FAD1022 CQ5: Rapid-Fire Drill Pack — Electromagnetism

Objective: Master all Electromagnetism concepts for CQ5.
Target: 3-4 min per problem (CQ5 style: 4 questions in 15 min).
Total problems: 28
Estimated time: ~100 min

Cheat Sheet (Memorize First)

Constants for CQ5

Constant Symbol Value
Permeability of free space $\mu_0$ $4\pi \times 10^{-7}\ \text{T·m·A}^{-1}$

Electromagnetic Induction

Law Formula Description
Faraday's Law $\mathcal{E} = -N\frac{d\Phi_B}{dt}$ Induced EMF
Lenz's Law Direction opposes change Determines current direction
Flux $\Phi_B = BA\cos\theta$ Magnetic flux
Motional EMF $\mathcal{E} = BLv$ Conductor moving in B-field

Self & Mutual Inductance

Concept Formula
Self-inductance $L = \frac{N\Phi_B}{I}$
EMF across inductor $\mathcal{E}_L = -L\frac{dI}{dt}$
Energy in inductor $U = \frac{1}{2}LI^2$
Mutual inductance $M = \frac{N_2\Phi_{21}}{I_1}$
Induced EMF (mutual) $\mathcal{E}_2 = -M\frac{dI_1}{dt}$

Transformers (Ideal)

Quantity Formula
Voltage ratio $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
Current ratio $\frac{I_s}{I_p} = \frac{N_p}{N_s}$
Power (ideal) $P_p = P_s$
Impedance transformation $Z_{\text{in}} = (\frac{N_p}{N_s})^2 Z_L$

RL Circuits

Quantity Formula
Time constant $\tau = L/R$
Current growth $I(t) = \frac{V}{R}(1 - e^{-t/\tau})$
Current decay $I(t) = I_0 e^{-t/\tau}$
Inductor voltage $V_L = V e^{-t/\tau}$ (growth)

Part A: Faraday's Law & Magnetic Flux (6 problems)

Target: 3 min per problem.

Set A1 — Flux and Induced EMF (3 problems)

  1. A coil with $N = 100$ turns and area $A = 25\ \text{cm}^2$ is placed perpendicular to a uniform magnetic field. The field increases from $B = 0$ to $B = 0.40\ \text{T}$ in $\Delta t = 2.0\ \text{s}$. Calculate the average induced EMF.

  2. A square loop of side $a = 15\ \text{cm}$ is in a magnetic field $B = 0.30\ \text{T}$ directed perpendicular to the plane of the loop. The loop is rotated by $90°$ in $\Delta t = 0.50\ \text{s}$ so the plane becomes parallel to the field. Calculate the average induced EMF.

  3. A coil with $N = 50$ turns and radius $r = 4.0\ \text{cm}$ is placed in a uniform magnetic field that changes at a rate $\frac{dB}{dt} = 0.20\ \text{T/s}$. The coil's axis is parallel to the field. Calculate the induced EMF.

Score: ___/3

Set A2 — Lenz's Law Applications (3 problems)

  1. A bar magnet with its north pole facing a conducting loop is moved toward the loop. Determine the direction of the induced current in the loop (clockwise or counterclockwise when viewed from the magnet).

  2. A conducting loop is pulled out of a region of uniform magnetic field directed into the page. Determine the direction of the induced current.

  3. A coil is placed near a solenoid carrying increasing current (clockwise when viewed from the coil). Determine the direction of the induced current in the coil.

Score: ___/3

Part B: Motional EMF (6 problems)

Target: 3 min per problem.

Set B1 — Basic Motional EMF (3 problems)

  1. A conductor of length $L = 25\ \text{cm}$ moves perpendicular to a uniform magnetic field $B = 0.50\ \text{T}$ with velocity $v = 4.0\ \text{m/s}$. Calculate the motional EMF.

  2. An airplane with wingspan $L = 40\ \text{m}$ flies horizontally at $v = 250\ \text{m/s}$ where the vertical component of Earth's magnetic field is $B = 5.0 \times 10^{-5}\ \text{T}$. Calculate the induced EMF between the wingtips.

  3. A metal rod of length $L = 30\ \text{cm}$ slides on conducting rails at $v = 2.0\ \text{m/s}$ in a uniform magnetic field $B = 0.40\ \text{T}$ perpendicular to the rails. The circuit has resistance $R = 5.0\ \Omega$. Calculate: (a) The induced EMF (b) The induced current (c) The force required to maintain constant velocity

Score: ___/3

Set B2 — Sliding Rod Problems (3 problems)

  1. A conducting rod of length $L = 50\ \text{cm}$ slides without friction on conducting rails connected to a resistor $R = 10\ \Omega$. A uniform magnetic field $B = 0.60\ \text{T}$ is perpendicular to the plane. The rod is pulled with constant force $F = 0.15\ \text{N}$. Find: (a) The terminal velocity of the rod (b) The power dissipated in the resistor

  2. A rod of mass $m = 0.10\ \text{kg}$ and length $L = 40\ \text{cm}$ slides on vertical conducting rails in a horizontal magnetic field $B = 0.50\ \text{T}$. The rails are connected by resistance $R = 2.0\ \Omega$. Find the terminal velocity of the rod as it falls under gravity.

