Question

$A$ metallic ring of radius $a$ and resistance $R$ is held fixed with its axis along a spatially uniform magnetic field whose magnitude is $B_0$ sin $\omega t$. Gravity is neglected. Then,

• A. the current in the ring oscillates with a frequency of $2\omega$
• B. the joule heating loss in the ring is proportional to $a^2$
• C. the force per unit length on the ring will be proportional to $B_0^2$
• D. the net force on the ring is non-zero

Answer 1

Answer: (C)

Induced emf in ring is
$E=\frac{d}{dt}{\varphi }_B=\frac{d}{dt}BA$
$=\frac{d}{dt}{B}_0\sin \omega t\cdot 2\pi a^2$
$=2B_0\pi a^2\cdot \omega \cdot \cos \omega t$
Current in loop, $I=\frac{E}{R}=\frac{2B_0\pi a^2\omega }{R}\cdot \cos \omega t$
So, current oscillates with a frequency $\omega$.
Heat loss per unit time $=I^{2}R$
$=\frac{4B_0^2{\pi }^2{a}^4{\omega }^2}{R^2}{\cos }^2\omega t\cdot R$
$=\frac{4B_0^2{\pi }^2{a}^4{\omega }^2}{R}\cdot {\cos }^2\omega t$
$\therefore$ Heat loss $\propto {\alpha }^4$
Force on a small segment $dl$ of ring is
$F=Bidl=dl\times B_0\sin \omega t\cdot \frac{2B_0\pi a^2\omega }{R}\cos \omega t$
$\therefore$ Force per unit length on loop is
$\frac{F}{dl}=\frac{2B_0^2\pi a^2\omega }{R}\sin \omega t\cos \omega t$
$\therefore$ Force per unit length $\propto B_0^2$
Also, net force on loop is zero.