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 −3 µC and −5 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 = −3 µCq₂ = −5 µCr = 40 cm

F = N

2.Coulomb’s law & electric fieldsHard

An electron enters the uniform field between two parallel plates moving parallel to them at 1.5 × 10⁷ m/s. The plates are 3 cm long and the field between them is 1100 N/C. Electron charge 1.602 × 10⁻¹⁹ C, mass 9.109 × 10⁻³¹ kg; ignore gravity. How far y is it deflected sideways when it leaves the plates? Give the answer in mm.

v₀ = 1.5 × 10⁷ m/sL = 3 cmE = 1100 N/Ce = 1.602 × 10⁻¹⁹ Cm_e = 9.109 × 10⁻³¹ kg

y = mm

3.Series & parallel resistorsMedium

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

V₁ = 10 VR₁ = 4 ΩR₂ = 12 ΩR₃ = 15 ΩR₄ = 6 Ω

I_R1 = A

4.Series & parallel resistorsHard

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

V₁ = 15 VR₁ = 4 ΩR₂ = 10 ΩR₃ = 8 ΩR₄ = 6 ΩR₅ = 8 ΩR₆ = 12 Ω

I_R1 = A

5.RC & RL transientsMedium

The source Vₛ charges the capacitor through R₁, and R₂ is connected in parallel with C. The capacitor is uncharged when the switch closes at t = 0. Find the time constant τ of the circuit after the switch closes.

V_s = 12 VR₁ = 10 kΩR₂ = 3.3 kΩC = 22 µF

τ = ms

6.RC & RL transientsHard

Just before the switch closes at t = 0, the capacitor holds 6 V with the polarity shown. Find the capacitor voltage v_C at t = 500 ms.

V_s = 20 VR = 10 kΩC = 100 µFv_C(0) = 6 Vt = 500 ms

v_C = V

7.Magnetic fields & inductionMedium

Two long parallel wires 20 cm apart carry currents of 20 A and 10 A in the same direction. Find the force per metre of length F/L between them.

I₁ = 20 AI₂ = 10 Ad = 20 cmμ₀ = 4π × 10⁻⁷ T·m/A

F/L = µN/m

8.Magnetic fields & inductionHard

A coil of 500 turns, 10 cm × 25 cm, rotates in a uniform field of 0.3 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 333.2 V?

N = 500A = 10 cm × 25 cmB = 0.3 TE_rms = 333.2 V

n = rpm

9.AC impedance & phasorsMedium

A series RLC circuit is connected to a 12 V (rms), 200 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 12 V (rms)f = 200 HzR = 25 ΩL = 200 mHC = 2.2 µF

φ = °

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), 50 Hz supply. Find the rms current I.

V = 24 V (rms)f = 50 HzR₁ = 25 ΩR₂ = 40 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′ = (−3 + −5)/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: y = 0.3869 mm

Time between the plates

t = L/v₀ = 0.03/1.5 × 10⁷ = 2 × 10⁻⁹ s

Sideways acceleration

a = e·E/m = 1.602 × 10⁻¹⁹·1100/9.109 × 10⁻³¹ = 1.935 × 10¹⁴ m/s²

Deflection

y = ½·a·t² = ½·1.935 × 10¹⁴·(2 × 10⁻⁹)² = 0.0003869 m = 0.3869 mm

Answer

y = 0.3869 mm

The electron is pushed towards the positive plate, against the field direction.

3.Series & parallel resistors

Answer: I_R1 = 0.8594 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 + 6 = 21 Ω

R₂ ∥ 21 = 12·21/(12+21) = 7.636 Ω

R_eq = R₁ + 7.636 = 4 + 7.636 = 11.64 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 = 10/11.64 = 0.8594 A

Answer

I_R1 = 0.8594 A

4.Series & parallel resistors

Answer: I_R1 = 1.566 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 + 12 = 20 Ω

R₄ ∥ 20 = 4.615 Ω

R₃ + 4.615 = 12.62 Ω

R₂ ∥ 12.62 = 5.578 Ω

R_eq = R₁ + 5.578 = 4 + 5.578 = 9.578 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 = 15/9.578 = 1.566 A

Answer

I_R1 = 1.566 A

5.RC & RL transients

Answer: τ = 54.59 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.

Thévenin equivalent seen by the capacitor

Short the source to find R_th; open the capacitor to find the voltage it will settle at.

R_th = R₁ ∥ R₂ = 10·3.3/(10 + 3.3) = 2.481 kΩ

v_C(∞) = 12·3.3/(10 + 3.3) = 2.977 V

Time constant

τ = R_th·C = 2.481·22 = 54.59 ms

Answer

τ = 54.59 ms

6.RC & RL transients

Answer: v_C = 11.51 V

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 = 10 kΩ · 100 µF = 1000 ms

v_C(t) = 20 + (6 − 20)·e^−t/τ

Voltage at the given time

v_C(500) = 20 + (−14)·e^−500/1000 = 11.51 V

Answer

v_C = 11.51 V

7.Magnetic fields & induction

Answer: F/L = 200 µN/m

Magnetic forces: a straight wire carrying I in a field B feels F = BIL·sin θ, and a moving charge F = |q|vB·sin θ, where θ is the angle between the current (or velocity) and B. The force is perpendicular to both: point the fingers of the right hand along I or v, curl them towards B, and the thumb gives F for a positive charge; reverse it for a negative charge. Because F is perpendicular to v, a charge moving across a uniform field moves on a circle: |q|vB = mv²/r gives r = mv/(|q|B) and the period T = 2πm/(|q|B), which does not depend on the speed. Two parallel wires attract for currents in the same direction: F/L = μ₀I₁I₂/(2πd).

Force per metre

F/L = μ₀I₁I₂/(2πd) = 2 × 10⁻⁷·20·10/0.2 = 200 µN/m

Answer

F/L = 200 µN/m

Currents in the same direction attract.

8.Magnetic fields & induction

Answer: n = 1200 rpm

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).

Peak emf

E₀ = √2·E_rms = 471.2 V

Angular speed

ω = E₀/(NBA) = 471.2/(500·0.3·0.025) = 125.7 rad/s

Revolutions per minute

n = 60·125.7/(2π) = 1200 rpm

Answer

n = 1200 rpm

9.AC impedance & phasors

Answer: φ = −77.24 °

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·200 = 1257 rad/s

X_L = ωL = 1257·0.2 = 251.3 Ω

X_C = 1/(ωC) = 1/(1257·2.2·10⁻⁶) = 361.7 Ω

Impedance

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

|Z| = √(R² + (X_L − X_C)²) = √(25² + (−110.4)²) = 113.2 Ω

Phase angle

φ = arctan((X_L − X_C)/R) = arctan(−110.4/25) = −77.24°

Answer

φ = −77.24°

10.AC impedance & phasors

Answer: I = 0.3809 A

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 = 37.16 − j10.27 Ω

Total impedance

Z = 25 + 37.16 − j10.27 = 62.16 − j10.27 Ω

|Z| = √(62.16² + 10.27²) = 63 Ω

Current

I = V/|Z| = 24/63 = 0.3809 A

Answer

I = 0.3809 A

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