FAD1022 CQ4: Rapid-Fire Drill Pack — Magnetism

Objective: Master all Magnetism concepts for CQ4.
Target: 3-4 min per problem (CQ4 style: 4 questions in 15 min).
Total problems: 28
Estimated time: ~100 min

Cheat Sheet (Memorize First)

Constants for CQ4

Constant Symbol Value
Permeability of free space $\mu_0$ $4\pi \times 10^{-7}\ \text{T·m·A}^{-1}$
Elementary charge $e$ $1.60 \times 10^{-19}\ \text{C}$
Electron mass $m_e$ $9.11 \times 10^{-31}\ \text{kg}$

Magnetic Field Formulas (Ampere's Law)

Source Formula Notes
Long straight wire $B = \frac{\mu_0 I}{2\pi r}$ $r >$ wire radius
Inside wire ($r < R$) $B = \frac{\mu_0 I r}{2\pi R^2}$ Uniform current density
Center of circular loop $B = \frac{\mu_0 I}{2R}$ Single loop
Solenoid (ideal) $B = \mu_0 n I$ $n = N/L$ turns per meter
Toroid $B = \frac{\mu_0 N I}{2\pi r}$ Inside toroid only

Lorentz Force

Situation Formula
Force on moving charge $\vec{F} = q\vec{v} \times \vec{B}$
Magnitude $F =
Force on current element $d\vec{F} = I d\vec{l} \times \vec{B}$
Force on straight wire $F = ILB\sin\theta$

Circular Motion in B-field

Quantity Formula
Cyclotron radius $r = \frac{mv}{
Period $T = \frac{2\pi m}{
Angular frequency $\omega = \frac{

Torque on Current Loop

Quantity Formula
Torque $\tau = NIAB\sin\theta$
Magnetic moment $\vec{\mu} = NIA\hat{n}$
Potential energy $U = -\vec{\mu} \cdot \vec{B} = -\mu B\cos\theta$

Right-Hand Rules

Rule Application
RHR #1 Thumb: velocity ($v$), Fingers: $B$, Palm: $F$ on positive charge
RHR #2 (Ampere) Thumb: current, Fingers curl: $B$ direction

Part A: Magnetic Field from Currents (8 problems)

Target: 3 min per problem.

Set A1 — Long Straight Wire (4 problems)

  1. A long straight wire carries $I = 5.0\ \text{A}$. Calculate the magnetic field at $r = 10\ \text{cm}$ from the wire.

  2. At what distance from a wire carrying $I = 8.0\ \text{A}$ is the magnetic field $B = 2.0 \times 10^{-5}\ \text{T}$?

  3. Two long parallel wires are separated by $d = 20\ \text{cm}$. Wire 1 carries $I_1 = 3.0\ \text{A}$ into the page, wire 2 carries $I_2 = 4.0\ \text{A}$ out of the page. Find the net magnetic field at the midpoint between them.

  4. Three long parallel wires are arranged at the corners of an equilateral triangle with side $a = 30\ \text{cm}$. Each carries $I = 2.0\ \text{A}$ in the same direction. Find the net magnetic field at the center of the triangle.

Score: ___/4

Set A2 — Ampere's Law Applications (4 problems)

  1. A long cylindrical wire of radius $R = 2.0\ \text{mm}$ carries $I = 5.0\ \text{A}$ uniformly distributed. Calculate the magnetic field at: (a) $r = 1.0\ \text{mm}$ from the center (b) $r = 3.0\ \text{mm}$ from the center

  2. A coaxial cable has inner radius $a = 1.0\ \text{mm}$ carrying $I = 3.0\ \text{A}$ outward, and outer radius $b = 3.0\ \text{mm}$ carrying the same current inward. Find $B$ at: (a) $r = 0.5\ \text{mm}$ (b) $r = 2.0\ \text{mm}$ (c) $r = 4.0\ \text{mm}$

  3. A solenoid has $N = 500$ turns, length $L = 25\ \text{cm}$, and carries $I = 2.0\ \text{A}$. Calculate the magnetic field inside the solenoid.

