An aluminum ring of radius r 1 = 5.00 cm r 2 = 3.00 cm as shown in the figure be

Question

An aluminum ring of radius

r1 = 5.00 cm

r2 = 3.00 cm

as shown in the figure below. Assume the axial component of the field produced by the solenoid is one-half as strong over the area of the end of the solenoid as at the center of the solenoid. Also assume the solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing at a rate of 270 A/s.

(a) What is the induced current in the ring?
____A

(b) At the center of the ring, what is the magnitude of the magnetic field produced by the induced current in the ring?
____µT

(c) At the center of the ring, what is the direction of the magnetic field produced by the induced current in the ring?to the left,to the right,upward, or downward

Explanation / Answer

given

r1 = 5 cm = 0.05 m

R = 3.35*10^-4 ohms

n = 1060 turns/m

r2 = 3 cm = 0.03 m

dI/dt = 270 A/s

a) magnetic field due to solenoid, B = mue*n*I

induced emf = A*dB/dt

= pi*r2^2*mue*n*dI/dt

= pi*0.03^2*4*pi*10^-7*1060*270

= 1.02*10^-3 volts

induced current = induced emf/R

= 1.02*10^-3/(3.35*10^-4)

= 3.04 A

b) B = mue*I/(2*pi*r2)

= 4*pi*10^-7*3.04/(2*pi*0.05)

= 1.22*10^-5 T

c) to the left (according to Lenz's law)

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