A brief note on Eddy Currents and Inductance . Read it before a test. Try each question at the end before you open its answer.
Eddy currents What they are: currents that swirl inside a solid conductor when the flux through it changes. By Lenz's law they oppose the change. A moving conductor is braked, and the work done against the drag becomes heat. Magnetic brakes and damping use this. A disc of radius a a a and thickness t t t in a field changing at d B d t \dfrac{dB}{dt} d t d B loses P = π t a 4 8 ρ ( d B d t ) 2 P = \dfrac{\pi t a^4}{8\rho}\left(\dfrac{dB}{dt}\right)^2 P = 8 ρ π t a 4 ( d t d B ) 2 . Thin insulated sheets and slots break up the loops. In a sheet, P V = π 2 f 2 B m 2 d 2 6 ρ \dfrac{P}{V} = \dfrac{\pi^2 f^2 B_m^2 d^2}{6\rho} V P = 6 ρ π 2 f 2 B m 2 d 2 , and the loss goes as d 2 d^2 d 2 . Other uses: the induction furnace and cooker, and the metal detector. Self-inductance Definition: N ϕ = L I N\phi = LI N ϕ = L I . The unit is the henry, 1 H = 1 V s/A 1\ \text{H} = 1\ \text{V s/A} 1 H = 1 V s/A . Emf: ε = − L d I d t \varepsilon = -L\dfrac{dI}{dt} ε = − L d t d I . It opposes the change of the current, not the current. Long solenoid: L = μ r μ 0 N 2 A l L = \dfrac{\mu_r\mu_0 N^2 A}{l} L = l μ r μ 0 N 2 A . It goes as N 2 N^2 N 2 . Breaking a circuit quickly gives a large emf, and a spark. Series: L 1 + L 2 L_1 + L_2 L 1 + L 2 . Parallel: 1 L = 1 L 1 + 1 L 2 \dfrac{1}{L} = \dfrac{1}{L_1} + \dfrac{1}{L_2} L 1 = L 1 1 + L 2 1 . Mutual inductance and energy Mutual inductance: N 2 ϕ 2 = M I 1 N_2\phi_2 = MI_1 N 2 ϕ 2 = M I 1 and ε 2 = − M d I 1 d t \varepsilon_2 = -M\dfrac{dI_1}{dt} ε 2 = − M d t d I 1 . Coil inside a long solenoid: M = μ 0 n N 2 π r 2 2 M = \mu_0 nN_2\pi r_2^2 M = μ 0 n N 2 π r 2 2 . A wider coil gets only the flux of the solenoid. Coupling: M = k L 1 L 2 M = k\sqrt{L_1L_2} M = k L 1 L 2 with 0 ≤ k ≤ 1 0 \leq k \leq 1 0 ≤ k ≤ 1 . In series, L = L 1 + L 2 ± 2 M L = L_1 + L_2 \pm 2M L = L 1 + L 2 ± 2 M . Energy stored: U = 1 2 L I 2 U = \dfrac{1}{2}LI^2 U = 2 1 L I 2 . In the field: u = B 2 2 μ 0 u = \dfrac{B^2}{2\mu_0} u = 2 μ 0 B 2 in each cubic metre. A worked example A coil of 0.15 0.15 0.15 H carries 4.0 4.0 4.0 A. The current falls to zero steadily in 0.025 0.025 0.025 s. Find the energy stored and the emf.
The energy. U = 1 2 L I 2 = 1 2 × 0.15 H × ( 4.0 A ) 2 = 1.2 J U = \dfrac{1}{2}LI^2 = \dfrac{1}{2} \times 0.15\ \text{H} \times (4.0\ \text{A})^2 = 1.2\ \text{J} U = 2 1 L I 2 = 2 1 × 0.15 H × ( 4.0 A ) 2 = 1.2 J . The emf. ∣ ε ∣ = L Δ I Δ t = 0.15 H × 4.0 A 0.025 s = 24 V |\varepsilon| = L\dfrac{\Delta I}{\Delta t} = 0.15\ \text{H} \times \dfrac{4.0\ \text{A}}{0.025\ \text{s}} = 24\ \text{V} ∣ ε ∣ = L Δ t Δ I = 0.15 H × 0.025 s 4.0 A = 24 V . Practise with new numbers
Check yourself Sheets of a core are made half as thick. By what factor does the eddy-current loss change? Stuck? Show a hint Show the answerA coil has a steady current. What is the self-induced emf? Stuck? Show a hint Show the answerThe current in an inductor is made three times larger. By what factor does the stored energy change? Stuck? Show a hint Show the answer