Solving for Current in a Coil: A Troubleshooting Guide

[MeKnos]

Homework Statement

A 21.8 cm diameter coil consists of 18 turns of circular copper wire 1.70 mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.15E-3 T/s. Determine the current in the loop.

Homework Equations

V=IR
V=B/T * A * N
permitivity of copper= 1.72e-8
R=p * L / A'

The Attempt at a Solution

V = 8.15e-3 * (.218/2)^2 * pi * 18 turns = 0.005475 V

L = N * pi * coil diameter = 18 * pi * .218 = 12.32 m
A'= pi * (.0017/2)^2 = 2.269e-6 m^2
R = 1.72e-8 * (12.32 m/2.269e-6 m^2) = 0.0934 ohms

I = V/R = 0.005475 V / 0.0934 ohms = 0.0586 A

But that's wrong... where'd I go wrong?

 

[lukas86]

It's been a long time since I've done any questions like this, but the equation "V=B/T * A * N" you use but you do not use the 'T' anywhere. You have "V = 8.15e-3 * (.218/2)^2 * pi * 18 turns" but unless I'm missing something, I don't see 'T' in there or a value of 'T' given in the question.

 

[MeKnos]

Isn't B/T = 8.15e-3 Tesla per second?

If not, what do I do in that case.

 

[lukas86]

Ya, I ignored that for a second when doing the calculation and got the same 0.005...V that you did, units cancelled. I'll wait for somebody else's reply because I am a bit confused now too haha. Do you have a correct answer for that?

 

[MeKnos]

Unfortunately no. I looked at other similar problems, but they all had the process I'm doing, but I couldn't compare answers.

 

[MeKnos]

Anyone?