FAD1022: Rapid-Fire Drill Pack — CQ4 Magnetism
Objective: Achieve mechanical fluency in computing magnetic fields, magnetic forces, torque, flux, and induced EMF for standard exam archetypes.
Target: 2–3 minutes per problem. If you stall >3 minutes, skip and mark it.
Total problems: 64
Estimated time: 2.5–3.5 hours
Cheat Sheet (Memorize First)
| Symbol | Meaning |
|---|---|
| $\odot$ | Out of the page |
| $\otimes$ | Into the page |
Constants
- Permeability of free space: $\mu_0 = 4\pi \times 10^{-7};\text{T}\cdot\text{m}\cdot\text{A}^{-1}$
- $1,\text{G} = 10^{-4},\text{T}$
Magnetic Field Formulas
| Geometry | Formula | Notes |
|---|---|---|
| Long straight wire (outside) | $B = \dfrac{\mu_0 I}{2\pi r}$ | $r \ge R$ |
| Long straight wire (inside) | $B = \dfrac{\mu_0 I r}{2\pi R^2}$ | $r < R$; enclosed current scales with area |
| Circular loop (centre) | $B = \dfrac{\mu_0 N I}{2r}$ | No $\pi$ in the numerator! |
| Solenoid (inside) | $B = \mu_0 n I = \dfrac{\mu_0 N I}{L}$ | Uniform; $n = N/L$ |
Forces & Motion
| Quantity | Formula |
|---|---|
| Lorentz force | $\vert\vec{F}_B\vert = \vert q\vert vB\sin\theta$ |
| Cyclotron radius | $r = \dfrac{mv}{qB}$ |
| Period | $T = \dfrac{2\pi m}{qB}$ (independent of $v$) |
| Angular frequency | $\omega = \dfrac{qB}{m}$ |
| Velocity selector | $v = \dfrac{E}{B}$ |
| Mass spectrometer | $m = \dfrac{qB'r}{v} = \dfrac{qB'Br}{E}$ |
| Force on a wire | $F = ILB\sin\theta$ |
| Force between parallel wires | $\dfrac{F}{L} = \dfrac{\mu_0 I_1 I_2}{2\pi d}$ |
Wire-force mnemonic: Same flow → attract; opposite flow → repel.
Torque, Flux & Induction
| Quantity | Formula |
|---|---|
| Torque on loop | $\tau = NIAB\sin\theta$ (any planar shape) |
| Dipole moment | $\mu = NIA$ |
| Magnetic flux | $\Phi_B = BA\cos\theta$ (normal to area) |
| Faraday’s law | $\varepsilon = -N\dfrac{d\Phi_B}{dt}$ |
| Inductor EMF | $\varepsilon = -L\dfrac{dI}{dt}$ |
| Energy in inductor | $U = \tfrac{1}{2}LI^2$ |
Torque trap: $\theta$ is the angle between the normal $\vec{A}$ and $\vec{B}$, not the plane and $\vec{B}$. Max torque when the plane is parallel to $\vec{B}$.
Part A: B‑Field Calculations
Target: 2 min per problem.
Set A1 — Long Straight Wire (6 problems)
Apply $B = \mu_0 I / (2\pi r)$. Watch your units (cm → m).
- A long straight wire carries $I = 8,\text{A}$. Calculate the magnetic field at a perpendicular distance of $5.0,\text{cm}$.
- Find the magnitude of $B$ at $r = 2.0,\text{cm}$ from a wire carrying $I = 15,\text{A}$.
- The magnetic field $2.0,\text{cm}$ from a wire is $4.0 \times 10^{-5},\text{T}$. Determine the current.
- At what distance from a wire carrying $I = 25,\text{A}$ is the field $5.0 \times 10^{-5},\text{T}$?
- Two long parallel wires carry $I_1 = 5,\text{A}$ and $I_2 = 5,\text{A}$, both out of the page, separated by $20,\text{cm}$. Find the resultant magnetic field at the midpoint.
- Two wires lie on the $x$‑axis at $x = 0$ ($I = 8,\text{A}$, out) and $x = 40,\text{cm}$ ($I = 12,\text{A}$, out). Calculate the net $B$ at $x = 10,\text{cm}$ (magnitude and direction).
