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 −3 µC. They are touched together and then put back 20 cm apart. Find the magnitude of the force F between them now. Use k = 8.99 × 10⁹ N·m²/C².

q₁ = −2 µCq₂ = −3 µCr = 20 cm

F = N

2.Coulomb’s law & electric fieldsHard

Three positive point charges sit at the corners of a right triangle: q₃ = 4 µC is at the right-angle corner, q₁ = 2 µC is 5 cm from it along one side and q₂ = 10 µC is 3 cm from it along the other side. Find the magnitude of the net electric force F on q₃.

q₁ = 2 µCq₂ = 10 µCq₃ = 4 µCr₁₃ = 5 cmr₂₃ = 3 cm

F = N

3.Series & parallel resistorsMedium

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 10 VR₁ = 4 ΩR₂ = 20 ΩR₃ = 6 ΩR₄ = 10 Ω

R_eq = Ω

4.Series & parallel resistorsHard

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 15 VR₁ = 5 ΩR₂ = 30 ΩR₃ = 3 ΩR₄ = 2 ΩR₅ = 3 ΩR₆ = 10 Ω

R_eq = Ω

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 = 9 VR = 33 ΩL = 220 mH

W(∞) = mJ

6.RC & RL transientsHard

The switch across R₂ has been open for a long time, so the inductor current is steady. At t = 0 the switch closes and shorts out R₂. Find the inductor current i at t = 1 ms.

V_s = 15 VR₁ = 33 ΩR₂ = 10 ΩL = 33 mHt = 1 ms

i = A

7.Magnetic fields & inductionMedium

A wire carries a current to the left through a uniform magnetic field pointing into the page. Click the direction of the magnetic force on the wire.

I: to the leftB: into the page

1up the page
2to the right
3down the page
4to the left

8.Magnetic fields & inductionHard

The current in a 5 mH inductor changes steadily from 0.5 A to 3.5 A in 10 ms. Find the magnitude of the self-induced emf ε.

L = 5 mH0.5 A → 3.5 AΔt = 10 ms

ε = V

9.AC impedance & phasorsMedium

A series RLC circuit is connected to a 230 V (rms), 1000 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 230 V (rms)f = 1000 HzR = 60 ΩL = 150 mHC = 47 µF

φ = °

10.AC impedance & phasorsHard

A series RLC circuit is connected to a 100 V (rms), 500 Hz supply. Find the rms voltage V_C across the capacitor.

V = 100 V (rms)f = 500 HzR = 30 ΩL = 20 mHC = 10 µF

V_C = V

3Solutions

1.Coulomb’s law & electric fields

Answer: F = 1.405 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 + −3)/2 = −2.5 µC

New force

F = 8.99 × 10⁹·(2.5 × 10⁻⁶)²/0.2² = 1.405 N

Answer

F = 1.405 N

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

2.Coulomb’s law & electric fields

Answer: F = 400.6 N

Each force on q₃

F₁ = k·2·4 × 10⁻¹²/0.05² = 28.77 N

F₂ = k·10·4 × 10⁻¹²/0.03² = 399.6 N

Vector sum

F = √(28.77² + 399.6²) = 400.6 N

Answer

F = 400.6 N

3.Series & parallel resistors

Answer: R_eq = 12.89 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 + 10 = 16 Ω

R₂ ∥ 16 = 20·16/(20+16) = 8.889 Ω

R_eq = R₁ + 8.889 = 4 + 8.889 = 12.89 Ω

Answer

R_eq = 12.89 Ω

4.Series & parallel resistors

Answer: R_eq = 9.088 Ω

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 = 3 + 10 = 13 Ω

R₄ ∥ 13 = 1.733 Ω

R₃ + 1.733 = 4.733 Ω

R₂ ∥ 4.733 = 4.088 Ω

R_eq = R₁ + 4.088 = 5 + 4.088 = 9.088 Ω

Answer

R_eq = 9.088 Ω

5.RC & RL transients

Answer: W(∞) = 8.182 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 = 220 mH / 33 Ω = 6.667 ms

i(∞) = V_s/R = 9/33 = 0.2727 A

Stored energy

W(∞) = ½·L·i(∞)² = ½·220·0.2727² = 8.182 mJ

Answer

W(∞) = 8.182 mJ

6.RC & RL transients

Answer: i = 0.4157 A

Initial value, final value, time constant

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

i(0) = 15/(33 + 10) = 0.3488 A

i(∞) = 15/33 = 0.4545 A

τ = L/R = 33/33 = 1 ms

i(t) = 0.4545 + (0.3488 − 0.4545)·e^−t/τ

Current at the given time

i(1) = 0.4545 + (−0.1057)·e^−1/1 = 0.4157 A

Answer

i = 0.4157 A

7.Magnetic fields & induction

Answer: down the page

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).

Right-hand rule

Point the fingers along the current or velocity and curl them towards B; the thumb gives the force on a positive charge.

I: to the left, B: into the page

Direction of the force

down the page

Answer

F: down the page

8.Magnetic fields & induction

Answer: ε = 1.5 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).

Self-induced emf

The emf opposes the change of current (Lenz), so its sign is opposite to ΔI; the size is L·|ΔI|/Δt.

ε = L·ΔI/Δt = 0.005·3/0.01 = 1.5 V

Answer

ε = 1.5 V

9.AC impedance & phasors

Answer: φ = 86.34 °

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·1000 = 6283 rad/s

X_L = ωL = 6283·0.15 = 942.5 Ω

X_C = 1/(ωC) = 1/(6283·47·10⁻⁶) = 3.386 Ω

Impedance

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

|Z| = √(R² + (X_L − X_C)²) = √(60² + (939.1)²) = 941 Ω

Phase angle

φ = arctan((X_L − X_C)/R) = arctan(939.1/60) = 86.34°

Answer

φ = 86.34°

10.AC impedance & phasors

Answer: V_C = 73.79 V

Reactances

ω = 2πf = 2π·500 = 3142 rad/s

X_L = ωL = 3142·0.02 = 62.83 Ω

X_C = 1/(ωC) = 1/(3142·10·10⁻⁶) = 31.83 Ω

Impedance

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

|Z| = √(R² + (X_L − X_C)²) = √(30² + (31)²) = 43.14 Ω

Current

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

I = V/|Z| = 100/43.14 = 2.318 A

Voltage across the capacitor

V_C = I·X_C = 2.318·31.83 = 73.79 V

Answer

V_C = 73.79 V

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