Answer
a. 4.68mV.
b. The end marked “a” is at a higher potential.
Work Step by Step
a. $L = \frac{\epsilon}{|di/dt|}$
$$|\epsilon| = L|\frac{di}{dt}|=(0.260H)(0.0180A/s)=4.68mV$$
b. The end marked “a” is at a higher potential. By Lenz’s Law, the emf would tend to drive current (through an external circuit) from right to left. If the coil were replaced by a battery, the positive end would be on the left.
