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?

Explanation / Answer

Here ,

a)

dB/dt = u0*n*dI/dt/2

dB/dt = u0*1050 * 270 /2

dB/dt = 0.178 T/s

Now,

induced current = Area * dB/dt /R

induced current = pi*(0.03)^2 * 0.178 /3.85 *10^-4

induced current = 1.31 A

b)

Bin = u0*Iin/(2r1)

Bin = u0 * 1.31 /(2 * 0.05)

Bin = 1.64 *10^-5 T

the induced magnetic field is 1.64 *10^-5 T

c)

as the field due to solenoid is to the left and it is incresasing ,

Now, due to induced current ,

induced Magnetic field is to the right

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