Induced Current Problem(Electromagnetism)

[pjwasz]

Homework Statement

A small, 1.60-mm-diameter circular loop with R = 1.40×10−2Ω is at the center of a large 120-mm-diameter circular loop. Both loops lie in the same plane. The current in the outer loop changes from 1 A to -1 A in 8.00×10−2 s.

What is the induced current in the inner loop?

Homework Equations

I = ε/R
d$\Phi$/dt = ε
$\Phi$ = BAcos(0) = BA
B = (μ_0)(I)/2r

The Attempt at a Solution

First, I got (μ_0)(I)($\pi$)(r)(1/2) = $\Phi$, using $\pi$r^2 as A.

Then, d$\Phi$/dt = (μ_0)($\pi$)(r)(1/2)(dI/dt)

So...I = ((μ_0)($\pi$)(r)(1/2)(dI/dt))/R

Which is ((4*10^(-7))($\pi$)^2(.03)(2/.08))/.014.

My answer I got was 2.11 *10^-4 A.Please help, thank you. :)

 

[barryj]

It looks to me like you did not consider the smaller area of the inside loop.

 

[barryj]

I think you need to multiply your answer by the area of the small loop and you will get what I did.

 

[rude man]

Homework Statement

A small, 1.60-mm-diameter circular loop with R = 1.40×10−2Ω is at the center of a large 120-mm-diameter circular loop. Both loops lie in the same plane. The current in the outer loop changes from 1 A to -1 A in 8.00×10−2 s.

What is the induced current in the inner loop?

Homework Equations

I = ε/R
d$\Phi$/dt = ε
$\Phi$ = BAcos(0) = BA
B = (μ_0)(I)/2r

The Attempt at a Solution

First, I got (μ_0)(I)($\pi$)(r)(1/2) = $\Phi$, using $\pi$r^2 as A.

Where did this come from?
You need to compute B in the middle of the large loop. Computing flux thruout the area of the larger loop is both nearly impossible and pointless. Your small loop sees only the B field near the center of the large loop, and fortunately this is easy to calculate using Biot-Savart law.

EDIT: as barryj pointed out, you already had the correct value for B in your equations list. So use it!

 

[barryj]

I think he did compute B in the center . It seems he did not consider the smaller area of the inner loop.

 

[barryj]

I think this is incorrect "First, I got (μ_0)(I)(π)(r)(1/2) = Φ, using πr^2 as A"
You have to consider the R of the big loop and the r of the small loop.

Big R = .06m and little r = .0008m

Phi = uI/2R X (pi)r^2

 

[pjwasz]

Ah, I see. So what I was doing was calculated the flux due to the magnetic field of the big loop, but through the big loop itself. Instead, I need to calculate the flux due to the magnetic field of the big loop, but through the small loop.

...Good news, that worked! Thanks guys, I get what I was doing wrong! :)