Score: ___/6
Set A2 — Circular Loops & Solenoids (6 problems)
Apply $B = \mu_0 NI/(2r)$ or $B = \mu_0 nI$. Remember: the loop formula has no $\pi$ in the numerator.
- A single circular loop of radius $12,\text{cm}$ carries $I = 5.0,\text{A}$ counter‑clockwise. Find the magnetic field at its centre.
- A flat coil has $N = 50$ turns, radius $8.0,\text{cm}$, and carries $I = 2.0,\text{A}$. What is $B$ at the centre?
- A solenoid of length $0.50,\text{m}$ has $1000$ turns and carries $I = 0.80,\text{A}$. Determine the field inside.
- The interior field of a solenoid is $5.0,\text{mT}$. If the winding density is $n = 2000;\text{turns/m}$, find the current.
- A circular coil ($N = 20$, $r = 5.0,\text{cm}$) produces $B = 4.0 \times 10^{-4},\text{T}$ at its centre. Calculate the current.
- Two concentric flat coils share the same centre. Coil 1: $r = 10,\text{cm}$, $N = 10$, $I_1 = 2.0,\text{A}$ clockwise. Coil 2: $r = 15,\text{cm}$, $N = 5$, $I_2 = 3.0,\text{A}$ counter‑clockwise. Find the net magnetic field at the centre (magnitude and direction).
Score: ___/6
Set A3 — Superposition, Ampère’s Law & Direction (6 problems)
Use Ampère’s law for cylindrical wires and vector superposition for multiple sources.
- A cylindrical wire of radius $R = 3.0,\text{mm}$ carries $I = 12,\text{A}$. Find $B$ at the surface.
- For the same wire, calculate $B$ at a point $1.0,\text{mm}$ from the centre.
- For the same wire, calculate $B$ at $5.0,\text{mm}$ from the centre.
- A uniform solid wire carries current $I$. What is the ratio $B(r = R/2) / B(r = R)$?
- Two parallel wires on the $x$‑axis carry $8,\text{A}$ (out) at $x = 0$ and $12,\text{A}$ (out) at $x = 40,\text{cm}$. Find the net $B$ (magnitude and direction) at $x = 10,\text{cm}$.
- A long straight wire carries current toward the east. State the direction of $\vec{B}$ at a point directly north of the wire.
Score: ___/6
Part B: Forces on Charges & Wires
Target: 2–2.5 min per problem.
Set B1 — Lorentz Force (6 problems)
Use $|F_B| = |q|vB\sin\theta$ and the right‑hand rule (reverse for negative charges).
- A proton ($q = +e$) travels at $v = 4.0 \times 10^{6},\text{m/s}$ perpendicular to $B = 0.50,\text{T}$. Calculate the magnetic force magnitude.
- An electron moves at $5.0 \times 10^{7},\text{m/s}$ at $30^{\circ}$ to a field $B = 2.0,\text{mT}$. Find the force magnitude.
- A particle with $q = 3.2 \times 10^{-19},\text{C}$ experiences $F = 1.6 \times 10^{-14},\text{N}$ while moving at $v = 2.0 \times 10^{6},\text{m/s}$ perpendicular to $\vec{B}$. Determine $B$.
- A proton moves north; the magnetic field points east. What is the direction of the magnetic force on the proton?
- An electron moves upward in a region where $\vec{B}$ points out of the page. Determine the direction of the magnetic force.
- An alpha particle ($q = +2e$, $m = 6.64 \times 10^{-27},\text{kg}$) moves at $1.0 \times 10^{6},\text{m/s}$ perpendicular to $B = 0.20,\text{T}$. Find (a) the force and (b) the radius of its circular path.
Score: ___/6
Set B2 — Circular Motion & Applications (6 problems)
Apply $r = mv/(qB)$, $T = 2\pi m/(qB)$, the velocity selector, and the mass spectrometer.
- A proton ($m = 1.67 \times 10^{-27},\text{kg}$) moves at $v = 2.0 \times 10^{6},\text{m/s}$ perpendicular to $B = 0.30,\text{T}$. Find the orbit radius.
- For the proton in problem 25, calculate the period of revolution.
- An electron travels in a circle of radius $5.0,\text{cm}$ inside $B = 4.0,\text{mT}$. What is its speed?
- In a velocity selector, $E = 2.0 \times 10^{4},\text{V/m}$ and $B = 0.40,\text{T}$. What speed is selected?
