# What is the induced current in the ring?

An aluminum ring of radius r1 and resistance R is placed around the top of a long air-core solenoid with n turns per meter and smaller radius r2 as shown in Figure P31.7. Assume that the axial component of the field produced by the solenoid over the area of the end of the solenoid is half as strong as at the center of the solenoid. Assume that the solenoid produces negligible field outside its cross-sectional area. The current in the solenoid is increasing at a rate of ΔI / Δt. Figure P31.7

(a) What is the induced current in the ring? (Use any variable or symbol stated above along with the following as necessary: μ0 and π.)

Iring =  ΔI / Δt

(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: μ0 and π. Do not use Iring in your answer.)

B =  ΔI / Δt

(c) What is the direction of this field?

Bring points downward. Bring points upward. Bring points to the left. Bring points to the right. Bring points out of the page. Bring points into the page. Bring has a magnitude of zero.

asked Feb 28, 2012 in Physics

Using Ohm's law we know that 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)*μ0*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)*μ0*n)*(pi*(r2)^2)) dI/dt   (*notice the I was moved to d/dt)

And so

Current = ((1/2)*μ0*n)*(pi*(r2)^2) / R  dI/dt

Part B:

B  = (1/2)*(1/r1)*μ0*n*I   (n=1 for the ring) so we get:

B  = (1/2)*(1/r1)*μ0*I    and we know I from part A, so plugging in I, we get:

Part C: The induced current and field will behave in such a way to oppose the change, so in this case it Points upward

answered Mar 3, 2012 by ~Expert~ (3,020 points)
edited Mar 8, 2012 by kirby

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