A long solenoid with a radius of 8 cm has 3500 turns/m and a 30 A current. A sho

Question

A long solenoid with a radius of 8 cm has 3500 turns/m and a 30 A current. A short circular coil 20 turns and a radius of 15 cm is placed around the center of the solenoid. At t = 0 s, the current begins to decrease so that 0.5 s later, the current is zero, a) What is the initial magnetic field in the solenoid? b) What is the rate of change of the magnetic field, assuming that the current decreases at a constant rate? c) What is the induced voltage in the ring surrounding the solenoid? d) If the resistance of the ring is 0.35 ohm, what is the current in the ring? Which direction does it flow?

Explanation / Answer

a)

magnetic field B1 = uo*n*I

B1 = 4*pi*10^-7*3500*30

B1 = 0.132 T

(b)

rate of change = (B2-B1)/t = (0-0.132/0.5 = -0.264 T/s

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c)

magnetic flux = N*B*A

A = pi*r^2

r = radius of solenoid = 0.08 m

induced voltage = -rate of change in flux

induced voltage = - N*A*dB/dt

induced voltage = 20*pi*0.08^2*0.264 = 0.11 V

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(d)

induced current i = induced voltage /resistance

induced current i = 0.11/0.35 = 0.314 A

direction clockwise

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