- Ions with $q = +e$ exit the selector in problem 28 and enter a second field $B' = 0.80,\text{T}$, striking a detector $8.0,\text{cm}$ from entry. Determine the ion mass.
- A proton enters a uniform $B = 0.50,\text{T}$ at $60^{\circ}$ to the field with speed $v = 1.0 \times 10^{7},\text{m/s}$. Calculate the radius of the helical (circular) component of its motion.
Score: ___/6
Set B3 — Force on Wires & Between Wires (6 problems)
Use $F = ILB\sin\theta$ and $F/L = \mu_0 I_1 I_2 / (2\pi d)$. Remember: same direction → attract.
- A wire of length $L = 0.50,\text{m}$ carries $I = 10,\text{A}$ perpendicular to $B = 0.40,\text{T}$. Find the force.
- A $20,\text{cm}$ segment carries $I = 5.0,\text{A}$ at $30^{\circ}$ to $B = 0.20,\text{T}$. Calculate the force.
- Two parallel wires are $5.0,\text{cm}$ apart. Wire 1 carries $8,\text{A}$ and wire 2 carries $12,\text{A}$ in the same direction. Over a $2.0,\text{m}$ length, what is the force magnitude? Is it attractive or repulsive?
- Three long parallel wires lie on a line: wire A ($5,\text{A}$, up) at $x = 0$, wire B ($10,\text{A}$, down) at $x = 10,\text{cm}$, wire C ($8,\text{A}$, up) at $x = 20,\text{cm}$. Calculate the resultant force per unit length on wire A (magnitude and direction).
- Two parallel wires carry $I_1 = 3,\text{A}$ (up) and $I_2 = 5,\text{A}$ (down), separated by $8.0,\text{cm}$. Locate the point along the line joining them where the net magnetic field is zero.
- A horizontal wire of length $0.30,\text{m}$ and mass $5.0,\text{g}$ is suspended in a uniform $B = 0.25,\text{T}$ directed into the page. What current (magnitude and direction) will keep the wire in equilibrium with zero tension?
Score: ___/6
Part C: Torque, Flux & Induction
Target: 2–2.5 min per problem.
Set C1 — Torque on Current Loops (6 problems)
Use $\tau = NIAB\sin\theta$. Remember that $\theta$ is measured from the normal to the plane.
- A rectangular coil ($N = 50$, sides $4.0,\text{cm} \times 6.0,\text{cm}$, $I = 3.0,\text{A}$) is placed with its plane parallel to $B = 0.80,\text{T}$. Find the torque.
- A circular coil ($N = 100$, diameter $10,\text{cm}$, $I = 0.50,\text{A}$) has its normal at $30^{\circ}$ to $B = 2.0,\text{T}$. Calculate the torque.
- A single square loop of side $15,\text{cm}$ carries $I = 4.0,\text{A}$ in $B = 0.60,\text{T}$. What is the maximum torque?
- A coil ($N = 20$, $I = 2.0,\text{A}$, $A = 10,\text{cm}^2$) experiences $\tau = 12,\text{mN}\cdot\text{m}$ when its normal makes $45^{\circ}$ with $B$. Find $B$.
- A motor coil has $N = 200$ turns, $A = 30,\text{cm}^2$, and sits in $B = 0.50,\text{T}$. What current is required for maximum torque $\tau_{\max} = 1.5,\text{N}\cdot\text{m}$?
- A circular coil of radius $8.0,\text{cm}$ lies in the $xy$‑plane, carrying $2.0,\text{A}$ clockwise (viewed from $+z$). If $B = 0.40,\text{T}$ points along $+x$, determine the direction of the torque vector.
Score: ___/6
Set C2 — Magnetic Flux & Faraday’s Law (5 problems)
Use $\Phi_B = BA\cos\theta$ and $\varepsilon = -N,\Delta\Phi/\Delta t$.
- A loop of area $0.020,\text{m}^2$ is perpendicular to a uniform $B = 0.30,\text{T}$. Calculate the magnetic flux.
- The loop in problem 43 is rotated so that the angle between its normal and $\vec{B}$ is $60^{\circ}$. What is the new flux?
- A $100$‑turn coil of area $5.0,\text{cm}^2$ is perpendicular to a field that increases from $0.10,\text{T}$ to $0.50,\text{T}$ in $2.0,\text{ms}$. Find the magnitude of the average induced emf.
