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Unformatted text preview: €™s are b Îµ 1 = âˆ’ L1 + M di
g dt b and Îµ 2 = âˆ’ L2 + M di
g dt . Therefore, the total emf across both coils is b Îµ = Îµ 1 + Îµ 2 = âˆ’ 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 Îµ 1 = âˆ’ L1 âˆ’ M b di
g dt . b di
g dt . Similarly, the emf across coil 2 is Îµ 2 = âˆ’ L2 âˆ’ M
The total emf across both coils is b Îµ = âˆ’ L1 + L2 âˆ’ 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 â€“ 2M.
78. Taking the derivative of Eq. 3041, we have
di d âŽ¡ Îµ
Îµ âˆ’ t /Ï„ L Îµ âˆ’ t /Ï„ L
âŽ¤
e
= âŽ¢ (1 âˆ’ e âˆ’ t /Ï„ L ) âŽ¥ =
=e
.
dt dt âŽ£ R
L
âŽ¦ RÏ„ L 1195
With Ï„L = L/R (Eq. 3042), L = 0.023 H and Îµ = 12 V, t = 0.00015 s, and di/dt = 280 A/s,
t
we obtain eâˆ’ /Ï„L = 0.537. Taking the natural log and rearranging leads to R = 95.4 Î©.
79. (a) When switch S is just closed, V1 = Îµ and i1 = Îµ/R1 = 10 V/5.0 Î© = 2.0 A.
(b) Since now ÎµL = Îµ, we have i2 = 0.
(c) is = i1 + i2 = 2.0 A + 0 = 2.0 A.
(d) Since VL = Îµ, V2 = Îµ â€“ ÎµL = 0.
(e) VL = Îµ = 10 V.
di2 VL Îµ 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.
 Fall '08
 schuller
 Magnetism, Work

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