Question

A highly conducting ring of radius $R$ is perpendicular to and concentric with the axis of a long solenoid as shown in figure. The ring has a narrow gap of width $d$ in its circumference. The solenoid has cross-sectional area $A$ and a uniform internal field of magnitude $B_{0}$. Now beginning at $t=0$, the solenoid current is steadily increased so that the field magnitude at any time $t$ is given by $B(t)=B_{0}+\alpha t$ where $\alpha > 0$. Assuming that no charge can flow across the gap, the end of ring which has excess of positive charge and the magnitude of induced e.m.f. in the ring are respectively

• A. $X, A a$
• B. $X \pi R^{2} a$
• C. $Y, p A^{2} \alpha$
• D. $Y, \pi R^{2} a$

Answer 1

Answer: (A)

Since the current is increasing, so inward magnetic flux linked with the ring also increasing (as viewed from left side). Hence induced current in the ring is anticlockwise, so end $X$ will be positive. Induced emf,
$|e|=A \frac{d B}{d t}=A \frac{d}{d t}\left(B_{0}+\alpha t\right)$
$\Rightarrow |e|=A \alpha$