Induced Current problem from review

  • #1

Homework Statement


A 40 turn circular coil of a 4.0 cm radius and a total resistance of 0.20 Ω is placed in a uniform magnetic field directed perpendicular to the plane of the coil. The magnitude of the magnetic field varies with time as B = 50*sin(10π*t) mT where t is measured in s. What is the induced current in the coil at 0.10 s?

Homework Equations


I believe these are the equations needed:
ε = -N*d(ΦB)/dt
I = V/R
ΦB = ∫B*dA


The Attempt at a Solution


The way I did it was start off with I = V/R, then knowing that V is the emf, I plugged in ε = -N*d(ΦB)/dt.

Then I used the flux equation, but modified it since we were given dB (or what I assume is dB) and I thus get I = (-N* ∫dB*A)/ R.

From there I get I = (-N*50*sin(10π*t)*10-3 *πr2)/R

I solve this, but I don't get the correct answer (which is known since this is from an older exam that I'm doing as additional practice for my fnal).

Am I missing something key here, or did I improperly set up the mag. field equation?
 

Answers and Replies

  • #2
rude man
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Then I used the flux equation, but modified it since we were given dB (or what I assume is dB) and I thus get I = (-N* ∫dB*A)/ R.
Why did you bring dB into the picture? The problem does not allude to dB in any way.
This problem could have been made really interesting if B = 0 for t < 0. But you don't want to consider that possibility. Just assume the B field has been there for a long time and then when the field is zero and starting to increase (B=0, dB/dt > 0) that defines t=0. Then you want the current 0.1sec. later.
 

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