A very long, straight solenoid with a cross-sectional area of 1.90 cm2 is wound with 89.7 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.
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.
n = 89.7 turns/cm = 8970 turns/metre
The field inside the long solenoid is given by B = μ₀ni
B = 4πx10⁻⁷ x 8970 x 0.178t² = 2.0064x10⁻³ t²
dB/dt = 4.0128 x10⁻³ t
A = 1.90 cm² = 1.90 x10⁻⁴ m²
|Emf| = rate of change of flux linkage
|Emf| = d(NAB)/dt = NA dB/dt
= 5 x 1.9 x10⁻⁴ x 4.0128 x10⁻³ t
= 3.812x10⁻⁶ t
If T is the time at which the current = 3.2A
3.2 = 0.178T²
T = 4.24 s
|emf| = 3.812 x10⁻⁶ T
= 3.812 x x10⁻⁶ x4.24
= 1.62 x10⁻⁵ V
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