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
$E=\frac{d}{d t} \phi_{B}=\frac{d}{d t} B A$
$=\frac{d}{d t} B_{0} \sin \omega t \cdot 2 \pi a^{2}$
$=2 B_{0} \pi a^{2} \cdot \omega \cdot \cos \omega t$
Current in loop, $I=\frac{E}{R}=\frac{2 B_{0} \pi a^{2} \omega}{R} \cdot \cos \omega t$
So, current oscillates with a frequency $\omega$.
Heat loss per unit time $= I^{2}R$
$=\frac{4 B_{0}^{2} \pi^{2} a^{4} \omega^{2}}{R^{2}} \cos ^{2} \omega t \cdot R$
$=\frac{4 B_{0}^{2} \pi^{2} a^{4} \omega^{2}}{R} \cdot \cos ^{2} \omega t$
$\therefore $ Heat loss $\propto \alpha^{4}$
Force on a small segment $dl$ of ring is
$F=B i d l=d l \times B_{0} \sin \omega t \cdot \frac{2 B_{0} \pi a^{2} \omega}{R} \cos \omega t$
$\therefore $ Force per unit length on loop is
$\frac{F}{d l}=\frac{2 B_{0}^{2} \pi a^{2} \omega}{R} \sin \omega t \cos \omega t$
$\therefore $ Force per unit length $\propto B_{0}^{2}$
Also, net force on loop is zero.