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 )*t*2. 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

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

________________________________

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

_________________________________

|emf| = 4.2436x10⁻⁶ T

= 4.2436 x x10⁻⁶ x4.239

= 1.798x10⁻⁵ V

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