Physics 2: Electricity and Magnetism: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Coulomb’s law & electric fields

Coulomb’s law: F = k·|q₁q₂|/r²
Field of a point charge: E = k·|q|/r²
Constant: k = 8.99 × 10⁹ N·m²/C²
Force on a charge: F = q·E

Series & parallel resistors

Resistors in series: R_eq = R₁ + R₂ + …
Resistors in parallel: 1/R_eq = 1/R₁ + 1/R₂ + …
Two in parallel: R_eq = R₁R₂ / (R₁ + R₂)
Ohm's law: V = I·R
Power: P = V·I = I²R = V²/R

Method

Start at the end of the network furthest from the terminals.
Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
Redraw the circuit after each step.
Repeat until one resistor is left between the terminals.

RC & RL transients

Time constant: τ = RC | τ = L/R
Response: x(t) = x(∞) + [x(0) − x(∞)]·e^−t/τ
Capacitor at DC: open circuit; v_C cannot jump
Inductor at DC: short circuit; i_L cannot jump
Settled: t ≈ 5τ

Method

Find the initial value just after switching: a capacitor voltage and an inductor current cannot jump.
Find the final value long after switching: a capacitor acts as an open circuit, an inductor as a short circuit.
Find the time constant: τ = RC or τ = L/R, using the resistance seen by the capacitor or inductor.
Use x(t) = x(∞) + [x(0) − x(∞)]·e^−t/τ.

Magnetic fields & induction

Long straight wire: B = μ₀I/(2πr)
Centre of a flat coil: B = μ₀NI/(2R)
Long solenoid: B = μ₀nI, n = N/L
Constant: μ₀ = 4π × 10⁻⁷ T·m/A
Force on a wire: F = BIL·sin θ
Force on a charge: F = |q|vB·sin θ
Circular motion: r = mv/(|q|B), T = 2πm/(|q|B)
Parallel wires: F/L = μ₀I₁I₂/(2πd)
Flux: Φ = BA·cos θ
Faraday’s law: ε = −N·ΔΦ/Δt
Moving rod: ε = BLv
Generator: E₀ = NBAω, E_rms = E₀/√2
Inductance: ε = L·ΔI/Δt, L = μ₀N²A/l, W = ½LI²
Mutual inductance: ε₂ = M·di₁/dt, M = k√(L₁L₂)
Velocity selector and spectrometer: v = E/B, r = √(2mV/|q|)/B

Method

Convert every quantity to SI units: cm to m, cm² to m², mT and µT to T.
Pick the right formula: B = μ₀I/(2πr) for a wire, μ₀NI/(2R) for the centre of a coil, μ₀nI inside a solenoid; F = BIL·sin θ or |q|vB·sin θ for forces.
For induction, find the change of flux ΔΦ = Δ(BA·cos θ) and use ε = N·ΔΦ/Δt, or ε = BLv for a moving rod and NBAω for a rotating coil.
Find directions separately with the right-hand rules, and Lenz’s law for induced currents.

AC impedance & phasors

Impedances: Z_R = R, Z_L = jωL, Z_C = 1/(jωC)
Angular frequency: ω = 2πf
Magnitude and angle: |Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law: V = I·Z

Method

Convert the frequency: ω = 2πf.
Find each reactance: X_L = ωL and X_C = 1/(ωC).
Add impedances like resistors, with Z = R + jX.
Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.

2Practice problems

1.Coulomb’s law & electric fieldsMedium

Three point charges lie on a straight line: −5 µC on the left, +3 µC on the right 20 cm away, and −1 µC between them, 12 cm from the left charge. Find the magnitude of the net electric force F on the middle charge. Use k = 8.99 × 10⁹ N·m²/C².

q_L = −5 µCq_R = +3 µCq_M = −1 µCr_L = 12 cmr_R = 8 cm

F = N

2.Coulomb’s law & electric fieldsHard

Three positive point charges sit at the corners of a right triangle: q₃ = 5 µC is at the right-angle corner, q₁ = 5 µC is 8 cm from it along one side and q₂ = 10 µC is 6 cm from it along the other side. Find the magnitude of the net electric force F on q₃.

q₁ = 5 µCq₂ = 10 µCq₃ = 5 µCr₁₃ = 8 cmr₂₃ = 6 cm

F = N

3.Series & parallel resistorsMedium

Find the current I_R1 delivered by the source, which flows through R₁.