  3. A square conducting loop of side $a = 20\ \text{cm}$ and resistance $R = 5.0\ \Omega$ is pulled out of a region of uniform field $B = 0.30\ \text{T}$ at constant velocity $v = 0.50\ \text{m/s}$. Calculate: (a) The induced current while leaving the field (b) The power dissipated (c) The force required to pull the loop

Score: ___/3

Part C: Self-Inductance & RL Circuits (6 problems)

Target: 3-4 min per problem.

Set C1 — Self-Inductance (3 problems)

  1. A solenoid with $N = 400$ turns, length $l = 20\ \text{cm}$, and cross-sectional area $A = 4.0\ \text{cm}^2$ carries current $I = 3.0\ \text{A}$. Calculate: (a) The self-inductance (b) The magnetic energy stored

  2. The current in a $50\ \text{mH}$ inductor changes at $\frac{dI}{dt} = 0.20\ \text{A/s}$. Calculate the induced EMF.

  3. An inductor stores $U = 0.50\ \text{J}$ when carrying $I = 4.0\ \text{A}$. Calculate: (a) The inductance (b) The energy stored when $I = 2.0\ \text{A}$

Score: ___/3

Set C2 — RL Circuits (3 problems)

  1. An RL circuit has $R = 20\ \Omega$, $L = 0.40\ \text{H}$, and is connected to $V = 12\ \text{V}$. Calculate: (a) The time constant (b) The current after $t = 2\tau$ (c) The final steady-state current

  2. In the circuit of problem 16, how long does it take for the current to reach $90%$ of its final value?

  3. An RL circuit with $R = 50\ \Omega$ and $L = 0.50\ \text{H}$ carries steady current $I_0 = 2.0\ \text{A}$. The battery is disconnected. Calculate: (a) The time constant (b) The current after $t = 5\ \text{ms}$ (c) The voltage across the inductor at $t = 0^+$

Score: ___/3

Part D: Mutual Inductance & Transformers (6 problems)

Target: 3-4 min per problem.

Set D1 — Mutual Inductance (3 problems)

  1. Two coils have mutual inductance $M = 0.30\ \text{H}$. The current in coil 1 changes from $I_1 = 2.0\ \text{A}$ to $I_1 = 5.0\ \text{A}$ in $\Delta t = 0.10\ \text{s}$. Calculate the average induced EMF in coil 2.

  2. Two coaxial solenoids have $N_1 = 500$, $N_2 = 200$, common area $A = 10\ \text{cm}^2$, and length $l = 25\ \text{cm}$. Calculate the mutual inductance.

  3. Two coils have $L_1 = 0.40\ \text{H}$, $L_2 = 0.90\ \text{H}$, and coupling coefficient $k = 0.80$. Calculate: (a) The mutual inductance (b) The induced EMF in coil 2 when current in coil 1 changes at $\frac{dI_1}{dt} = -0.50\ \text{A/s}$

Score: ___/3

Set D2 — Transformers (3 problems)

  1. An ideal transformer has $N_p = 200$ turns and $N_s = 800$ turns. The primary is connected to $V_p = 120\ \text{V}_{\text{rms}}$. Calculate: (a) The secondary voltage (b) The secondary current if a $40\ \Omega$ resistor is connected (c) The primary current

  2. A step-down transformer is used to convert $V_p = 2400\ \text{V}$ to $V_s = 120\ \text{V}$. If $N_p = 1000$ turns, find $N_s$. If the load draws $I_s = 10\ \text{A}$, find $I_p$.

  3. An audio transformer has turns ratio $N_s/N_p = 4:1$ and is connected to a speaker with $R_L = 8.0\ \Omega$. What is the impedance seen by the primary?

Score: ___/3

Part E: AC Generator & Applications (4 problems)

Target: 3-4 min per problem.

  1. A rectangular coil with $N = 100$ turns has dimensions $w = 10\ \text{cm}$ and $h = 15\ \text{cm}$. It rotates at $f = 60\ \text{Hz}$ in a uniform field $B = 0.40\ \text{T}$. Calculate: (a) The maximum magnetic flux through the coil (b) The maximum induced EMF (c) Write the expression for instantaneous EMF

  2. An AC generator produces $V_{\text{rms}} = 120\ \text{V}$ at $f = 60\ \text{Hz}$. The coil has $N = 200$ turns and area $A = 0.02\ \text{m}^2$. Calculate the magnetic field strength.

