Question

A solenoid of lenght 35 cm has 380 turns of radius 2.6 cm. A tightly wound coil with 16 turns of radius 4.8 cm is placed around the solenoid. The axes of the coil and solenoid coincide. Find the emf induced in the coil if the current in the solenoid varies according to I=5.4sin(100πt) A.

Answer 1

oppose the change, Vcoil = -Ldi/dt, where L is the self
inductance of the coil. Similarly, if the time varying
magnetic flux of one coil links another coil, then it will
induce a voltage in the 2nd coil equal to V’coil = -MdI/dt
where I is the current in the 1st coil and M is the mutual
inductance of the two coils.

The magnetic field of the solenoid(away from the edges) is
given by: B = μoNI/L. The flux through the short coil is
φ = BA, where the area of the short coil is A.

Then, by Faraday’s law of induction:

emf: V12 = -N2 dφ/dt = -N2 AdB/dt = -(N2)A μo(N1)(dI/dt)/L
where,
N1 = 380, N2 = 16, L = 2.06 m, d = 0.048 m,
A = πd²/4 = π(0.048²)/4 = 1*10^-3 m²

μo = 4πx10^-7 weber/A-m, dI/dt = 4.7/0.75 = 6.267 A/s

V12 = -(4π*10^-7)(380)(16)(1x10^-3)(6.267)/0.2...

V12= 2.39*10^-3volts