  4. A toroid has $N = 1000$ turns and carries $I = 5.0\ \text{A}$. The mean radius is $r = 15\ \text{cm}$. Calculate the magnetic field inside the toroid.

Score: ___/4

Part B: Lorentz Force on Moving Charges (6 problems)

Target: 3-4 min per problem.

Set B1 — Force on Charged Particles (3 problems)

  1. A proton ($q = +e$) moves with $v = 5.0 \times 10^6\ \text{m/s}$ perpendicular to a uniform magnetic field $B = 0.50\ \text{T}$. Calculate: (a) The magnitude of the magnetic force (b) The direction of the force (using right-hand rule) (c) The radius of the circular path

  2. An electron travels at $v = 2.0 \times 10^7\ \text{m/s}$ at $\theta = 30°$ to a uniform magnetic field $B = 0.20\ \text{T}$. Calculate: (a) The force on the electron (b) The radius of the helical path (c) The pitch of the helix (distance along $B$ for one revolution)

  3. A charged particle moves undeflected through crossed electric and magnetic fields. $E = 2.0 \times 10^4\ \text{V/m}$ and $B = 0.10\ \text{T}$, both perpendicular to each other and to the particle's velocity. Calculate: (a) The velocity of the particle (b) If the particle is an electron, what is the radius of its path when the E-field is turned off?

Score: ___/3

Set B2 — Circular Motion in Magnetic Fields (3 problems)

  1. An electron is accelerated through $V = 500\ \text{V}$ and then enters a uniform magnetic field $B = 0.20\ \text{T}$ perpendicular to its velocity. Find: (a) The electron's speed (b) The radius of its circular path (c) The period of revolution

  2. A proton moves in a circular path of radius $r = 15\ \text{cm}$ in a magnetic field $B = 0.30\ \text{T}$. Calculate: (a) The speed of the proton (b) The kinetic energy in eV (c) The frequency of revolution (cyclotron frequency)

  3. In a mass spectrometer, ions with charge $+e$ and mass $m$ are accelerated through voltage $V$ and then enter a magnetic field $B$. Show that the radius is $r = \frac{1}{B}\sqrt{\frac{2mV}{e}}$. Calculate $r$ for protons with $V = 1000\ \text{V}$ and $B = 0.50\ \text{T}$.

Score: ___/3

Part C: Force on Current-Carrying Wires (6 problems)

Target: 3 min per problem.

Set C1 — Force on Wires (3 problems)

  1. A straight wire of length $L = 50\ \text{cm}$ carries $I = 3.0\ \text{A}$ perpendicular to a uniform magnetic field $B = 0.40\ \text{T}$. Calculate the force on the wire.

  2. A wire carries $I = 5.0\ \text{A}$ along the x-axis from $x = 0$ to $x = 2.0\ \text{m}$ in a uniform magnetic field $\vec{B} = (0.30\hat{j} + 0.40\hat{k})\ \text{T}$. Calculate the force on the wire.

  3. A square loop of side $a = 10\ \text{cm}$ carries $I = 2.0\ \text{A}$. The loop is in a uniform magnetic field $B = 0.50\ \text{T}$ parallel to the plane of the loop. Calculate the net force on the loop.

Score: ___/3

Set C2 — Force Between Parallel Wires (3 problems)

  1. Two long parallel wires separated by $d = 15\ \text{cm}$ carry currents $I_1 = 4.0\ \text{A}$ and $I_2 = 6.0\ \text{A}$ in the same direction. Calculate the force per unit length between them.

  2. For the wires in problem 18, if the currents are in opposite directions, what is the force per unit length? Is it attractive or repulsive?

  3. Three long parallel wires are arranged at the corners of an equilateral triangle with side $a = 20\ \text{cm}$. Each carries $I = 5.0\ \text{A}$, with $I_1$ and $I_2$ going into the page, and $I_3$ coming out. Calculate the net force per unit length on wire 3.

Score: ___/3

Part D: Torque on Current Loops (4 problems)

Target: 3-4 min per problem.

