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 +6 µC and a charge of −5 µC are 20 cm apart. Point P lies on the line between them, 12 cm from the first charge. Find the magnitude of the electric field E at P.

q₁ = +6 µCq₂ = −5 µCr₁ = 12 cmr₂ = 8 cm

E = N/C

2.Coulomb’s law & electric fieldsHard

An electron enters the uniform field between two parallel plates moving parallel to them at 4 × 10⁷ m/s. The plates are 6 cm long and the field between them is 3200 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 = 4 × 10⁷ m/sL = 6 cmE = 3200 N/Ce = 1.602 × 10⁻¹⁹ Cm_e = 9.109 × 10⁻³¹ kg

y = mm

3.Series & parallel resistorsMedium

Find the current I_R2 through R₂.

V₁ = 30 VR₁ = 5 ΩR₂ = 5 ΩR₃ = 3 ΩR₄ = 10 Ω

I_R2 = A

4.Series & parallel resistorsHard

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 36 VR₁ = 15 ΩR₂ = 4 ΩR₃ = 5 ΩR₄ = 25 ΩR₅ = 30 ΩR₆ = 4 Ω

R_eq = Ω

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 capacitor voltage v_C at t = 100 ms.

V_s = 15 VR₁ = 4.7 kΩR₂ = 1 kΩC = 100 µFt = 100 ms

v_C = V

6.RC & RL transientsHard

The switch has been closed for a long time, so the circuit is in steady state. At t = 0 the switch opens. Find the capacitor voltage v_C at t = 300 ms.

V_s = 20 VR₁ = 3.3 kΩR₂ = 6.8 kΩC = 68 µFt = 300 ms

v_C = V

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

A short coil of 100 turns is wound tightly round the middle of a long solenoid of 1000 turns, length 20 cm and diameter 2 cm. The current in the solenoid changes at 10 A/s. Find the emf ε induced in the short coil.

N₁ = 1000N₂ = 100l = 20 cm⌀ = 2 cmdi/dt = 10 A/sμ₀ = 4π × 10⁻⁷ T·m/A

ε = mV

9.AC impedance & phasorsMedium

A series RLC circuit with L = 10 mH must resonate at 1000 Hz. Find the capacitance C needed.

f₀ = 1000 HzL = 10 mH

C = µF

10.AC impedance & phasorsHard

A series RLC circuit is connected to a 230 V (rms), 100 Hz supply. Find the rms current I.

V = 230 V (rms)f = 100 HzR = 80 ΩL = 20 mHC = 10 µF

I = A

3Solutions

1.Coulomb’s law & electric fields

Answer: E = 1.077 × 10⁷ 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·6 × 10⁻⁶/0.12² = 3.746 × 10⁶ N/C

E₂ = k·5 × 10⁻⁶/0.08² = 7.023 × 10⁶ N/C

Directions and sum

The first field points away from the positive charge and the second towards the negative charge: both point the same way.

E = |3.746 × 10⁶ + 7.023 × 10⁶| = 1.077 × 10⁷ N/C

Answer

E = 1.077 × 10⁷ N/C

2.Coulomb’s law & electric fields

Answer: y = 0.6331 mm

Time between the plates

t = L/v₀ = 0.06/4 × 10⁷ = 1.5 × 10⁻⁹ s

Sideways acceleration

a = e·E/m = 1.602 × 10⁻¹⁹·3200/9.109 × 10⁻³¹ = 5.628 × 10¹⁴ m/s²

Deflection

y = ½·a·t² = ½·5.628 × 10¹⁴·(1.5 × 10⁻⁹)² = 0.0006331 m = 0.6331 mm

Answer

y = 0.6331 mm

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

3.Series & parallel resistors

Answer: I_R2 = 2.516 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 = 3 + 10 = 13 Ω

R₂ ∥ 13 = 5·13/(5+13) = 3.611 Ω

R_eq = R₁ + 3.611 = 5 + 3.611 = 8.611 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 = 30/8.611 = 3.484 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 = 30 − 3.484·5 = 12.58 V

I_R2 = V_A/R₂ = 12.58/5 = 2.516 A

Answer

I_R2 = 2.516 A

4.Series & parallel resistors

Answer: R_eq = 18.32 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 = 30 + 4 = 34 Ω

R₄ ∥ 34 = 14.41 Ω

R₃ + 14.41 = 19.41 Ω

R₂ ∥ 19.41 = 3.316 Ω

R_eq = R₁ + 3.316 = 15 + 3.316 = 18.32 Ω

Answer

R_eq = 18.32 Ω

5.RC & RL transients

Answer: v_C = 1.849 V

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 = 4.7·1/(4.7 + 1) = 0.8246 kΩ

v_C(∞) = 15·1/(4.7 + 1) = 2.632 V

Time constant

τ = R_th·C = 0.8246·100 = 82.46 ms

Charging curve

v_C(t) = 2.632·(1 − e^−100/82.46) = 1.849 V

Answer

v_C = 1.849 V

6.RC & RL transients

Answer: v_C = 7.038 V

Initial capacitor voltage

In steady state the capacitor is an open circuit, so R₁ and R₂ form a voltage divider. The capacitor voltage cannot jump when the switch opens.

v_C(0) = 20·6.8/(3.3 + 6.8) = 13.47 V

Time constant after the switch opens

τ = R₂·C = 6.8·68 = 462.4 ms

Discharge

v_C(t) = 13.47·e^−300/462.4 = 7.038 V

Answer

v_C = 7.038 V

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 = 1.974 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 × 10⁻⁷·1000·100·0.0003142/0.2 = 197.4 µH

Induced emf

ε = M·di/dt = 0.0001974·10 = 1.974 mV

Answer

ε = 1.974 mV

9.AC impedance & phasors

Answer: C = 2.533 µF

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·1000 = 6283 rad/s

C = 1/(ω₀²·L) = 1/(6283²·0.01) = 2.533 µF

Answer

C = 2.533 µF

10.AC impedance & phasors

Answer: I = 1.377 A

Reactances

ω = 2πf = 2π·100 = 628.3 rad/s

X_L = ωL = 628.3·0.02 = 12.57 Ω

X_C = 1/(ωC) = 1/(628.3·10·10⁻⁶) = 159.2 Ω

Impedance

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

|Z| = √(R² + (X_L − X_C)²) = √(80² + (−146.6)²) = 167 Ω

Current

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

I = V/|Z| = 230/167 = 1.377 A

Answer

I = 1.377 A

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