  3. A metal disk of radius $R = 10\ \text{cm}$ rotates at $\omega = 100\ \text{rad/s}$ in a uniform magnetic field $B = 0.20\ \text{T}$ parallel to the rotation axis. Calculate the EMF between the center and rim.

  4. Explain how an eddy current brake works and why it does not work on a stationary object.

Score: ___/4

Final Scorecard

Part Topic Problems Raw Score
A — Faraday's Law Sets A1-A2 6 ___/6
B — Motional EMF Sets B1-B2 6 ___/6
C — Inductance & RL Sets C1-C2 6 ___/6
D — Mutual & Transformers Sets D1-D2 6 ___/6
E — AC Generator Set E 4 ___/4
TOTAL 28 ___/28

Proficiency Benchmarks for CQ5

Level Score Meaning
Exam-Ready $\ge 25/28$ (89%+) Ready for CQ5. Focus on speed.
Solid $22-24/28$ (79-86%) Good grasp. Review missed concepts.
Developing $17-21/28$ (61-75%) Drill weak areas again.
Needs Work $< 17/28$ (<61%) Re-study cheat sheet, retry tutorial.

Speed Benchmarks

  • < 80 min: Excellent speed for CQ5 conditions.
  • 80–100 min: Good. Practice under time pressure.
  • > 100 min: Need more pattern recognition drills.

Error Log Template

After grading, list every wrong problem with a one-word reason:

Problem Reason

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Part A — Faraday's Law & Flux

  1. $\mathcal{E} = -N\frac{\Delta\Phi}{\Delta t} = -100\frac{(0.40)(25 \times 10^{-4})}{2.0} = -5.0 \times 10^{-2}\ \text{V} = -50\ \text{mV}$ (magnitude: $50\ \text{mV}$)

  2. $\Delta\Phi = BA(\cos 90° - \cos 0°) = 0 - (0.30)(0.15)^2 = -6.75 \times 10^{-3}\ \text{Wb}$ $$\mathcal{E} = -\frac{\Delta\Phi}{\Delta t} = \frac{6.75 \times 10^{-3}}{0.50} = 1.35 \times 10^{-2}\ \text{V} = 13.5\ \text{mV}$$

  3. $\mathcal{E} = -NA\frac{dB}{dt} = -50(\pi \times 0.04^2)(0.20) = -5.03 \times 10^{-2}\ \text{V} = -50.3\ \text{mV}$

  4. By Lenz's law, the induced current creates a field opposing the approaching north pole. So current is counterclockwise (creating north pole facing the magnet).

  5. Induced current creates field into the page to oppose the decreasing flux. Current is clockwise.

  6. Increasing clockwise current in solenoid creates field pointing away from coil (into solenoid). Coil sees increasing flux away from it. Induced current creates field toward coil, so counterclockwise.

Part B — Motional EMF

  1. $\mathcal{E} = BLv = (0.50)(0.25)(4.0) = 0.50\ \text{V}$

  2. $\mathcal{E} = BLv = (5.0 \times 10^{-5})(40)(250) = 0.50\ \text{V}$

  3. (a) $\mathcal{E} = BLv = (0.40)(0.30)(2.0) = 0.24\ \text{V}$ (b) $I = \mathcal{E}/R = 0.24/5.0 = 48\ \text{mA}$ (c) $F = ILB = (0.048)(0.30)(0.40) = 5.76 \times 10^{-3}\ \text{N} = 5.76\ \text{mN}$

  4. (a) At terminal velocity: $F = BIL = B(\frac{BLv}{R})L = \frac{B^2L^2v}{R}$ $$v = \frac{FR}{B^2L^2} = \frac{(0.15)(10)}{(0.60)^2(0.50)^2} = \frac{1.5}{0.09} = 16.7\ \text{m/s}$$ (b) $P = Fv = 0.15 \times 16.7 = 2.5\ \text{W}$

  5. At terminal velocity: $mg = BIL = B(\frac{BLv}{R})L = \frac{B^2L^2v}{R}$ $$v = \frac{mgR}{B^2L^2} = \frac{(0.10)(9.8)(2.0)}{(0.50)^2(0.40)^2} = \frac{1.96}{0.04} = 49\ \text{m/s}$$

  6. (a) $\mathcal{E} = Ba v = (0.30)(0.20)(0.50) = 0.030\ \text{V}$, $I = \mathcal{E}/R = 6.0\ \text{mA}$ (b) $P = I^2R = (0.006)^2(5.0) = 1.8 \times 10^{-4}\ \text{W} = 0.18\ \text{mW}$ (c) $F = BIa = (0.30)(0.006)(0.20) = 3.6 \times 10^{-4}\ \text{N} = 0.36\ \text{mN}$