  1. A rectangular coil with $N = 50$ turns has dimensions $w = 8.0\ \text{cm}$ and $h = 12\ \text{cm}$. It carries $I = 2.0\ \text{A}$ in a uniform magnetic field $B = 0.30\ \text{T}$. Calculate: (a) The magnetic moment (b) The maximum torque (c) The torque when the plane of the coil makes $\theta = 30°$ with $\vec{B}$

  2. A circular coil with $N = 100$ turns and radius $R = 5.0\ \text{cm}$ carries $I = 1.5\ \text{A}$. The coil is free to rotate in a uniform field $B = 0.20\ \text{T}$. Calculate: (a) The magnetic moment (b) The maximum torque (c) The potential energy when $\vec{\mu}$ is antiparallel to $\vec{B}$

  3. A square loop of side $a = 5.0\ \text{cm}$ carries $I = 3.0\ \text{A}$. The loop is in a uniform field $B = 0.25\ \text{T}$ with the plane of the loop perpendicular to the field. Calculate: (a) The torque on the loop (b) The potential energy (c) The work required to rotate the loop by $90°$ so the plane is parallel to $\vec{B}$

  4. An AC motor uses a coil with $N = 200$ turns, area $A = 40\ \text{cm}^2$, in a field $B = 0.50\ \text{T}$. The coil carries $I = 3.0\ \text{A}$ and rotates at $60\ \text{Hz}$. Write the expression for instantaneous torque $\tau(t)$ and calculate the average torque.

Score: ___/4

Part E: Gauss's Law for Magnetism (4 problems)

Target: 2-3 min per problem.

  1. State Gauss's law for magnetism in integral form and explain its physical significance.

  2. A magnetic dipole is placed at the center of a spherical Gaussian surface. What is the net magnetic flux through the surface? Explain.

  3. A cube is placed in a uniform magnetic field $\vec{B} = B_0\hat{k}$. Calculate the magnetic flux through: (a) The top face (normal in $+z$ direction) (b) The bottom face (normal in $-z$ direction) (c) One side face (normal perpendicular to $z$) (d) What is the net flux through the entire cube?

  4. Explain why magnetic monopoles cannot exist based on Gauss's law for magnetism.

Score: ___/4

Final Scorecard

Part Topic Problems Raw Score
A — B-field from currents Sets A1-A2 8 ___/8
B — Lorentz force (charges) Sets B1-B2 6 ___/6
C — Force on wires Sets C1-C2 6 ___/6
D — Torque on loops Set D 4 ___/4
E — Gauss's law for B Set E 4 ___/4
TOTAL 28 ___/28

Proficiency Benchmarks for CQ4

Level Score Meaning
Exam-Ready $\ge 25/28$ (89%+) Ready for CQ4. 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 CQ4 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 — Magnetic Field from Currents

  1. $B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7})(5.0)}{2\pi(0.10)} = 1.0 \times 10^{-5}\ \text{T}$

  2. $r = \frac{\mu_0 I}{2\pi B} = \frac{(4\pi \times 10^{-7})(8.0)}{2\pi(2.0 \times 10^{-5})} = 8.0 \times 10^{-2}\ \text{m} = 8.0\ \text{cm}$

  3. Midpoint: $B_1 = \frac{\mu_0(3.0)}{2\pi(0.10)}$ (clockwise around wire 1, so down), $B_2 = \frac{\mu_0(4.0)}{2\pi(0.10)}$ (counter-clockwise, so also down). $B_{\text{net}} = B_1 + B_2 = 1.4 \times 10^{-5}\ \text{T}$ downward

  4. At center, distance to each wire: $r = a/\sqrt{3} = 0.173\ \text{m}$. By symmetry, fields cancel. $B_{\text{net}} = 0$

  5. (a) Inside: $B = \frac{\mu_0 I r}{2\pi R^2} = \frac{(4\pi \times 10^{-7})(5.0)(0.001)}{2\pi(0.002)^2} = 2.5 \times 10^{-4}\ \text{T}$ (b) Outside: $B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7})(5.0)}{2\pi(0.003)} = 3.33 \times 10^{-4}\ \text{T}$

