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

Three point charges lie on a straight line: +10 µC on the left, −4 µC on the right 10 cm away, and +2 µC between them, 7.5 cm from the left charge. Find the magnitude of the net electric force F on the middle charge. Use k = 8.99 × 10⁹ N·m²/C².

q_L = +10 µCq_R = −4 µCq_M = +2 µCr_L = 7.5 cmr_R = 2.5 cm

F = N

2.Coulomb’s law & electric fieldsHard

Charges of +5 µC and +5 µC are 6 cm apart. Point P lies 3 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 = +5 µCq₂ = +5 µCd = 6 cmh = 3 cm

E = N/C

3.Series & parallel resistorsMedium

Find the current I_R2 through R₂.

V₁ = 15 VR₁ = 15 ΩR₂ = 10 ΩR₃ = 5 ΩR₄ = 15 Ω

I_R2 = A

4.Series & parallel resistorsHard

Find the current I_R2 through R₂.

V₁ = 24 VR₁ = 5 ΩR₂ = 4 ΩR₃ = 25 ΩR₄ = 25 ΩR₅ = 3 ΩR₆ = 2 Ω

I_R2 = 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 = 3.3 kΩR₂ = 4.7 kΩC = 47 µF

τ = ms

6.RC & RL transientsHard

Just before the switch closes at t = 0, the capacitor holds 30 V with the polarity shown. Find the capacitor voltage v_C at t = 80 ms.

V_s = 5 VR = 3.3 kΩC = 47 µFv_C(0) = 30 Vt = 80 ms

v_C = V

7.Magnetic fields & inductionMedium

A wire carries a current down the page through a uniform magnetic field pointing out of the page. Click the direction of the magnetic force on the wire.

I: down the pageB: out of the page

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

8.Magnetic fields & inductionHard

A short coil of 50 turns is wound tightly round the middle of a long solenoid of 2000 turns, length 40 cm and diameter 4 cm. The current in the solenoid changes at 100 A/s. Find the emf ε induced in the short coil.

N₁ = 2000N₂ = 50l = 40 cm⌀ = 4 cmdi/dt = 100 A/sμ₀ = 4π × 10⁻⁷ T·m/A

ε = mV

9.AC impedance & phasorsMedium

A series RLC circuit is connected to a 10 V (rms), 60 Hz supply. Find the rms current I.

V = 10 V (rms)f = 60 HzR = 40 ΩL = 100 mHC = 100 µF

I = A

10.AC impedance & phasorsHard

A series RLC circuit has R = 100 Ω, L = 50 mH and C = 0.022 µF. Find its bandwidth B (the width of the band between the half-power frequencies), in Hz.

R = 100 ΩL = 50 mHC = 0.022 µF

B = Hz

3Solutions

1.Coulomb’s law & electric fields

Answer: F = 147 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.

Force from each charge

Right is positive.

F_L = +k·10·2 × 10⁻¹²/0.075² = 31.96 N

F_R = +k·4·2 × 10⁻¹²/0.025² = 115.1 N

Net force

F = 31.96 + (115.1) = 147 N → |F| = 147 N

Answer

F = 147 N

The net force points to the right.

2.Coulomb’s law & electric fields

Answer: E = 3.532 × 10⁷ N/C

Distance and field of one charge

r = √(3² + 3²) = 4.243 cm

E₁ = k·5 × 10⁻⁶/0.04243² = 2.497 × 10⁷ N/C

Components

The components along the line joining the charges cancel.

E = 2·E₁·(h/r) = 2·2.497 × 10⁷·0.7071 = 3.532 × 10⁷ N/C

Answer

E = 3.532 × 10⁷ N/C

The field points along the bisector, away from the charges.

3.Series & parallel resistors

Answer: I_R2 = 0.4615 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 + 15 = 20 Ω

R₂ ∥ 20 = 10·20/(10+20) = 6.667 Ω

R_eq = R₁ + 6.667 = 15 + 6.667 = 21.67 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 = 15/21.67 = 0.6923 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 = 15 − 0.6923·15 = 4.615 V

I_R2 = V_A/R₂ = 4.615/10 = 0.4615 A

Answer

I_R2 = 0.4615 A

4.Series & parallel resistors

Answer: I_R2 = 2.478 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 = 3 + 2 = 5 Ω

R₄ ∥ 5 = 4.167 Ω

R₃ + 4.167 = 29.17 Ω

R₂ ∥ 29.17 = 3.518 Ω

R_eq = R₁ + 3.518 = 5 + 3.518 = 8.518 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 = 24/8.518 = 2.818 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 − 2.818·5 = 9.912 V

I_R2 = V_A/R₂ = 9.912/4 = 2.478 A

Answer

I_R2 = 2.478 A

5.RC & RL transients

Answer: τ = 91.12 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 = 3.3·4.7/(3.3 + 4.7) = 1.939 kΩ

v_C(∞) = 12·4.7/(3.3 + 4.7) = 7.05 V

Time constant

τ = R_th·C = 1.939·47 = 91.12 ms

Answer

τ = 91.12 ms

6.RC & RL transients

Answer: v_C = 19.93 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) = 30 V, v_C(∞) = V_s = 5 V

τ = R·C = 3.3 kΩ · 47 µF = 155.1 ms

v_C(t) = 5 + (30 − 5)·e^−t/τ

Voltage at the given time

v_C(80) = 5 + (25)·e^−80/155.1 = 19.93 V

Answer

v_C = 19.93 V

7.Magnetic fields & induction

Answer: to the left

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: down the page, B: out of the page

Direction of the force

to the left

Answer

F: to the left

8.Magnetic fields & induction

Answer: ε = 39.48 mV

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).

Mutual inductance

M = μ₀N₁N₂A/l = 4π × 10⁻⁷·2000·50·0.001257/0.4 = 394.8 µH

Induced emf

ε = M·di/dt = 0.0003948·100 = 39.48 mV

Answer

ε = 39.48 mV

9.AC impedance & phasors

Answer: I = 0.2408 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·60 = 377 rad/s

X_L = ωL = 377·0.1 = 37.7 Ω

X_C = 1/(ωC) = 1/(377·100·10⁻⁶) = 26.53 Ω

Impedance

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

|Z| = √(R² + (X_L − X_C)²) = √(40² + (11.17)²) = 41.53 Ω

Current

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

I = V/|Z| = 10/41.53 = 0.2408 A

Answer

I = 0.2408 A

10.AC impedance & phasors

Answer: B = 318.3 Hz

Resonant frequency

f₀ = 1/(2π·√(0.05·0.022·10⁻⁶)) = 4799 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/100)·√(0.05/(0.022·10⁻⁶)) = 15.08

Bandwidth

B = f₀/Q = 4799/15.08 = 318.3 Hz

Answer

B = 318.3 Hz

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