- A loop of area $0.040,\text{m}^2$ is oriented with its normal perpendicular to a field that is changing at $0.50,\text{T/s}$. What is the induced emf?
- A $50$‑turn coil ($A = 8.0,\text{cm}^2$) is perpendicular to a uniform field that rises linearly from $0$ to $0.60,\text{T}$ in $0.50,\text{s}$. Calculate the average induced emf.
Score: ___/5
Set C3 — Lenz’s Law & Inductance Basics (5 problems)
Apply Lenz’s law for direction and $\varepsilon = -L,\Delta I/\Delta t$ or $U = \tfrac12 LI^2$ for inductors.
- A bar magnet’s north pole is pushed toward a conducting loop. Determine the direction of the induced current as seen from the magnet side.
- A loop sits in a uniform magnetic field directed into the page. The loop is shrinking while the field stays constant. State the direction of the induced current.
- An inductor $L = 40,\text{mH}$ carries a current increasing at $5.0,\text{A/s}$. Find the magnitude of the induced emf.
- The current through a $25,\text{mH}$ inductor drops from $3.0,\text{A}$ to $1.0,\text{A}$ in $4.0,\text{ms}$. Calculate the average induced emf.
- An inductor stores $0.20,\text{J}$ when carrying $2.0,\text{A}$. Determine its inductance.
Score: ___/5
Part D: Mixed & Reverse Problems
Target: 3 min per problem.
Set D1 — Mixed Concept Problems (6 problems)
- A proton moves in a circular path of radius $8.0,\text{cm}$ inside $B = 0.50,\text{T}$. Calculate its kinetic energy in electron‑volts.
- A $200$‑turn circular coil of radius $5.0,\text{cm}$ carries $I = 0.40,\text{A}$ and is placed in $B = 0.25,\text{T}$ with its normal at $30^{\circ}$ to the field. Find (a) the dipole moment, (b) the torque, and (c) the field at the centre of the coil due to its own current.
- Two long parallel wires A ($6,\text{A}$, up) and B ($9,\text{A}$, up) are $10,\text{cm}$ apart. A third wire C ($4,\text{A}$, up) is placed between them, $4.0,\text{cm}$ from A. Find the net force per unit length on C.
- Inside a solenoid ($L = 0.40,\text{m}$, $N = 800$, $I = 2.0,\text{A}$) a proton moves perpendicular to the axis with $v = 5.0 \times 10^{5},\text{m/s}$. Find the radius of its path.
- A rectangular loop ($8.0,\text{cm} \times 12,\text{cm}$) is half inside and half outside a uniform $B = 0.50,\text{T}$ directed into the page. The field collapses to zero in $0.10,\text{s}$. Calculate the average induced emf.
- A $150$‑turn coil of area $12,\text{cm}^2$ rotates from $\theta = 0^{\circ}$ (normal parallel to $B = 0.40,\text{T}$) to $\theta = 90^{\circ}$ in $0.05,\text{s}$. Find the average induced emf.
Score: ___/6
Set D2 — Reverse Engineering (6 problems)
- A long straight wire produces $B = 2.0 \times 10^{-5},\text{T}$ at $r = 4.0,\text{cm}$. Determine the current.
- A proton in a uniform magnetic field has a period of $2.0;\mu\text{s}$. Calculate the field strength.
- A velocity selector is designed to pass ions with $v = 2.0 \times 10^{5},\text{m/s}$. If $B = 0.50,\text{T}$, what electric field is required?
- Two parallel wires $6.0,\text{cm}$ apart attract with a force per unit length of $4.0 \times 10^{-5},\text{N/m}$. If $I_1 = 3.0,\text{A}$, find $I_2$ (same direction).
- A coil of $N = 80$ turns and area $6.0,\text{cm}^2$ experiences a maximum torque of $0.096,\text{N}\cdot\text{m}$ in $B = 0.40,\text{T}$. Find the current.
- A solenoid of length $0.50,\text{m}$ must produce $B = 2.0,\text{mT}$ when carrying $I = 5.0,\text{A}$. How many total turns are needed?
