Question

A very long, straight solenoid with a cross-sectional area of 2.06 cm2 is wound with 92.1 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to i(t)= ( 0.178 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.

Q1:

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 ?

Answer 1

the equation, e=N*(dB/dt) "

This should be e=N*A*(dB/dt)

n = 92.1 turns/cm = 9210 turns/metre

The field inside the long solenoid is given by B = μ₀ni
B = 4πx10⁻⁷ x 9210 x 0.178t² = 2.060x10⁻³ t²

dB/dt = 4.12x10⁻³ t
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A = 2.06cm² = 2.06x10⁻⁴ m²

|Emf| = rate of change of flux linkage

|Emf| = d(NAB)/dt = NA dB/dt
= 5 x 2.06x10⁻⁴ x 4.12x10⁻³ t
= 4.2436x10⁻⁶ t
_______________________________

If T is the time at which the current = 3.2A
3.2 = 0.178T²
T = 4.239s
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|emf| = 4.2436x10⁻⁶ T
= 4.2436 x x10⁻⁶ x4.239
= 1.798x10⁻⁵ V