Question 7.54: A circular wire loop (radius r, resistance R) encloses a reg...

\text { (a) } ε =-\frac{d \Phi}{d t}=-\pi r^{2} \frac{d B}{d t}=-\alpha \pi r^{2}=I R \Rightarrow \text { (in magnitude) } I=\frac{\pi \alpha r^{2}}{R} . If B is out of the page, Lenz’s law says the current is clockwise.

(b) Inside the shaded region, for a circle of radius s, apply Faraday’s law:

\oint E \cdot d l =E 2 \pi s=-\pi s^{2} \alpha \Rightarrow E =-\frac{\alpha s}{2} \hat{\phi}=-\frac{\alpha s}{2}(-\sin \phi \hat{ x }+\cos \phi \hat{ y })=\frac{\alpha}{2}(s \sin \phi \hat{ x }-s \cos \phi \hat{ y })=\frac{\alpha}{2}(y \hat{ x }-x \hat{ y }) .

Along the line from P to Q, d l=d x \hat{ x }, \text { and } y=r / \sqrt{2}, \text { so } V=-\int E \cdot d l =-\frac{\alpha}{2} \int y d x=-\frac{\alpha}{2} \frac{r}{\sqrt{2}}(r \sqrt{2}) .

Thus P is at the higher voltage, and the meter reads \frac{\alpha r^{2}}{2} .

That’s the simplest way to do it. But you might instead regard the 3/4 circle+chord as a circuit, and use Kirchho↵’s rule (the total emf around a closed loop is zero): V+I R_{1}-\alpha A_{1}=0, \text { where } R_{1}=3 / 4 R is the resistance of the curved portion, and A_{1}=(3 / 4) \pi r^{2}+r^{2} / 2=\left(r^{2} / 4\right)(3 \pi+2) is the area (3/4 of the circle, plus the triangle). Then 

V=\alpha \frac{r^{2}}{4}(3 \pi+2)-\frac{\pi r^{2} \alpha}{R} \frac{3}{4} R=\frac{\alpha r^{2}}{4}(3 \pi+2-3 \pi)=\frac{\alpha r^{2}}{2} .

Or we could do the same thing, using the small loop at the top:  -V+I R_{2}-\alpha A_{2}=0, \text { where } R_{2}=(1 / 4) R and A_{2}=(1 / 4) \pi r^{2}-r^{2} / 2=\left(r^{2} / 4\right)(\pi-2) .

V=\frac{\pi \alpha r^{2}}{R} \frac{R}{4}-\alpha \frac{r^{2}}{4}(\pi-2)=\frac{\alpha r^{2}}{4}(\pi-\pi+2)=\frac{\alpha r^{2}}{2} .