A solenoid has a cross-sectional area of 5.8 x 10-4 m2, consists of 330 turns pe

Question

A solenoid has a cross-sectional area of 5.8 x 10-4 m2, consists of 330 turns per meter, and carries a current of 0.41 A. A 19-turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a 2.1- resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of 0.072 s. Find the average current induced in the coil.


having trouble following the other posts on here. please use my numbers and only post if you are confident. thank you!

Explanation / Answer

The e.m.f. induced in the secondary coil is given by
e = - M dI / dt
Current induced is
i = e / R
The coefficient of mutual inductance M for an air core solenoid is given by

M = o N1N2 A / L
where N1 and N2 are the no. of turns in the primary and secondary. A is their area of cross section. L is the length of the primary ( solenoid )

o = 4**(10^(-7))
N1 / L = 330
N2 = 19
A = 5.8*(10^(-4)) m^2
dI = 0.41 A
dt = 0.072 s
R = 2.1 ohm

M = 4.57 x 10^-6 H
e = 26.02 x10^-6 V
i = 12.392 x10^-6 A = 0.0124mA

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