Question

A 3.91 m -long coil containing 225 loops is wound on an iron core (average ?=1850?0) along with a second coil of 115 loops. The loops of each coil have a radius of 2.90 cm . If the current in the first coil drops uniformly from 12.0 A to zero in 99.0 ms

Determine the mutual inductance M.

Determine the emf induced in the second coil.

Answer 1

magnetic field produce by first coil B1 = u*(N1/L1)*I

magnetic flux through second coil = N2*B1*A2 = N2*u*(N1/L1)*I*pi*r2^2

magnetic flux through second coil = (u*N1*N2*I*pi*r^2)L1

magnetic flux through second coil = M*I

(u*N1*N2*I*pi*r^2)/L1 = M*I

(u*N1*N2*pi*r^2)/L1 = M

u = 1850*uo = 1850*4*pi*10^-7

mutual inductance M = 1850*4*pi*10^-7*225*115*pi*(2.9*10^-2)^2/3.91 = 0.0406 H

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induced emf = rate of change in flux

emf , e = -M*dI/dt

dI = I2 - I1 = 0- 12 = -12

dt = 99 ms = 0.099 s

e = 0.0406*12/0.099 =s