Part C — Self-Inductance & RL

  1. (a) $L = \frac{\mu_0 N^2 A}{l} = \frac{(4\pi \times 10^{-7})(400)^2(4.0 \times 10^{-4})}{0.20} = 4.02 \times 10^{-4}\ \text{H} = 0.40\ \text{mH}$ (b) $U = \frac{1}{2}LI^2 = 0.5(4.02 \times 10^{-4})(9) = 1.81 \times 10^{-3}\ \text{J} = 1.81\ \text{mJ}$

  2. $\mathcal{E} = -L\frac{dI}{dt} = -(50 \times 10^{-3})(0.20) = -0.010\ \text{V} = -10\ \text{mV}$

  3. (a) $L = \frac{2U}{I^2} = \frac{2(0.50)}{16} = 0.0625\ \text{H} = 62.5\ \text{mH}$ (b) $U = \frac{1}{2}(0.0625)(4) = 0.125\ \text{J}$

  4. (a) $\tau = L/R = 0.40/20 = 0.020\ \text{s} = 20\ \text{ms}$ (b) $I = \frac{12}{20}(1 - e^{-2}) = 0.60(0.865) = 0.519\ \text{A}$ (c) $I_{\text{final}} = V/R = 12/20 = 0.60\ \text{A}$

  5. $0.90 = 1 - e^{-t/\tau}$ → $e^{-t/\tau} = 0.10$ → $t = \tau\ln(10) = 2.30\tau = 46\ \text{ms}$

  6. (a) $\tau = L/R = 0.50/50 = 0.010\ \text{s} = 10\ \text{ms}$ (b) $I = I_0 e^{-t/\tau} = 2.0e^{-0.5} = 2.0(0.607) = 1.21\ \text{A}$ (c) $V_L = L\frac{dI}{dt}|_{t=0} = I_0 R = 2.0 \times 50 = 100\ \text{V}$

Part D — Mutual Inductance & Transformers

  1. $\mathcal{E} = -M\frac{\Delta I}{\Delta t} = -0.30(\frac{5.0-2.0}{0.10}) = -0.30(30) = -9.0\ \text{V}$

  2. $M = \frac{\mu_0 N_1 N_2 A}{l} = \frac{(4\pi \times 10^{-7})(500)(200)(10 \times 10^{-4})}{0.25} = 5.03 \times 10^{-4}\ \text{H} = 0.50\ \text{mH}$

  3. (a) $M = k\sqrt{L_1 L_2} = 0.80\sqrt{(0.40)(0.90)} = 0.80(0.60) = 0.48\ \text{H}$ (b) $\mathcal{E}_2 = -M\frac{dI_1}{dt} = -(0.48)(-0.50) = +0.24\ \text{V}$

  4. (a) $V_s = V_p \times \frac{N_s}{N_p} = 120 \times \frac{800}{200} = 480\ \text{V}$ (b) $I_s = V_s/R = 480/40 = 12\ \text{A}$ (c) $I_p = I_s \times \frac{N_s}{N_p} = 12 \times 4 = 48\ \text{A}$

  5. $N_s = N_p \times \frac{V_s}{V_p} = 1000 \times \frac{120}{2400} = 50$ turns $$I_p = I_s \times \frac{V_s}{V_p} = 10 \times \frac{120}{2400} = 0.50\ \text{A}$$

  6. $Z_{\text{in}} = (\frac{N_p}{N_s})^2 R_L = (\frac{1}{4})^2(8.0) = 0.50\ \Omega$

Part E — AC Generator & Applications

  1. (a) $\Phi_{\text{max}} = BA = (0.40)(0.10 \times 0.15) = 6.0 \times 10^{-3}\ \text{Wb} = 6.0\ \text{mWb}$ (b) $\mathcal{E}_{\text{max}} = NBA\omega = 100(0.40)(0.015)(2\pi \times 60) = 226\ \text{V}$ (c) $\mathcal{E}(t) = 226\sin(120\pi t)\ \text{V}$

  2. $V_{\text{rms}} = \frac{NBA\omega}{\sqrt{2}}$ → $B = \frac{V_{\text{rms}}\sqrt{2}}{NA(2\pi f)} = \frac{120\sqrt{2}}{200(0.02)(120\pi)} = 0.112\ \text{T}$

  3. $\mathcal{E} = \frac{1}{2}B\omega R^2 = 0.5(0.20)(100)(0.10)^2 = 0.10\ \text{V} = 100\ \text{mV}$

  4. Eddy current brake: When a conducting plate moves through a magnetic field, changing flux induces eddy currents. By Lenz's law, these currents create a magnetic field opposing the motion, producing a braking force. It does not work on stationary objects because there is no changing flux to induce currents.

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