  6. (a) Inside inner: $B = \frac{\mu_0 I r}{2\pi a^2} = \frac{(4\pi \times 10^{-7})(3.0)(0.0005)}{2\pi(0.001)^2} = 3.0 \times 10^{-4}\ \text{T}$ (b) Between: $B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7})(3.0)}{2\pi(0.002)} = 3.0 \times 10^{-4}\ \text{T}$ (c) Outside: Net enclosed current = 0, so $B = 0$

  7. $n = N/L = 500/0.25 = 2000\ \text{turns/m}$. $B = \mu_0 n I = (4\pi \times 10^{-7})(2000)(2.0) = 5.03 \times 10^{-3}\ \text{T}$

  8. $B = \frac{\mu_0 N I}{2\pi r} = \frac{(4\pi \times 10^{-7})(1000)(5.0)}{2\pi(0.15)} = 6.67 \times 10^{-3}\ \text{T}$

Part B — Lorentz Force on Charges

  1. (a) $F = qvB = (1.6 \times 10^{-19})(5.0 \times 10^6)(0.50) = 4.0 \times 10^{-13}\ \text{N}$ (b) Using RHR: for $+q$, $v$ in some direction, $B$ perpendicular, force perpendicular to both (c) $r = \frac{mv}{qB} = \frac{(1.67 \times 10^{-27})(5.0 \times 10^6)}{(1.6 \times 10^{-19})(0.50)} = 0.104\ \text{m} = 10.4\ \text{cm}$

  2. (a) $F = |e|vB\sin\theta = (1.6 \times 10^{-19})(2.0 \times 10^7)(0.20)(0.5) = 3.2 \times 10^{-13}\ \text{N}$ (b) Perpendicular component: $v_\perp = v\sin\theta = 1.0 \times 10^7\ \text{m/s}$, $r = \frac{mv_\perp}{|e|B} = 2.85 \times 10^{-4}\ \text{m}$ (c) $v_\parallel = v\cos\theta = 1.73 \times 10^7\ \text{m/s}$, $T = \frac{2\pi r}{v_\perp} = \frac{2\pi m}{|e|B} = 1.79 \times 10^{-10}\ \text{s}$, pitch = $v_\parallel T = 3.09 \times 10^{-3}\ \text{m}$

  3. (a) $v = E/B = (2.0 \times 10^4)/(0.10) = 2.0 \times 10^5\ \text{m/s}$ (b) $r = \frac{mv}{|e|B} = \frac{(9.11 \times 10^{-31})(2.0 \times 10^5)}{(1.6 \times 10^{-19})(0.10)} = 1.14 \times 10^{-5}\ \text{m}$

  4. (a) $\frac{1}{2}m_e v^2 = eV$ → $v = \sqrt{\frac{2eV}{m_e}} = \sqrt{\frac{2(1.6 \times 10^{-19})(500)}{9.11 \times 10^{-31}}} = 1.33 \times 10^7\ \text{m/s}$ (b) $r = \frac{m_e v}{eB} = \frac{(9.11 \times 10^{-31})(1.33 \times 10^7)}{(1.6 \times 10^{-19})(0.20)} = 3.78 \times 10^{-4}\ \text{m}$ (c) $T = \frac{2\pi m_e}{eB} = \frac{2\pi(9.11 \times 10^{-31})}{(1.6 \times 10^{-19})(0.20)} = 1.79 \times 10^{-10}\ \text{s}$

  5. (a) $v = \frac{|e|Br}{m_p} = \frac{(1.6 \times 10^{-19})(0.30)(0.15)}{1.67 \times 10^{-27}} = 4.31 \times 10^6\ \text{m/s}$ (b) $KE = \frac{1}{2}m_p v^2 = 1.55 \times 10^{-14}\ \text{J} = \frac{1.55 \times 10^{-14}}{1.6 \times 10^{-19}} = 9.69 \times 10^4\ \text{eV}$ (c) $f = \frac{|e|B}{2\pi m_p} = \frac{(1.6 \times 10^{-19})(0.30)}{2\pi(1.67 \times 10^{-27})} = 4.58 \times 10^6\ \text{Hz}$

