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: −3 µC on the left, −10 µC on the right 20 cm away, and +5 µC between them, 8 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 = −3 µCq_R = −10 µCq_M = +5 µCr_L = 8 cmr_R = 12 cm

F = N

2.Coulomb’s law & electric fieldsHard

An electron enters the uniform field between two parallel plates moving parallel to them at 2.5 × 10⁷ m/s. The plates are 6 cm long and the field between them is 3800 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 = 2.5 × 10⁷ m/sL = 6 cmE = 3800 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 = 12 VR₁ = 4 ΩR₂ = 10 ΩR₃ = 15 ΩR₄ = 4 Ω

I_R1 = A

4.Series & parallel resistorsHard

Find the equivalent resistance R_eq seen by the source V₁.

V₁ = 36 VR₁ = 2 ΩR₂ = 2 ΩR₃ = 4 ΩR₄ = 3 ΩR₅ = 30 ΩR₆ = 3 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 = 6 VR = 22 ΩL = 10 mH

W(∞) = mJ

6.RC & RL transientsHard

The switch across R₂ has been closed for a long time, so the inductor current is steady. At t = 0 the switch opens. How long after t = 0 does the inductor current reach 1.34 A? Find t.

V_s = 20 VR₁ = 10 ΩR₂ = 47 ΩL = 100 mH

t = ms

7.Magnetic fields & inductionMedium

Two long parallel wires 6 cm apart carry currents of 2 A and 12 A in opposite directions. Find the magnitude of the magnetic flux density B at point P, halfway between them.

I₁ = 2 AI₂ = 12 Ad = 6 cmμ₀ = 4π × 10⁻⁷ T·m/A

B = µT

8.Magnetic fields & inductionHard

A coil of 150 turns, 10 cm × 5 cm, rotates at 1200 rpm in a uniform field of 0.1 T about an axis perpendicular to the field. Find the peak emf E₀.

N = 150A = 10 cm × 5 cmB = 0.1 Tn = 1200 rpm

E₀ = V

9.AC impedance & phasorsMedium

A resistor and a capacitor are connected in parallel across a 12 V (rms), 200 Hz supply. Find the rms current I.

V = 12 V (rms)f = 200 HzR = 30 ΩC = 10 µF

I = A

10.AC impedance & phasorsHard

A series RLC circuit is connected to a 100 V (rms), 200 Hz supply. Find the rms voltage V_C across the capacitor.

V = 100 V (rms)f = 200 HzR = 30 ΩL = 150 mHC = 10 µF

V_C = V

3Solutions

1.Coulomb’s law & electric fields

Answer: F = 10.14 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·3·5 × 10⁻¹²/0.08² = −21.07 N

F_R = +k·10·5 × 10⁻¹²/0.12² = 31.22 N

Net force

F = −21.07 + (31.22) = 10.14 N → |F| = 10.14 N

Answer

F = 10.14 N

The net force points to the right.

2.Coulomb’s law & electric fields

Answer: y = 1.925 mm

Time between the plates

t = L/v₀ = 0.06/2.5 × 10⁷ = 2.4 × 10⁻⁹ s

Sideways acceleration

a = e·E/m = 1.602 × 10⁻¹⁹·3800/9.109 × 10⁻³¹ = 6.683 × 10¹⁴ m/s²

Deflection

y = ½·a·t² = ½·6.683 × 10¹⁴·(2.4 × 10⁻⁹)² = 0.001925 m = 1.925 mm

Answer

y = 1.925 mm

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

3.Series & parallel resistors

Answer: I_R1 = 1.137 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 + 4 = 19 Ω

R₂ ∥ 19 = 10·19/(10+19) = 6.552 Ω

R_eq = R₁ + 6.552 = 4 + 6.552 = 10.55 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 = 12/10.55 = 1.137 A

Answer

I_R1 = 1.137 A

4.Series & parallel resistors

Answer: R_eq = 3.543 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 + 3 = 33 Ω

R₄ ∥ 33 = 2.75 Ω

R₃ + 2.75 = 6.75 Ω

R₂ ∥ 6.75 = 1.543 Ω

R_eq = R₁ + 1.543 = 2 + 1.543 = 3.543 Ω

Answer

R_eq = 3.543 Ω

5.RC & RL transients

Answer: W(∞) = 0.3719 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 = 10 mH / 22 Ω = 0.4545 ms

i(∞) = V_s/R = 6/22 = 0.2727 A

Stored energy

W(∞) = ½·L·i(∞)² = ½·10·0.2727² = 0.3719 mJ

Answer

W(∞) = 0.3719 mJ

6.RC & RL transients

Answer: t = 0.8968 ms

Initial value, final value, time constant

Before t = 0 the closed switch bypasses R₂; afterwards R₁ and R₂ are both in the loop.

i(0) = 20/10 = 2 A

i(∞) = 20/(10 + 47) = 0.3509 A

τ = L/R = 100/57 = 1.754 ms

i(t) = 0.3509 + (2 − 0.3509)·e^−t/τ

Solve for the time

e^−t/τ = (1.34 − 0.3509)/(2 − 0.3509) = 0.5998

t = 1.754·ln(1.667) = 0.8968 ms

Answer

t = 0.8968 ms

7.Magnetic fields & induction

Answer: B = 93.33 µT

Magnetic field of currents: around a long straight wire the field lines are circles (right-hand grip rule) with B = μ₀I/(2πr). At the centre of a flat circular coil of N turns, B = μ₀NI/(2R). Inside a long solenoid the field is uniform, B = μ₀nI with n = N/L turns per metre, whatever the radius. μ₀ = 4π × 10⁻⁷ T·m/A. Work in SI units (metres and amperes), and add the fields of several sources as vectors.

Field of each wire at P

B₁ = 2 × 10⁻⁷·2/0.03 = 13.33 µT

B₂ = 2 × 10⁻⁷·12/0.03 = 80 µT

Combine

B = B₁ + B₂ = 93.33 µT

Answer

B = 93.33 µT

8.Magnetic fields & induction

Answer: E₀ = 9.425 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).

Angular speed and area

ω = 2π·1200/60 = 125.7 rad/s

A = 0.005 m²

Peak emf

E₀ = NBAω = 150·0.1·0.005·125.7 = 9.425 V

Answer

E₀ = 9.425 V

9.AC impedance & phasors

Answer: I = 0.4275 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|.

Reactance

X_C = 1/(ωC) = 1/(2π·200·10·10⁻⁶) = 79.58 Ω

Branch currents

Both branches are connected straight across the supply, so each sees the full voltage V.

I_R = V/R = 12/30 = 0.4 A

I_X = V/X_C = 12/79.58 = 0.1508 A

Total current

The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.

I = √(I_R² + I_X²) = √(0.4² + 0.1508²) = 0.4275 A

Answer

I = 0.4275 A

10.AC impedance & phasors

Answer: V_C = 70.44 V

Reactances

ω = 2πf = 2π·200 = 1257 rad/s

X_L = ωL = 1257·0.15 = 188.5 Ω

X_C = 1/(ωC) = 1/(1257·10·10⁻⁶) = 79.58 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 + (108.9)²) = 113 Ω

Current

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

I = V/|Z| = 100/113 = 0.8852 A

Voltage across the capacitor

V_C = I·X_C = 0.8852·79.58 = 70.44 V

Answer

V_C = 70.44 V

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