Faraday's Law — Induced EMF
V_emf = −N · dΦ/dt
N=turns, Φ=flux per turn (Wb), −ve sign = Lenz's law (opposes the change)
Tutorial 9 — All Problems
Tutorial 9 · Problem 1 — Faraday's Law (square coil)
Given: N=200 turns, square side=18cm, B changes from 0→0.50T in 0.80s. R=2Ω.
A = (0.18)² = 0.0324 m²
ΔΦ = A·ΔB = 0.0324 × 0.50 = 0.0162 Wb
V_emf = N·|ΔΦ/Δt| = 200 × 0.0162/0.80
V_emf = 4.05 V
(a) If open circuit: I = 0 (no path for current)
(b) If R=2Ω connected: I = V/R = 4.05/2 = 2.025 A
Tutorial 9 · Problem 2 — Lenz's Law direction (metal ring near solenoid)
Given: Metal ring near solenoid. Find induced current direction at three instants.
(a) Switch closes: Flux = 0 → increasing to left. Induced current must oppose increase → creates B pointing right → current flows counterclockwise viewed from right.
(b) After several seconds (steady state): Flux is constant → dΦ/dt = 0 → Induced current = ZERO.
(c) Switch opens: Flux decreasing from left. Induced current maintains flux → creates B pointing left → current flows clockwise viewed from right.
Tutorial 9 · Problem 3 — Solenoid inductance & self-induced EMF
Given: N=300 turns, l=25cm=0.25m, A=4cm²=4×10⁻⁴m², air core. di/dt = −50 A/s (decreasing).
(A) L = μ₀N²A/l = (4π×10⁻⁷ × 300² × 4×10⁻⁴) / 0.25
L = (4π×10⁻⁷ × 90000 × 4×10⁻⁴) / 0.25
L = 4.52×10⁻⁸ × 90000 / 0.25 = 0.181×10⁻³ H
(B) V_L = −L·(di/dt) = −0.181×10⁻³ × (−50) = +9.05×10⁻³ V
L = 0.181 mH | V_L = 9.05 mV (positive: opposes the decreasing current)
Tutorial 9 · Problem 4 — Inductor + bulb at different frequencies
Given: AC source + series inductor + light bulb. Frequency adjusted, voltage constant. When brightest?
At high frequency: Z_L = jωL → ∞ → inductor acts as open circuit → no current → bulb doesn't glow.
At low frequency: Z_L → 0 → inductor acts as short circuit → maximum current → bulb glows brightest.
Answer: (b) Low frequencies — inductor impedance is minimum at low ω
Tutorial 9 · Problem 5 — Inductor + capacitor + bulb
Given: AC source + series LC + light bulb. When brightest?
At high frequency: L = open, C = short → net open → no current.
At low frequency: L = short, C = open → net open → no current.
At resonance (ω₀ = 1/√LC): L and C impedances cancel → maximum current → brightest.
Answer: (b) Low frequencies (for this circuit configuration) — check which element dominates in the specific circuit
Tutorial 9 · Problem 6 — Find L_eq for parallel/series combination
Given: L₁=5H, L₂=15H, L₃=10H, L₄=20H, L₅=30H (see circuit for connections).
L_eq1 = L₂//L₃//L₄ = 1/(1/10 + 1/20 + 1/30) = 5.45 mH
L_eq = L₁ + L_eq1 = 5 + 5.45
L_eq = 10.45 mH
Tutorial 9 · Problem 7 — DC steady state with L and C
Given: V=5V DC, find R_eq and I at steady state.
DC steady state: L → short circuit (v_L = 0), C → open circuit (i_C = 0)
After removing C (open) and replacing L with wire: 0//4 = 0Ω
R_eq = 5Ω
R_eq = 5 Ω | I = V/R_eq = 5/5 = 1 A