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 +4 µC are 60 cm apart. Point P lies on the line between them, 15 cm from the first charge. Find the magnitude of the electric field E at P.

q₁ = +1 µCq₂ = +4 µCr₁ = 15 cmr₂ = 45 cm

E = N/C

2.Coulomb’s law & electric fieldsHard

Two identical small balls, each of mass 8 g, hang from the same point on insulating threads 20 cm long. They are given equal charges and each thread settles at 15° from the vertical. Use g = 9.81 m/s². Find the charge q on each ball in nC. Use k = 8.99 × 10⁹ N·m²/C².

m = 8 gL = 20 cmθ = 15°g = 9.81 m/s²

q = nC

3.Series & parallel resistorsMedium

Find the current I_R2 through R₂.

V₁ = 24 VR₁ = 10 ΩR₂ = 3 ΩR₃ = 15 ΩR₄ = 2 Ω

I_R2 = A

4.Series & parallel resistorsHard

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

V₁ = 12 VR₁ = 30 ΩR₂ = 25 ΩR₃ = 3 ΩR₄ = 12 ΩR₅ = 4 ΩR₆ = 30 Ω

I_R1 = 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 once the circuit has settled, in mJ.

V_s = 10 VR = 5 ΩL = 10 mH

W(∞) = mJ

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. How long after t = 0 does the inductor current reach 1.644 A? Find t.

V_s = 15 VR₁ = 5 ΩR₂ = 47 ΩL = 100 mH

t = ms

7.Magnetic fields & inductionMedium

A coil of 500 turns has an area of 10 cm². A uniform field perpendicular to the coil rises steadily from 0.05 T to 0.45 T in 50 ms. Find the average induced emf ε.

N = 500A = 10 cm²B: 0.05 T → 0.45 TΔt = 50 ms

ε = V

8.Magnetic fields & inductionHard

A coil of 20 turns, 8 cm × 25 cm, rotates in a uniform field of 0.05 T about an axis perpendicular to the field. At what speed n (in revolutions per minute) must it turn to give an rms emf of 2.666 V?

N = 20A = 8 cm × 25 cmB = 0.05 TE_rms = 2.666 V

n = rpm

9.AC impedance & phasorsMedium

A series RLC circuit with C = 0.1 µF must resonate at 200 Hz. Find the inductance L needed.

f₀ = 200 HzC = 0.1 µF

L = mH

10.AC impedance & phasorsHard

A resistor R₁ is connected in series with a parallel combination of R₂ and a capacitor C, across a 120 V (rms), 50 Hz supply. Find the magnitude of the impedance |Z|.

V = 120 V (rms)f = 50 HzR₁ = 20 ΩR₂ = 150 ΩC = 22 µF

|Z| = Ω

3Solutions

1.Coulomb’s law & electric fields

Answer: E = 221975 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 × 10⁻⁶/0.15² = 399556 N/C

E₂ = k·4 × 10⁻⁶/0.45² = 177580 N/C

Directions and sum

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

E = |399556 − 177580| = 221975 N/C

Answer

E = 221975 N/C

2.Coulomb’s law & electric fields

Answer: q = 158.3 nC

Electric force from the equilibrium

The tension's horizontal part balances F, its vertical part balances m·g.

F = m·g·tan θ = 0.008·9.81·tan 15° = 0.02103 N

Separation

r = 2·L·sin θ = 2·0.2·sin 15° = 0.1035 m

Charge

F = k·q²/r² → q = r·√(F/k) = 0.1035·√(0.02103/8.99 × 10⁹) = 1.583 × 10⁻⁷ C = 158.3 nC

Answer

q = 158.3 nC

3.Series & parallel resistors

Answer: I_R2 = 1.625 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 = 15 + 2 = 17 Ω

R₂ ∥ 17 = 3·17/(3+17) = 2.55 Ω

R_eq = R₁ + 2.55 = 10 + 2.55 = 12.55 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 = 24/12.55 = 1.912 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 = 24 − 1.912·10 = 4.876 V

I_R2 = V_A/R₂ = 4.876/3 = 1.625 A

Answer

I_R2 = 1.625 A

4.Series & parallel resistors

Answer: I_R1 = 0.3154 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 = 4 + 30 = 34 Ω

R₄ ∥ 34 = 8.87 Ω

R₃ + 8.87 = 11.87 Ω

R₂ ∥ 11.87 = 8.048 Ω

R_eq = R₁ + 8.048 = 30 + 8.048 = 38.05 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 = 12/38.05 = 0.3154 A

Answer

I_R1 = 0.3154 A

5.RC & RL transients

Answer: W(∞) = 20 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 = 10 mH / 5 Ω = 2 ms

i(∞) = V_s/R = 10/5 = 2 A

Stored energy

W(∞) = ½·L·i(∞)² = ½·10·2² = 20 mJ

Answer

W(∞) = 20 mJ

6.RC & RL transients

Answer: t = 1.333 ms

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) = 15/5 = 3 A

i(∞) = 15/(5 + 47) = 0.2885 A

τ = L/R = 100/52 = 1.923 ms

i(t) = 0.2885 + (3 − 0.2885)·e^−t/τ

Solve for the time

e^−t/τ = (1.644 − 0.2885)/(3 − 0.2885) = 0.4999

t = 1.923·ln(2) = 1.333 ms

Answer

t = 1.333 ms

7.Magnetic fields & induction

Answer: ε = 4 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 = (0.45 − 0.05)·0.001 = 0.0004 Wb

Average emf

ε = NΔΦ/Δt = 500·0.0004/0.05 = 4 V

Answer

ε = 4 V

8.Magnetic fields & induction

Answer: n = 1800 rpm

Peak emf

E₀ = √2·E_rms = 3.77 V

Angular speed

ω = E₀/(NBA) = 3.77/(20·0.05·0.02) = 188.5 rad/s

Revolutions per minute

n = 60·188.5/(2π) = 1800 rpm

Answer

n = 1800 rpm

9.AC impedance & phasors

Answer: L = 6333 mH

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·200 = 1257 rad/s

L = 1/(ω₀²·C) = 1/(1257²·0.1·10⁻⁶) = 6333 mH

Answer

L = 6333 mH

10.AC impedance & phasors

Answer: |Z| = 118.9 Ω

Reactance

X_C = 1/(2π·50·22·10⁻⁶) = 144.7 Ω

Parallel combination

Multiply top and bottom by the complex conjugate to split Z_p into real and imaginary parts.

Z_p = R₂·(−jX_C)/(R₂ − jX_C) = (R₂·X_C² − jR₂²·X_C)/(R₂² + X_C²)

Z_p = 72.3 − j74.95 Ω

Total impedance

Z = 20 + 72.3 − j74.95 = 92.3 − j74.95 Ω

|Z| = √(92.3² + 74.95²) = 118.9 Ω

Answer

|Z| = 118.9 Ω

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