Score: ___/6
Final Scorecard
| Part | Sets | Problems | Raw Score |
|---|---|---|---|
| A — B‑Field Calculations | A1, A2, A3 | 18 | ___/18 |
| B — Forces on Charges & Wires | B1, B2, B3 | 18 | ___/18 |
| C — Torque, Flux & Induction | C1, C2, C3 | 16 | ___/16 |
| D — Mixed & Reverse | D1, D2 | 12 | ___/12 |
| TOTAL | 64 | ___/64 |
Proficiency Benchmarks
- 45/64 (70%) — Proficient. You can handle standard exam problems.
- 54/64 (85%) — Solid. Fast and accurate on most archetypes.
- 60/64 (94%) — Exam‑ready. Any remaining mistake is a careless slip.
Speed Benchmarks
- < 2.5 hours: Excellent mechanical fluency.
- < 3.5 hours: Good. Review the sets you scored lowest on.
- > 4 hours: Re‑drill the specific sets with the most errors tomorrow.
Error Log Template
After grading, list every wrong problem number with a one‑word reason:
| Problem | Reason |
|---|---|
| e.g. 7 | forgot no‑pi rule |
Re‑solve all wrong problems immediately with notes, then again in 24 hours without notes.
Answer Key
- $3.2 \times 10^{-5},\text{T}$
- $1.5 \times 10^{-4},\text{T}$
- $20,\text{A}$
- $10,\text{cm}$
- $0,\text{T}$ (fields cancel)
- $8.0 \times 10^{-6},\text{T}$ upward ($+y$)
- $2.62 \times 10^{-5},\text{T}$ (out of page)
- $7.85 \times 10^{-4},\text{T}$
- $2.01 \times 10^{-3},\text{T} \approx 2.0,\text{mT}$
- $\approx 2.0,\text{A}$
- $\approx 1.59,\text{A}$
- $2\pi \times 10^{-5},\text{T} \approx 6.28 \times 10^{-5},\text{T}$ into the page
- $8.0 \times 10^{-4},\text{T} = 0.80,\text{mT}$
- $2.67 \times 10^{-4},\text{T}$
- $4.8 \times 10^{-4},\text{T}$
- $1/2$
- $8.0 \times 10^{-6},\text{T}$ upward ($+y$)
- Out of the page ($\odot$)
- $3.2 \times 10^{-13},\text{N}$
- $8.0 \times 10^{-15},\text{N}$
- $2.5 \times 10^{-2},\text{T} = 25,\text{mT}$
- Into the page ($\otimes$)
- West ($-x$)
- (a) $6.4 \times 10^{-14},\text{N}$; (b) $10.4,\text{cm}$
- $6.96,\text{cm}$
- $2.18 \times 10^{-7},\text{s} \approx 0.22;\mu\text{s}$
- $3.51 \times 10^{7},\text{m/s}$
- $5.0 \times 10^{4},\text{m/s}$
- $6.4 \times 10^{-26},\text{kg}$
- $18,\text{cm}$
- $2.0,\text{N}$
- $0.10,\text{N}$
- $7.68 \times 10^{-4},\text{N}$; attractive
- $6.0 \times 10^{-5},\text{N/m}$ to the left
- $12,\text{cm}$ to the left of wire 1
- $0.65,\text{A}$ to the right
- $0.288,\text{N}\cdot\text{m}$
- $0.39,\text{N}\cdot\text{m}$
- $0.054,\text{N}\cdot\text{m}$
- $0.42,\text{T}$
- $5.0,\text{A}$
- $-y$ direction
- $6.0 \times 10^{-3},\text{Wb}$
- $3.0 \times 10^{-3},\text{Wb}$
- $10,\text{V}$
- $0,\text{V}$ (flux is zero because $\cos 90^{\circ}=0$)
- $48,\text{mV}$
- Counter‑clockwise (as viewed from the magnet side)
- Clockwise
- $0.20,\text{V}$
- $12.5,\text{V}$
- $0.10,\text{H} = 100,\text{mH}$
- $\approx 77,\text{keV}$
- (a) $0.628,\text{A}\cdot\text{m}^2$; (b) $0.079,\text{N}\cdot\text{m}$; (c) $1.0,\text{mT}$
- $0,\text{N/m}$ (forces cancel)
- $1.04,\text{m}$
- $24,\text{mV}$
- $1.44,\text{V}$
- $4.0,\text{A}$
- $33,\text{mT}$
- $1.0 \times 10^{5},\text{V/m}$
- $4.0,\text{A}$
- $5.0,\text{A}$
- $159$ turns (≈ $160$)