Homework 10 Solutions

I r r dt as the loop is crossing the boundary between

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Unformatted text preview: s are b ε 1 = − L1 + M di g dt b and ε 2 = − L2 + M di g dt . Therefore, the total emf across both coils is b ε = ε 1 + ε 2 = − L1 + L2 + 2 M di g dt which is exactly the emf that would be produced if the coils were replaced by a single coil with inductance Leq = L1 + L2 + 2M. (b) We imagine reversing the leads of coil 2 so the current enters at the back of coil rather than the front (as pictured in the diagram). Then the field produced by coil 2 at the site of coil 1 is opposite to the field produced by coil 1 itself. The fluxes have opposite signs. An increasing current in coil 1 tends to increase the flux in that coil, but an increasing current in coil 2 tends to decrease it. The emf across coil 1 is ε 1 = − L1 − M b di g dt . b di g dt . Similarly, the emf across coil 2 is ε 2 = − L2 − M The total emf across both coils is b ε = − L1 + L2 − 2 M di g dt . This is the same as the emf that would be produced by a single coil with inductance Leq = L1 + L2 – 2M. 78. Taking the derivative of Eq. 30-41, we have di d ⎡ ε ε − t /τ L ε − t /τ L ⎤ e = ⎢ (1 − e − t /τ L ) ⎥ = =e . dt dt ⎣ R L ⎦ Rτ L 1195 With τL = L/R (Eq. 30-42), L = 0.023 H and ε = 12 V, t = 0.00015 s, and di/dt = 280 A/s, t we obtain e− /τL = 0.537. Taking the natural log and rearranging leads to R = 95.4 Ω. 79. (a) When switch S is just closed, V1 = ε and i1 = ε/R1 = 10 V/5.0 Ω = 2.0 A. (b) Since now εL = ε, we have i2 = 0. (c) is = i1 + i2 = 2.0 A + 0 = 2.0 A. (d) Since VL = ε, V2 = ε – εL = 0. (e) VL = ε = 10 V. di2 VL ε 10 V = == = 2.0 A/s . dt L L 5.0 H (g) Af...
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This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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