V₁ = 15 VR₁ = 6 ΩR₂ = 8 ΩR₃ = 5 ΩR₄ = 4 Ω

I_R1 = A

4.Series & parallel resistorsHard

Find the current I_R1 delivered by the source, which flows through R₁.

V₁ = 12 VR₁ = 4 ΩR₂ = 3 ΩR₃ = 3 ΩR₄ = 30 ΩR₅ = 8 ΩR₆ = 30 Ω

I_R1 = A

5.RC & RL transientsMedium

The inductor carries no current before the switch closes at t = 0. How long after the switch closes does the current reach 75% of its final value? Find t.

V_s = 12 VR = 33 ΩL = 220 mH

t = ms

6.RC & RL transientsHard

The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the capacitor voltage v_C at t = 15 ms.

V_s = 24 VR₁ = 4.7 kΩR₂ = 3.3 kΩC = 10 µFt = 15 ms

v_C = V

7.Magnetic fields & inductionMedium

A square loop moves to the right into a region with a uniform magnetic field pointing into the page. Click the direction of the induced current in the loop as seen in the drawing, while the loop is entering the field.

B: into the pageentering the field

1clockwise
2counterclockwise

8.Magnetic fields & inductionHard

Two coils with self-inductances of 100 mH and 200 mH have a coupling coefficient of 0.8. The current in the first coil changes at 500 A/s. Find the emf ε induced in the second coil.

L₁ = 100 mHL₂ = 200 mHk = 0.8di/dt = 500 A/s

ε = V

9.AC impedance & phasorsMedium

A series RLC circuit with L = 20 mH must resonate at 50 Hz. Find the capacitance C needed.

f₀ = 50 HzL = 20 mH

C = µF

10.AC impedance & phasorsHard

A series RLC circuit is driven by a 12 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage V_L across the inductor.

V = 12 V (rms)R = 5 ΩL = 10 mHC = 2.2 µF

V_L = V

3Solutions

1.Coulomb’s law & electric fields

Answer: F = 7.336 N

Coulomb’s law: two point charges attract or repel with F = k·|q₁·q₂|/r², with k = 8.99 × 10⁹ N·m²/C². The field of a point charge is E = k·|q|/r², pointing away from a positive charge and towards a negative one. Forces and fields from several charges add as vectors.

Force from each charge

Right is positive.

F_L = +k·5·1 × 10⁻¹²/0.12² = 3.122 N

F_R = +k·3·1 × 10⁻¹²/0.08² = 4.214 N

Net force

F = 3.122 + (4.214) = 7.336 N → |F| = 7.336 N

Answer

F = 7.336 N

The net force points to the right.

2.Coulomb’s law & electric fields

Answer: F = 129.7 N

Each force on q₃

F₁ = k·5·5 × 10⁻¹²/0.08² = 35.12 N

F₂ = k·10·5 × 10⁻¹²/0.06² = 124.9 N

Vector sum

F = √(35.12² + 124.9²) = 129.7 N

Answer

F = 129.7 N

3.Series & parallel resistors

Answer: I_R1 = 1.466 A

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R₃ + R₄ = 5 + 4 = 9 Ω

R₂ ∥ 9 = 8·9/(8+9) = 4.235 Ω

R_eq = R₁ + 4.235 = 6 + 4.235 = 10.24 Ω

Current from the source

The source sees only R_eq, so Ohm’s law gives the current it delivers. That current flows through R₁.

