In the circuit in Fig. P29.47, an emf of 90.0 V is added in series with the capacitor and the resistor, and the capacitor is initially uncharged. The emf is placed between the capacitor and switch $S$, with the positive terminal of the emf adjacent to the capacitor. Otherwise, the two circuits are the same as in Problem 29.47. The switch is closed at $t =$ 0. When the current in the large circuit is 5.00 A, what are the magnitude and direction of the induced current in the small circuit?
and this question were asked to find the magnitude and direction of the induced current in the small circuit. So the current flow in a charging RC circuit is expressed as the following. I here is equal to V not the initial potential divided by are not multiplied by one minus epsilon. Excuse me. Not absolute e to the minus. I current over our C Uh huh. So therefore, taking the derivative of this d I with respect to t we find that this is equal to minus our potential, which is the induced Ian meth divided by are not squared times the capacitance see times the exponential e to the minus t. So excuse me, this isn't I. This is t divided by R c. Okay, well, the induced e m f is equal to in times the change in the magnetic flux defy d t well, defy d t. So this is still in defy d t is equal to you, not divided by two pi multiplied by the natural log Ellen of one plus oversee multiplied by d i d. T, which we just found to be equal to e divided by are not squared, see times I Okay, so therefore Oh, and I forgot to be here. Therefore I induced which is equal to he induced divided by I am So we need to find this are this are comes from the material It's going to be equal to 25 times 0.6 meters times resistive 81.0 homes for meter So this comes out to equal 15 owes. So now we can plug this value into our equation. So we find I induced is equal to and times the magnetic permeability of free space mu not a times B divided by to pie multiplied by the natural log of one plus the ratio a oversee multiplied by one over the initial resistance are not well supplied by the capacitance See times. The current I multiplied by one over Are we just found OK, so carrying this out, we find that this is equal to 1.83 times 10 to the minus three amps or 1.83 million amps, and it asks us also to find the direction So using the rent law the direction of induced current in the small loop is clockwise direction. That's the direction of induced current in the smaller. But here's clockwise is what we want, right? So we'll just go ahead and type out clockwise plot wise in boxes in as our solution to the question.