An aluminum ring of radius r_1 5.00_cm and a resistance of 3.45*10^4 Ohms is pla

Question

An aluminum ring of radius r_1 5.00_cm and a resistance of 3.45*10^4 Ohms is placed around one end of a long air-core solenoid with 1 040 turns per meter and radius 0-3.00 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. What is the induced current in the ring? At the center of the ring, what is the magnitude of the magnetic field produced by the induced current in the ring? At the center of the ring, what is the direction of the magnetic field produced by the induced current in the ring?

Explanation / Answer

The flux of magnetic filed passing through the ring would be:

= B0**r22   also the magnetic filed at edge os the solenoid (B0) is half of that at the center so B0 =  0*N*I0

where N is the number of turns per uint length and I0 is the current

(a) Hence using Faraday's law we have induced current I = (0*N**r22/R)*dI0/dt =

or I = 4*2*10-7*1040*9*10-4*270/3.45*10-4  = 2.89 A

(b) The magnetic field at the center of the ring is given by B0 = 0*I/2r1  = 4**10-7*2.89/2*0.05 = 3.63*10-5 T

(c) Using the Lenz law the direction of magnetic field produced by the induced current will be opposite to that produced by the solenoid at the given edge. Hence the direction would be to the left in this case.

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