00:01
In this question, we're asked to find the magnitude and direction of the induced current in the small circuit.
00:06
So the current flow in a charging rc circuit is expressed as the following.
00:13
I here is equal to v0, the initial potential divided by r0, multiplied by 1 minus epsilon, or excuse me, not epsilon e, to the minus i, current over rc so therefore taking the derivative of this di with respect to t we find that this is equal to minus our potential which is the induced emf divided by r not squared times the capacitance c times the exponential e to the minus t so excuse me this is an i this is t divided by rc.
01:17
Okay.
01:18
Well, the induced emf is equal to n times the change in the magnetic flux, d -fi, d -t.
01:37
Well, d -fi -d -t, so this is still n, d -fi -d -t is equal to mu -not divided by 2 -pi, multiplied by the natural log, ln, of 1, plus a over c multiplied by d -i -d -t, which we just found to be equal to e divided by r not squared c times i.
02:37
Okay...