I_R1 = V₁/R_eq = 15/10.24 = 1.466 A

Answer

I_R1 = 1.466 A

4.Series & parallel resistors

Answer: I_R1 = 1.817 A

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R₅ + R₆ = 8 + 30 = 38 Ω

R₄ ∥ 38 = 16.76 Ω

R₃ + 16.76 = 19.76 Ω

R₂ ∥ 19.76 = 2.605 Ω

R_eq = R₁ + 2.605 = 4 + 2.605 = 6.605 Ω

Current from the source

The source sees only R_eq, so Ohm’s law gives the current it delivers. That current flows through R₁.

I_R1 = V₁/R_eq = 12/6.605 = 1.817 A

Answer

I_R1 = 1.817 A

5.RC & RL transients

Answer: t = 9.242 ms

First-order transients: in a circuit with one capacitor or one inductor, every voltage and current moves exponentially from its initial value to its final value: x(t) = x(∞) + [x(0) − x(∞)]·e^−t/τ, with τ = RC or τ = L/R. After about 5τ the circuit has settled.

Time constant and final current

Long after the switch closes the inductor behaves like a short circuit, so the final current is set by R alone.

τ = L/R = 220 mH / 33 Ω = 6.667 ms

i(∞) = V_s/R = 12/33 = 0.3636 A

Solve for the time

0.75 = 1 − e^−t/τ → t = −τ·ln(1 − 0.75)

t = −6.667·ln(0.25) = 9.242 ms

Answer

t = 9.242 ms

6.RC & RL transients

Answer: v_C = 6.284 V

Initial capacitor voltage

In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.

v_C(0) = 24·3.3/(4.7 + 3.3) = 9.9 V

Time constant after the switch opens

τ = R₂·C = 3.3·10 = 33 ms

Discharge

v_C(t) = 9.9·e^−15/33 = 6.284 V

Answer

v_C = 6.284 V

7.Magnetic fields & induction

Answer: counterclockwise

Electromagnetic induction: the flux through a loop is Φ = BA·cos θ, with θ between B and the normal of the loop. Faraday's law: a change of flux induces emf = −N·dΦ/dt in a coil of N turns, and Lenz's law says the induced current flows so that its own field opposes the change. A rod moving across a field gives emf = BLv. A coil rotating at ω gives e = NBAω·sin ωt, peak NBAω. Self-inductance: emf = −L·di/dt, with L = μ₀N²A/l for a long solenoid, and the energy stored is W = ½LI². Two coupled coils: emf₂ = −M·di₁/dt with M = k√(L₁L₂).

Change of flux

The flux into the page increases, or the flux out of the page decreases.

Field of the induced current

The induced current opposes the change.

Induced field inside the loop: out of the page.

Direction of the current

Right-hand grip rule: a counterclockwise current gives a field out of the page inside the loop.

counterclockwise

Answer

counterclockwise

8.Magnetic fields & induction

Answer: ε = 56.57 V

Mutual inductance

M = 0.8·√(100·200) = 113.1 mH

Induced emf

ε = M·di/dt = 0.1131·500 = 56.57 V

Answer

ε = 56.57 V

9.AC impedance & phasors

Answer: C = 506.6 µF

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Resonance condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·50 = 314.2 rad/s

C = 1/(ω₀²·L) = 1/(314.2²·0.02) = 506.6 µF

Answer

C = 506.6 µF

10.AC impedance & phasors

Answer: V_L = 161.8 V

Resonance

At resonance the reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.01·2.2·10⁻⁶) = 6742 rad/s

I = V/R = 12/5 = 2.4 A

Voltage across the inductor

X_L = ω₀L = 6742·0.01 = 67.42 Ω

V_L = I·X_L = 2.4·67.42 = 161.8 V

Answer

V_L = 161.8 V

This is Q = 13.48 times the supply voltage.

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