  6. $\frac{1}{2}mv^2 = qV$ → $v = \sqrt{\frac{2qV}{m}}$, $r = \frac{mv}{qB} = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$ For protons: $r = \frac{1}{0.50}\sqrt{\frac{2(1.67 \times 10^{-27})(1000)}{1.6 \times 10^{-19}}} = 2\sqrt{2.09 \times 10^{-5}} = 2(4.57 \times 10^{-3}) = 9.14 \times 10^{-3}\ \text{m}$

Part C — Force on Current-Carrying Wires

  1. $F = ILB = (3.0)(0.50)(0.40) = 0.60\ \text{N}$

  2. $\vec{L} = 2.0\hat{i}\ \text{m}$, $\vec{F} = I\vec{L} \times \vec{B} = 5.0(2.0\hat{i}) \times (0.30\hat{j} + 0.40\hat{k}) = 10(0.30\hat{k} - 0.40\hat{j}) = (-4.0\hat{j} + 3.0\hat{k})\ \text{N}$

  3. Net force on any closed current loop in uniform $B$-field is zero. $F_{\text{net}} = 0$

  4. $F/L = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi \times 10^{-7})(4.0)(6.0)}{2\pi(0.15)} = 3.2 \times 10^{-5}\ \text{N/m}$ (attractive)

  5. Same magnitude but repulsive (opposite currents repel)

  6. Wire 1 attracts wire 3 with $F_{13}/L = \frac{\mu_0(5)(5)}{2\pi(0.20)} = 2.5 \times 10^{-5}\ \text{N/m}$ at $60°$ below horizontal Wire 2 repels wire 3 with same magnitude at $60°$ above horizontal By symmetry, horizontal components cancel, vertical add: $F_{\text{net}}/L = 2(2.5 \times 10^{-5})\cos(30°) = 4.33 \times 10^{-5}\ \text{N/m}$ upward

Part D — Torque on Current Loops

  1. (a) $\mu = NIA = 50(2.0)(0.08 \times 0.12) = 0.96\ \text{A·m}^2$ (b) $\tau_{\text{max}} = \mu B = 0.96(0.30) = 0.288\ \text{N·m}$ (c) $\tau = \mu B\sin\theta = 0.288\sin(60°) = 0.249\ \text{N·m}$

  2. (a) $\mu = NIA = 100(1.5)(\pi \times 0.05^2) = 1.18\ \text{A·m}^2$ (b) $\tau_{\text{max}} = 1.18(0.20) = 0.236\ \text{N·m}$ (c) $U = -\mu B\cos(180°) = +\mu B = 0.236\ \text{J}$

  3. (a) $\tau = 0$ (plane perpendicular to B means $\vec{\mu}$ is parallel to $\vec{B}$) (b) $U = -\mu B = -(3.0 \times 0.05^2)(0.25) = -1.88 \times 10^{-3}\ \text{J}$ (c) $W = \Delta U = 0 - (-1.88 \times 10^{-3}) = 1.88 \times 10^{-3}\ \text{J}$

  4. $\tau(t) = NIAB\sin(\omega t) = 200(3.0)(40 \times 10^{-4})(0.50)\sin(120\pi t) = 1.20\sin(120\pi t)\ \text{N·m}$ $$\tau_{\text{avg}} = \frac{2}{\pi}(1.20) = 0.764\ \text{N·m}$$

Part E — Gauss's Law for Magnetism

  1. $\oint \vec{B} \cdot d\vec{A} = 0$. Physical significance: Magnetic field lines form closed loops; there are no magnetic monopoles (sources or sinks of magnetic field).

  2. Net flux = 0. By Gauss's law for magnetism, the net magnetic flux through any closed surface is always zero.

  3. (a) $\Phi = BA = B_0 A$ (b) $\Phi = -B_0 A$ (normal opposite to field) (c) $\Phi = 0$ (normal perpendicular to field) (d) Net flux = $B_0 A - B_0 A + 0 + 0 + 0 + 0 = 0$

  4. If magnetic monopoles existed, they would be sources/sinks of magnetic field lines, similar to electric charges for electric field lines. Gauss's law for magnetism states that the net magnetic flux through any closed surface is always zero, which means there are no isolated magnetic charges (monopoles). Magnetic field lines always form closed loops.

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