An aluminum ring of radius r1 and resistance R is placed around one end of a lon

Question

An aluminum ring of radius r1 and resistance R is placed around one end of a long air-core solenoid with n turns per meter and smaller radius r2 as shown in the figure. Assume that the axial component of the field produced by the solenoid over the area of the end of the solenoid is one-half as strong as at the center of the solenoid. Also assume that the solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing at a rate of delta I/1t. (a) What is the induced current in the ring? (Use any variable or symbol stated above along with the following as necessary: mu 0 and pi.) (b) At the center of the ring, what is the magnetic field produced by the induced current in the ring? (Use any variable or symbol stated above along with the following as necessary: mu 0 and pi. Do not use I ring in your answer.) (c)What is the direction of this field?

Explanation / Answer

part a )

current = emf / resistance,

So we need to find emf in order to solve for current:

Emf = (B*A)d/dt (*where B=magnetic field, A=area)

B = (1/2)*muo*n*I (*where I=current, 1/2 because the field is half as strong form center to ends)

A = pi*(r2)^2

Plugging back into equation for emf we get:

Emf = ((1/2)*muo*n)*(pi*(r2)^2)) dI/dt   

And so

Current = ((1/2)*muo*n)*(pi*(r2)^2) / R dI/dt

part b )

B = (1/2)*(1/r1)*muo*n*I (n=1 for the ring) so we get:

B  = (1/2)*(1/r1)*muo*I

B = (muo)^2*n*pi*r^2 / 4r1*R

part c )

Bring point upwards

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