A 0.200-m long solenoid has a radius of 0.075 m and 4.50 x 104 turn s. The curre

Question

A 0.200-m long solenoid has a radius of 0.075 m and 4.50 x 104 turn s. The current in the solenoid changes at a rate of 5.0 A/s. A conducting loop of radius 0.0300 m is placed at the center of the solenoid with its axis the same as that of the solenoid as shown. solenoid loop What is the magnetic flux through the small loop when the current through the solenoid is a. 3.50 A? b. What is the mutual inductance of this combination? c. What is the induced emf in the loop? d. What is the induced emf in the loop if the loop is oriented so that its axis is perpendicular to the axis of the solenoid, instead of parallel. What is the self-induced emf in the solenoid due to the changing current? e.

Explanation / Answer

given length of solenoid, l = 0.2 m

radius R = 0.075 m

N = 4.5*10^4 turns

dI/dt = 5 A/s

r = 0.03 m

a. magnetic field inside a solenoid = uo*N*I/l

hecne flux through the coil = (uo*N*I/l)*pi*r^2 = (4*pi*10^-7*4.5*10^4*I/0.2)*pi*0.03^2

phi = 7.94437*I *10^-4 T m^2

i = 3.5 A

phi = 2.79803284770 mT m^2

b. mutual inductance = M

now, M = -d(phi)/dt / (dI/dt) = 7.94437*10^-4 T m^2/A = 7.94437*10^-4 H

c. induced EMF = d(phi)/dt = 39.72185*10^-4 V

d. if axis of loop is perpendicular to the axis of solenoid, phi = 0

hence induced =emf = 0

e. self induced emf = d(N*phi)/dt = 1787.48325 V

self inductance L = self induced emf/dI/dt = 357.49665 H

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