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

Two identical small metal spheres carry charges of −2 µC and +4 µC. They are touched together and then put back 5 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².

q₁ = −2 µCq₂ = +4 µCr = 5 cm

F = N

2.Coulomb’s law & electric fieldsHard

Charges of +4 µC and −4 µC are 6 cm apart. Point P lies 4 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 = +4 µCq₂ = −4 µCd = 6 cmh = 4 cm

E = N/C

3.Series & parallel resistorsMedium

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 10 VR₁ = 4 ΩR₂ = 15 ΩR₃ = 2 ΩR₄ = 15 Ω

R_eq = Ω

4.Series & parallel resistorsHard

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

V₁ = 9 VR₁ = 25 ΩR₂ = 8 ΩR₃ = 4 ΩR₄ = 5 ΩR₅ = 12 ΩR₆ = 25 Ω

I_R1 = A

5.RC & RL transientsMedium

The inductor carries no current before the switch closes at t = 0. Find the current i at t = 1 ms.

V_s = 24 VR = 47 ΩL = 10 mHt = 1 ms

i = A

6.RC & RL transientsHard

The switch across R₂ has been closed for a long time, so the inductor current is steady. At t = 0 the switch opens. Find the inductor current i at t = 1 ms.

V_s = 15 VR₁ = 15 ΩR₂ = 33 ΩL = 47 mHt = 1 ms

i = A

7.Magnetic fields & inductionMedium

A coil of 100 turns has an area of 150 cm². The coil is turned over (rotated through 180°) in a uniform field of 0.5 T, which is perpendicular to its plane, in 500 ms. Find the average induced emf ε.

N = 100A = 150 cm²B = 0.5 TΔt = 500 ms

ε = V

8.Magnetic fields & inductionHard

Two coils with self-inductances of 100 mH and 200 mH have a coupling coefficient of 0.6. The current in the first coil changes at 200 A/s. Find the emf ε induced in the second coil.

L₁ = 100 mHL₂ = 200 mHk = 0.6di/dt = 200 A/s

ε = V

9.AC impedance & phasorsMedium

A series RLC circuit is connected to a 230 V (rms), 500 Hz supply. Find the power factor cos φ of the circuit.

V = 230 V (rms)f = 500 HzR = 60 ΩL = 50 mHC = 47 µF

cos φ =

10.AC impedance & phasorsHard

A series RLC circuit is driven by a 5 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage V_L across the inductor.

V = 5 V (rms)R = 15 ΩL = 100 mHC = 1 µF

V_L = V

3Solutions

1.Coulomb’s law & electric fields

Answer: F = 3.596 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.

Charge after touching

q′ = (−2 + 4)/2 = 1 µC

New force

F = 8.99 × 10⁹·(1 × 10⁻⁶)²/0.05² = 3.596 N

Answer

F = 3.596 N

Both spheres now carry the same sign, so they repel.

2.Coulomb’s law & electric fields

Answer: E = 1.726 × 10⁷ N/C

Distance and field of one charge

r = √(3² + 4²) = 5 cm

E₁ = k·4 × 10⁻⁶/0.05² = 1.438 × 10⁷ N/C

Components

The components along the bisector cancel.

E = 2·E₁·(a/r) = 2·1.438 × 10⁷·0.6 = 1.726 × 10⁷ N/C

Answer

E = 1.726 × 10⁷ N/C

The field points parallel to the line from the positive towards the negative charge.

3.Series & parallel resistors

Answer: R_eq = 11.97 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 = 2 + 15 = 17 Ω

R₂ ∥ 17 = 15·17/(15+17) = 7.969 Ω

R_eq = R₁ + 7.969 = 4 + 7.969 = 11.97 Ω

Answer

R_eq = 11.97 Ω

4.Series & parallel resistors

Answer: I_R1 = 0.3093 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 = 12 + 25 = 37 Ω

R₄ ∥ 37 = 4.405 Ω

R₃ + 4.405 = 8.405 Ω

R₂ ∥ 8.405 = 4.099 Ω

R_eq = R₁ + 4.099 = 25 + 4.099 = 29.1 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 = 9/29.1 = 0.3093 A

Answer

I_R1 = 0.3093 A

5.RC & RL transients

Answer: i = 0.506 A

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 = 10 mH / 47 Ω = 0.2128 ms

i(∞) = V_s/R = 24/47 = 0.5106 A

Current at the given time

i(t) = i(∞)·(1 − e^−t/τ) = 0.5106·(1 − e^−1/0.2128) = 0.506 A

Answer

i = 0.506 A

6.RC & RL transients

Answer: i = 0.5601 A

Initial value, final value, time constant

Before t = 0 the closed switch bypasses R₂; afterwards R₁ and R₂ are both in the loop.

i(0) = 15/15 = 1 A

i(∞) = 15/(15 + 33) = 0.3125 A

τ = L/R = 47/48 = 0.9792 ms

i(t) = 0.3125 + (1 − 0.3125)·e^−t/τ

Current at the given time

i(1) = 0.3125 + (0.6875)·e^−1/0.9792 = 0.5601 A

Answer

i = 0.5601 A

7.Magnetic fields & induction

Answer: ε = 3 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 = 2·0.5·0.015 = 0.015 Wb

Average emf

ε = NΔΦ/Δt = 100·0.015/0.5 = 3 V

Answer

ε = 3 V

8.Magnetic fields & induction

Answer: ε = 16.97 V

Mutual inductance

M = 0.6·√(100·200) = 84.85 mH

Induced emf

ε = M·di/dt = 0.08485·200 = 16.97 V

Answer

ε = 16.97 V

9.AC impedance & phasors

Answer: cos φ = 0.3707

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|.

Reactances

X_L = 157.1 Ω, X_C = 6.773 Ω

Impedance

|Z| = √(60² + (150.3)²) = 161.8 Ω

Power factor

cos φ = R/|Z| = 60/161.8 = 0.3707

The circuit is inductive, so the power factor is lagging.

Answer

cos φ = 0.3707

10.AC impedance & phasors

Answer: V_L = 105.4 V

Resonance

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

ω₀ = 1/√(LC) = 1/√(0.1·1·10⁻⁶) = 3162 rad/s

I = V/R = 5/15 = 0.3333 A

Voltage across the inductor

X_L = ω₀L = 3162·0.1 = 316.2 Ω

V_L = I·X_L = 0.3333·316.2 = 105.4 V

Answer

V_L = 105.4 V

This is Q = 21.08 times the supply voltage.

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