Question

A very long, straight solenoid with a cross-sectional area of 1.80 cm2 is wound with 88.7 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)= ( 0.179 A/s2 )t2. A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid.

Part A

What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.2 A ?

Express your answer with the appropriate units.

Answer 1

Solution-

Using formula-

e=N*A*(dB/dt)

n = 88.7 turns/cm = 8870 turns/metre

So the field inside the long solenoid is given by B = μ₀ni
B = 4*3.14x10^-7 x 8870 x 0.179^2 = 3.57 *10^-4 t²

dB/dt = 7.14x10^-4 t

A = 1.80cm^2 = 1.80x10^-4 m²

|Emf| = rate of change of flux linkage

|Emf| = d(NAB)/dt = NA dB/dt
= 5 x 1.80x10^-4 x 3.593x10^-3 t
= 3.23x10^-6 t  

If T is the time at which the current = 3.2A
3.2 = 0.179T^2

=> T = sqrt(3.2/0.179)

=4.23

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|emf| = 3.23x10^-6 T
= 3.23 x10^-6 x4.23
= 1.36x10^-5 V