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 −10 µC and +2 µC. They are touched together and then put back 40 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².

q₁ = −10 µCq₂ = +2 µCr = 40 cm

F = N

2.Coulomb’s law & electric fieldsHard

Three positive point charges sit at the corners of a right triangle: q₃ = 8 µC is at the right-angle corner, q₁ = 1 µC is 4 cm from it along one side and q₂ = 1 µC is 10 cm from it along the other side. Find the magnitude of the net electric force F on q₃.

q₁ = 1 µCq₂ = 1 µCq₃ = 8 µCr₁₃ = 4 cmr₂₃ = 10 cm

F = N

3.Series & parallel resistorsMedium

Find the current I_R2 through R₂.

V₁ = 18 VR₁ = 3 ΩR₂ = 4 ΩR₃ = 5 ΩR₄ = 8 Ω

I_R2 = A

4.Series & parallel resistorsHard

Find the current I_R2 through R₂.

V₁ = 18 VR₁ = 3 ΩR₂ = 5 ΩR₃ = 5 ΩR₄ = 6 ΩR₅ = 8 ΩR₆ = 6 Ω

I_R2 = A

5.RC & RL transientsMedium

The inductor carries no current before the switch closes at t = 0. Find the current i at t = 5 ms.

V_s = 6 VR = 22 ΩL = 47 mHt = 5 ms

i = A

6.RC & RL transientsHard

Just before the switch closes at t = 0, the capacitor holds 6 V with the polarity shown. How long does it take for v_C to reach 17.2 V? Find t.

V_s = 20 VR = 3.3 kΩC = 68 µFv_C(0) = 6 V

t = ms

7.Magnetic fields & inductionMedium

A coil of 250 turns has an area of 200 cm². The coil sits in a uniform field of 0.3 T perpendicular to its plane, and the field is switched off in 10 ms. Find the average induced emf ε.

N = 250A = 200 cm²B = 0.3 TΔt = 10 ms

ε = V

8.Magnetic fields & inductionHard

A coil of 20 turns, 5 cm × 10 cm, rotates at 600 rpm in a uniform field of 0.1 T about an axis perpendicular to the field. Find the peak emf E₀.

N = 20A = 5 cm × 10 cmB = 0.1 Tn = 600 rpm

E₀ = V

9.AC impedance & phasorsMedium

A series RLC circuit is connected to a 50 V (rms), 500 Hz supply. Find the rms current I.

V = 50 V (rms)f = 500 HzR = 100 ΩL = 150 mHC = 100 µF

I = A

10.AC impedance & phasorsHard

A resistor R₁ is connected in series with a parallel combination of R₂ and a capacitor C, across a 24 V (rms), 60 Hz supply. Find the rms current I.

V = 24 V (rms)f = 60 HzR₁ = 25 ΩR₂ = 100 ΩC = 22 µF

I = A

3Solutions

1.Coulomb’s law & electric fields

Answer: F = 0.899 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′ = (−10 + 2)/2 = −4 µC

New force

F = 8.99 × 10⁹·(4 × 10⁻⁶)²/0.4² = 0.899 N

Answer

F = 0.899 N

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

2.Coulomb’s law & electric fields

Answer: F = 45.52 N

Each force on q₃

F₁ = k·1·8 × 10⁻¹²/0.04² = 44.95 N

F₂ = k·1·8 × 10⁻¹²/0.1² = 7.192 N

Vector sum

F = √(44.95² + 7.192²) = 45.52 N

Answer

F = 45.52 N

3.Series & parallel resistors

Answer: I_R2 = 2.272 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 + 8 = 13 Ω

R₂ ∥ 13 = 4·13/(4+13) = 3.059 Ω

R_eq = R₁ + 3.059 = 3 + 3.059 = 6.059 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 = 18/6.059 = 2.971 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 = 18 − 2.971·3 = 9.087 V

I_R2 = V_A/R₂ = 9.087/4 = 2.272 A

Answer

I_R2 = 2.272 A

4.Series & parallel resistors

Answer: I_R2 = 1.869 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 + 6 = 14 Ω

R₄ ∥ 14 = 4.2 Ω

R₃ + 4.2 = 9.2 Ω

R₂ ∥ 9.2 = 3.239 Ω

R_eq = R₁ + 3.239 = 3 + 3.239 = 6.239 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 = 18/6.239 = 2.885 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 = 18 − 2.885·3 = 9.345 V

I_R2 = V_A/R₂ = 9.345/5 = 1.869 A

Answer

I_R2 = 1.869 A

5.RC & RL transients

Answer: i = 0.2465 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 = 47 mH / 22 Ω = 2.136 ms

i(∞) = V_s/R = 6/22 = 0.2727 A

Current at the given time

i(t) = i(∞)·(1 − e^−t/τ) = 0.2727·(1 − e^−5/2.136) = 0.2465 A

Answer

i = 0.2465 A

6.RC & RL transients

Answer: t = 361.2 ms

Initial value, final value, time constant

The capacitor voltage cannot jump, so it starts at its stored value and heads for the source voltage.

v_C(0) = 6 V, v_C(∞) = V_s = 20 V

τ = R·C = 3.3 kΩ · 68 µF = 224.4 ms

v_C(t) = 20 + (6 − 20)·e^−t/τ

Solve for the time

17.2 = 20 + (−14)·e^−t/τ → e^−t/τ = (17.2 − 20)/(6 − 20) = 0.2

t = τ·ln((6 − 20)/(17.2 − 20)) = 224.4·ln(5) = 361.2 ms

Answer

t = 361.2 ms

7.Magnetic fields & induction

Answer: ε = 150 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 = 0.3·0.02 = 0.006 Wb

Average emf

ε = NΔΦ/Δt = 250·0.006/0.01 = 150 V

Answer

ε = 150 V

8.Magnetic fields & induction

Answer: E₀ = 0.6283 V

Angular speed and area

ω = 2π·600/60 = 62.83 rad/s

A = 0.005 m²

Peak emf

E₀ = NBAω = 20·0.1·0.005·62.83 = 0.6283 V

Answer

E₀ = 0.6283 V

9.AC impedance & phasors

Answer: I = 0.1045 A

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

ω = 2πf = 2π·500 = 3142 rad/s

X_L = ωL = 3142·0.15 = 471.2 Ω

X_C = 1/(ωC) = 1/(3142·100·10⁻⁶) = 3.183 Ω

Impedance

In series the impedances add as complex numbers: Z = R + j(X_L − X_C).

|Z| = √(R² + (X_L − X_C)²) = √(100² + (468.1)²) = 478.6 Ω

Current

The same current flows through every element of a series circuit.

I = V/|Z| = 50/478.6 = 0.1045 A

Answer

I = 0.1045 A

10.AC impedance & phasors

Answer: I = 0.2461 A

Reactance

X_C = 1/(2π·60·22·10⁻⁶) = 120.6 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 = 59.25 − j49.14 Ω

Total impedance

Z = 25 + 59.25 − j49.14 = 84.25 − j49.14 Ω

|Z| = √(84.25² + 49.14²) = 97.53 Ω

Current

I = V/|Z| = 24/97.53 = 0.2461 A

Answer

I = 0.2461 A

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