)(\tau/2) = \frac{1}{2}mv_0^2 = KE_{\text{initial}}\).
FRQ

F2: Faraday's Law — Solenoid & Coil

A long solenoid (radius \(r_1 = 2.0\ \text{cm}\), \(n = 1000\ \text{turns/m}\)) carries current \(I(t) = 5.0\sin(120\pi t)\ \text{A}\). A circular coil (radius \(r_2 = 5.0\ \text{cm}\), \(N = 20\ \text{turns}\)) is placed around the solenoid, coaxial with it.

  1. Find the magnetic field inside the solenoid as a function of time.
  2. Determine the magnetic flux through one turn of the outer coil.
  3. Find the induced EMF in the outer coil as a function of time.
  4. What is the maximum induced EMF?
Show Solution
(a): \(B(t) = \mu_0 n I(t) = (4\pi\times10^{-7})(1000)(5.0\sin(120\pi t)) = 2\pi\times10^{-3}\sin(120\pi t)\ \text{T}\).
(b): Field exists only inside solenoid (\(r_1 = 2\ \text{cm}\)). Flux through one turn of outer coil: \(\Phi_1 = B \cdot \pi r_1^2 = 2\pi\times10^{-3}\sin(120\pi t) \times \pi(0.02)^2 = 7.90\times10^{-6}\sin(120\pi t)\ \text{Wb}\).
(c): \(\mathcal{E} = -N\frac{d\Phi_1}{dt} = -20 \times 7.90\times10^{-6} \times 120\pi\cos(120\pi t) = -0.0596\cos(120\pi t)\ \text{V}\).
(d): \(\mathcal{E}_{\text{max}} = 0.0596\ \text{V} \approx 59.6\ \text{mV}\).
FRQ

F3: RL Circuit Analysis

A battery \(\mathcal{E} = 24\ \text{V}\), resistor \(R = 6.0\ \Omega\), and inductor \(L = 0.30\ \text{H}\) are connected in series. The switch is closed at \(t = 0\).

  1. Find the time constant of the circuit.
  2. Derive \(I(t)\) and find the current at \(t = 0.10\ \text{s}\).
  3. Find the voltage across the inductor as a function of time.
  4. Calculate the energy stored in the inductor at \(t \to \infty\).
  5. Sketch \(I(t)\) and \(V_L(t)\) on labeled axes.
Show Solution
(a): \(\tau = L/R = 0.30/6.0 = 0.050\ \text{s}\).
(b): Loop: \(\mathcal{E} - IR - L\,dI/dt = 0\). Solution: \(I(t) = (\mathcal{E}/R)(1 - e^{-t/\tau}) = 4.0(1 - e^{-t/0.050})\ \text{A}\). At \(t = 0.10\ \text{s}\): \(I = 4.0(1 - e^{-2}) = 4.0(0.865) = 3.46\ \text{A}\).
(c): \(V_L = L\,dI/dt = L \cdot (\mathcal{E}/L)e^{-t/\tau} = \mathcal{E}e^{-t/\tau} = 24e^{-t/0.050}\ \text{V}\).
(d): At \(t \to \infty\): \(I = \mathcal{E}/R = 4.0\ \text{A}\). \(U_L = \frac{1}{2}LI^2 = 0.5 \times 0.30 \times 16.0 = 2.4\ \text{J}\).
(e): \(I(t)\) rises asymptotically from 0 to 4.0 A. \(V_L(t)\) decays exponentially from 24 V to 0. At \(\tau = 0.05\ \text{s}\): \(I = 0.632 \times 4.0 = 2.53\ \text{A}\), \(V_L = 0.368 \times 24 = 8.83\ \text{V}\).
FRQ

F4: LC Oscillation

An LC circuit has \(C = 20\ \mu\text{F}\) and \(L = 5.0\ \text{mH}\). The capacitor is initially charged to \(V_0 = 50\ \text{V}\) and connected to the inductor at \(t = 0\).

  1. Find the angular frequency and period of oscillation.
  2. Write expressions for \(q(t)\) and \(I(t)\).
  3. Find the maximum current in the circuit.
  4. At what times is the energy entirely in the inductor?
Show Solution
(a): \(\omega = 1/\sqrt{LC} = 1/\sqrt{(5.0\times10^{-3})(20\times10^{-6})} = 1/\sqrt{10^{-7}} = 3162\ \text{rad/s}\). \(T = 2\pi/\omega = 2\pi/3162 = 1.99\times10^{-3}\ \text{s} \approx 2.0\ \text{ms}\).
(b): \(Q_0 = CV_0 = (20\times10^{-6})(50) = 1.0\times10^{-3}\ \text{C}\). \(q(t) = Q_0\cos(\omega t) = 1.0\times10^{-3}\cos(3162t)\ \text{C}\). \(I(t) = -dq/dt = \omega Q_0\sin(\omega t) = 3.16\sin(3162t)\ \text{A}\).
(c): \(I_{\text{max}} = \omega Q_0 = 3162 \times 1.0\times10^{-3} = 3.16\ \text{A}\). (Check: \(\frac{1}{2}LI_{\text{max}}^2 = \frac{1}{2}CV_0^2\) → works.)
(d): All energy in inductor when \(q = 0\). \(\cos(\omega t) = 0\) → \(\omega t = \pi/2, 3\pi/2, 5\pi/2, \ldots\) → \(t = T/4, 3T/4, 5T/4, \ldots = 0.50\ \text{ms}, 1.50\ \text{ms}, 2.50\ \text{ms}, \ldots\)
FRQ

F5: Eddy Current Braking

A rectangular metal plate of width \(w\) and mass \(m\) falls vertically through a horizontal magnetic field \(B\) that extends over region of height \(h\). The plate has resistance \(R\).

  1. Explain physically why the plate experiences a magnetic braking force.
  2. When the plate is partly in the field, derive the induced EMF in terms of its speed \(v\).
  3. Find the terminal velocity of the plate (assume \(h\) is large).
  4. Describe the energy conversion that occurs.
Show Solution
(a): As the plate enters the B field, the changing flux induces eddy currents in the plate. By Lenz's Law, these currents produce a magnetic field opposing the motion — creating an upward magnetic force (braking).
(b): Consider the leading edge entering: the bottom portion in the field acts like a sliding bar. \(\mathcal{E} = Bwv\). Eddy current: \(I \sim \mathcal{E}/R = Bwv/R\).
(c): Magnetic braking force: \(F_B = IwB = B^2 w^2 v/R\) (upward). At terminal velocity: \(F_B = mg\). \(v_T = mgR/(B^2 w^2)\).
(d): Gravitational potential energy → kinetic energy → electrical energy (via induction) → thermal energy (Joule heating in the plate's resistance). Energy is conserved: \(mgv_T = I^2 R\) at terminal velocity.

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