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

*Note: 3.452 is not the correct answer*

(b) At the center of the ring, what is the magnitude of the magnetic field produced by the induced current in the ring?
________ µT
*Note: 43.391 is not the correct answer*

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*920*9*10-4*270/2*3.55*10-4  = 1.24 A

(b) The magnetic field at the center of the ring is given by B0 = 0*I/2r1  = 4**10-7*1.24/2*0.05 = 4.859*10-5 T*(1 µT/10-6T)= 48.95 µT

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