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. $X,Aa$
• 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,
$|e|=A\frac{dB}{dt}=A\frac{d}{dt}\left(B_0+\alpha t\right)$
$\Rightarrow |e|=A\alpha$