PHYS7021 · Electricity, Magnetism & Wave Optics

1. First law: whenever the magnetic flux linked with a circuit changes, an emf is induced in it; it lasts only while the change lasts.
2. Second law: the magnitude of the induced emf equals the rate of change of flux linkage: e = −dΦ/dt (for N turns: e = −N dΦ/dt).
3. Explanation/experiment: moving a magnet towards a coil increases Φ and a galvanometer deflects; faster motion ⇒ larger deflection; the minus sign is Lenz's law — the induced current opposes the change, which is just conservation of energy.

∴ e = −dΦ/dt (Lenz gives the direction)

1. Induced emf around a loop = ∮E·dl; flux Φ = ∫B·dA.
2. Integral form: ∮E·dl = −d/dt ∫B·dA
3. Applying Stokes' theorem ⇒ differential form: ∇×E = −∂B/∂t (a changing B produces a curling E).

∴ ∮E·dl = −dΦ/dt ; ∇×E = −∂B/t

1. A coil made by folding the wire back on itself and winding the doubled wire, so that at every turn the current in the two strands flows in opposite directions.
2. Their magnetic fluxes cancel ⇒ net flux ≈ 0 ⇒ self-inductance L ≈ 0.
3. Use: resistance boxes and standard resistors, where the coil must introduce no inductive reactance (no back-emf, no phase shift).

∴ Double-wound coil with L ≈ 0 (no induction)

1. Self induction: the property of a coil by which it opposes any change in its own current; induced emf e = −L dI/dt, where L = NΦ/I.
2. When current rises, the induced emf opposes the rise; when current falls, it opposes the fall.
3. Analogy: mass (inertia) opposes change of velocity in mechanics; inductance opposes change of current in a circuit ⇒ "electrical inertia".

∴ L opposes change of current, like mass opposes change of velocity

1. Self inductance L: flux linkages per unit current, L = NΦ/I. Numerically, L equals the emf induced when dI/dt = 1 A/s. Unit: henry (H).
2. Mutual inductance M: flux linked in the secondary per unit current in the primary, M = N₂Φ₂/I₁; equals the emf induced in the secondary when the primary current changes at 1 A/s. Unit: henry.

∴ L = NΦ/I ; M = N₂Φ₂/I₁ (both in henry)

Given: N₁ = 800, L₁ = 400 mH, N₂ = 500 (same geometry).

1. For identical geometry L = μ₀n²Al ⇒ L ∝ N².
2. L₂ = L₁ × (N₂/N₁)² = 400 × (500/800)² = 400 × 0.3906.
3. = 156.25 mH.

∴ L₂ = 156.25 mH

Given: l = 1 m, r = 5 cm, N = 5000 ⇒ n = 5000 m⁻¹; A = π(0.05)² = 7.854×10⁻³ m².

1. (i) L = μ₀n²Al = 4π×10⁻⁷ × (5000)² × 7.854×10⁻³ × 1.
2. = 4π×10⁻⁷ × 2.5×10⁷ × 7.854×10⁻³ ≈ 0.247 H.
3. (ii) U = ½LI² = 0.5 × 0.247 × (2)² ≈ 0.49 J.

∴ (i) L ≈ 0.247 H (ii) U ≈ 0.49 J

Given: l = 0.4 m ⇒ n = 200/0.4 = 500 m⁻¹; A = π(0.04)² = 5.027×10⁻³ m².

1. L = μ₀n²Al = 4π×10⁻⁷ × (500)² × 5.027×10⁻³ × 0.4.
2. = 4π×10⁻⁷ × 2.5×10⁵ × 5.027×10⁻³ × 0.4 ≈ 6.3×10⁻⁴ H.

∴ L ≈ 6.3×10⁻⁴ H ≈ 0.63 mH

1. Same voltage V across both coils: V = L₁(dI₁/dt) + M(dI₂/dt) and V = L₂(dI₂/dt) + M(dI₁/dt).
2. Total current I = I₁ + I₂; solve the two equations for dI₁/dt and dI₂/dt.
3. Adding: dI/dt = V(L₁+L₂−2M)/(L₁L₂−M²).
4. Comparing with V = L_eq(dI/dt) ⇒ L_eq = (L₁L₂−M²)/(L₁+L₂−2M).

∴ L_eq = (L₁L₂ − M²)/(L₁ + L₂ − 2M) (proved)

Given: L₁ = 50, L₂ = 100, L_eq = 75 mH.

1. Since 75 < L₁+L₂ = 150, the fluxes oppose (series-opposing): L_eq = L₁ + L₂ − 2M.
2. 75 = 150 − 2M ⇒ 2M = 75.
3. M = 37.5 mH.

∴ M = 37.5 mH

1. While building up the current, the source must do work against the back-emf: dW = (L i) di.
2. W = ∫₀ⁱ L i di = L i²/2.
3. This work is stored in the magnetic field: U = ½Li² (energy density = B²/2μ₀).

∴ U = ½Li²