A 2.00 cm x 2.00 cm square loop of wire with resistance 1.30×10^−2 Ohms is parallel to a long straight wire. The near edge of the loop is 1.00 cm from the wire. The current in the wire is increasing at the rate of 110 A/s. What is the current in the loop?

Question

A 2.00 cm  x  2.00 cm square loop of wire with resistance 1.30×10^−2 Ohms is parallel to a long straight wire. The near edge of the loop is 1.00 cm from the wire. The current in the wire is increasing at the rate of 110 A/s.  What is the current in the loop?

bobpursley · Answer

You have to find the flux, and since it is not constant with distance, integrate BdA outward

EMF= d/dt flux= d/dt INT BdA =d/dt INT .02B(r)dr from .01 to .03m outward. You know B as a function of r. Then, take the d/dt of B which is a function of current.

Manda · Answer

Ok, so I did INT from .01 to .03 of mu_0*I/r which ended up being mu_0*I*(ln .03-1n .01). I'm confused why you only used .02 as the area. It doesn't give radius of the wire and area of the wire is .02^2. When taking the derivative then, d/dt of I is just dI/dt? so i should just substitute in the 110 A/s? Also, do you do anything with the resistance or is that just extra information?