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

A charge of +1 µC and a charge of +5 µC are 40 cm apart. Point P lies on the line between them, 10 cm from the first charge. Find the magnitude of the electric field E at P.

q₁ = +1 µCq₂ = +5 µCr₁ = 10 cmr₂ = 30 cm

E = N/C

2.Coulomb’s law & electric fieldsHard

Charges of +8 µC and +8 µC are 6 cm apart. Point P lies 6 cm from the midpoint between them, on the perpendicular bisector. Find the magnitude of the electric field E at P. Use k = 8.99 × 10⁹ N·m²/C².

q₁ = +8 µCq₂ = +8 µCd = 6 cmh = 6 cm

E = N/C

3.Series & parallel resistorsMedium

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 18 VR₁ = 20 ΩR₂ = 15 ΩR₃ = 6 ΩR₄ = 6 Ω

R_eq = Ω

4.Series & parallel resistorsHard

Find the current I_R2 through R₂.

V₁ = 6 VR₁ = 5 ΩR₂ = 6 ΩR₃ = 5 ΩR₄ = 12 ΩR₅ = 8 ΩR₆ = 3 Ω

I_R2 = A

5.RC & RL transientsMedium

The inductor carries no current before the switch closes at t = 0. Find the energy W stored in the inductor at t = 1 ms, in mJ.

V_s = 20 VR = 15 ΩL = 22 mHt = 1 ms

W = mJ

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 = 400 ms.

V_s = 24 VR₁ = 1 kΩR₂ = 4.7 kΩC = 220 µFt = 400 ms

v_C = V

7.Magnetic fields & inductionMedium

A coil of 250 turns has an area of 40 cm². The coil sits in a uniform field of 0.8 T perpendicular to its plane, and the field is switched off in 250 ms. Find the average induced emf ε.

N = 250A = 40 cm²B = 0.8 TΔt = 250 ms

ε = V

8.Magnetic fields & inductionHard

A short coil of 100 turns is wound tightly round the middle of a long solenoid of 1500 turns, length 40 cm and diameter 3 cm. The current in the solenoid changes at 200 A/s. Find the emf ε induced in the short coil.

N₁ = 1500N₂ = 100l = 40 cm⌀ = 3 cmdi/dt = 200 A/sμ₀ = 4π × 10⁻⁷ T·m/A

ε = mV

9.AC impedance & phasorsMedium

A resistor and a capacitor are connected in parallel across a 24 V (rms), 100 Hz supply. Find the magnitude of the impedance |Z|.

V = 24 V (rms)f = 100 HzR = 20 ΩC = 22 µF

|Z| = Ω

10.AC impedance & phasorsHard

A series RLC circuit has R = 5 Ω, L = 10 mH and C = 0.047 µF. Find its bandwidth B (the width of the band between the half-power frequencies), in Hz.

R = 5 ΩL = 10 mHC = 0.047 µF

B = Hz

3Solutions

1.Coulomb’s law & electric fields

Answer: E = 399556 N/C

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.

Field of each charge

E₁ = k·1 × 10⁻⁶/0.1² = 899000 N/C

E₂ = k·5 × 10⁻⁶/0.3² = 499444 N/C

Directions and sum

Both charges are positive: at P their fields point away from each charge, so in opposite directions.

E = |899000 − 499444| = 399556 N/C

Answer

E = 399556 N/C

2.Coulomb’s law & electric fields

Answer: E = 2.859 × 10⁷ N/C

Distance and field of one charge

r = √(3² + 6²) = 6.708 cm

E₁ = k·8 × 10⁻⁶/0.06708² = 1.598 × 10⁷ N/C

Components

The components along the line joining the charges cancel.

E = 2·E₁·(h/r) = 2·1.598 × 10⁷·0.8944 = 2.859 × 10⁷ N/C

Answer

E = 2.859 × 10⁷ N/C

The field points along the bisector, away from the charges.

3.Series & parallel resistors

Answer: R_eq = 26.67 Ω

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 = 6 + 6 = 12 Ω

R₂ ∥ 12 = 15·12/(15+12) = 6.667 Ω

R_eq = R₁ + 6.667 = 20 + 6.667 = 26.67 Ω

Answer

R_eq = 26.67 Ω

4.Series & parallel resistors

Answer: I_R2 = 0.435 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 + 3 = 11 Ω

R₄ ∥ 11 = 5.739 Ω

R₃ + 5.739 = 10.74 Ω

R₂ ∥ 10.74 = 3.849 Ω

R_eq = R₁ + 3.849 = 5 + 3.849 = 8.849 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 = 6/8.849 = 0.678 A

Current through R₂

Subtract the drop across R₁ to get the voltage at node A, which is the voltage across R₂.

V_A = V₁ − I_R1·R₁ = 6 − 0.678·5 = 2.61 V

I_R2 = V_A/R₂ = 2.61/6 = 0.435 A

Answer

I_R2 = 0.435 A

5.RC & RL transients

Answer: W = 4.778 mJ

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 = 22 mH / 15 Ω = 1.467 ms

i(∞) = V_s/R = 20/15 = 1.333 A

Current at the given time

i(t) = 1.333·(1 − e^−1/1.467) = 0.6591 A

Stored energy

W = ½·L·i² = ½·22·0.6591² = 4.778 mJ

Answer

W = 4.778 mJ

6.RC & RL transients

Answer: v_C = 13.44 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) = 24·4.7/(1 + 4.7) = 19.79 V

Time constant after the switch opens

τ = R₂·C = 4.7·220 = 1034 ms

Discharge

v_C(t) = 19.79·e^−400/1034 = 13.44 V

Answer

v_C = 13.44 V

7.Magnetic fields & induction

Answer: ε = 3.2 V

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 per turn

ΔΦ = 0.8·0.004 = 0.0032 Wb

Average emf

ε = NΔΦ/Δt = 250·0.0032/0.25 = 3.2 V

Answer

ε = 3.2 V

8.Magnetic fields & induction

Answer: ε = 66.62 mV

Mutual inductance

M = μ₀N₁N₂A/l = 4π × 10⁻⁷·1500·100·0.0007069/0.4 = 333.1 µH

Induced emf

ε = M·di/dt = 0.0003331·200 = 66.62 mV

Answer

ε = 66.62 mV

9.AC impedance & phasors

Answer: |Z| = 19.28 Ω

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|.

Reactance

X_C = 1/(ωC) = 1/(2π·100·22·10⁻⁶) = 72.34 Ω

Branch currents

Both branches are connected straight across the supply, so each sees the full voltage V.

I_R = V/R = 24/20 = 1.2 A

I_X = V/X_C = 24/72.34 = 0.3318 A

Total current

The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.

I = √(I_R² + I_X²) = √(1.2² + 0.3318²) = 1.245 A

Impedance

|Z| = V/I = 24/1.245 = 19.28 Ω

Answer

|Z| = 19.28 Ω

10.AC impedance & phasors

Answer: B = 79.58 Hz

Resonant frequency

f₀ = 1/(2π·√(0.01·0.047·10⁻⁶)) = 7341 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/5)·√(0.01/(0.047·10⁻⁶)) = 92.25

Bandwidth

B = f₀/Q = 7341/92.25 = 79.58 Hz

Answer

B = 79.58 Hz

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