Induced current

1. Apr 10, 2006

Punchlinegirl

An aluminum ring of radius 5 cm and resistance 0.003 ohms is placed around the center of a long air-core solenoid with 1000 turns per meter and a smaller radius of 3 cm. If the current in the solenoid is increasing at a constant rate of 270 A/s, what is the induced current in the ring?

B= $$\mu*n*I$$
change in B/change in time = $$\mu$$ n *change in current/change in time
= $$4 \pi e-7)(1000)(270)$$ = .339
then change in flux/change in time= A*change in B/change in t
A= $$\pi*r^2$$
So A= (.03)^2 *3.14
then multiply that by .339 to get 9.58 e -4.
Then I divided this by 3 e -4 to get the current and found that it was 3.19..

Last edited: Apr 10, 2006
2. Apr 10, 2006

Hootenanny

Staff Emeritus
I think the question wants you to use Faraday's law. Incidently, thats how I would go about it.

-Hoot

3. Apr 10, 2006

Punchlinegirl

So if I use Faraday's Law, would I do
change in flux= B*cos (theta)A
Where B= $$\mu$$ I* n
and would A be the big area minus the small? $$\pi$$ (.05^2)-(.03^2)?