Induced Current in Aluminum Ring Around Solenoid

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 4K views
Punchlinegirl
Messages
221
Reaction score
0
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= [tex]\mu*n*I[/tex]
change in B/change in time = [tex]\mu[/tex] n *change in current/change in time
= [tex]4 \pi e-7)(1000)(270)[/tex] = .339
then change in flux/change in time= A*change in B/change in t
A= [tex]\pi*r^2[/tex]
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..
which wasn't right.. can someone please help me? Thanks in advance!
 
Last edited:
Physics news on Phys.org
So if I use Faraday's Law, would I do
change in flux= B*cos (theta)A
Where B= [tex]\mu[/tex] I* n
and would A be the big area minus the small? [tex]\pi[/tex] (.05^2)-(.03^2)?