Question

Two identical conducting rings $\text{A}$ & $\text{B}$ of radius $R$ are in pure rolling over a horizontal conducting plane with same speed of center of mass $v$ but in opposite direction. A constant horizontal magnetic field $B$ exists in the space pointing inside the plane of paper. The potential difference between the topmost points of the two rings is

• A. Zero
• B. $2BvR$
• C. $4BvR$
• D. Noe of these

Answer 1

The correct option is C $4BvR$

Considering an imaginary rod of length $2R$ between the bottom and topmost point of the ring $\text{B}$ as shown in the figure


Since, the ring is in pure rolling motion, we can say that,
$v=\omega R$

As, the end $\text{P}$ of the rod is stationary at the given instant and all the particles of the imaginary rod are moving in direction perpendicular to the length of the rod, we can say that rod $\text{PQ}$ is rotating about point $P$.

$\therefore$ Emf induced between points $\text{P}$ and $\text{Q}$ is

$V_Q-V_P=\frac{B\omega l^2}{2}=\frac{B\omega (2R{)}^2}{2}=2B\omega R^2$

$V_Q=2BvR[\because v=\omega R\text{and}{V}_P=0]$


$\therefore V_B=2BvR$

Where $V_B$ is the potential of the topmost point of ring $\text{B}$.

Similarly, emf at the top point of the ring $\text{A}$ is

$V_A=-2BvR$

Here, the negative sign indicates that the topmost point of ring $\text{A}$ is at negative potential.

$\Rightarrow V_B-V_A=2BvR-(-2BvR)$

$\therefore V_B-V_A=4BvR$

Hence, option $\left(\text{C}\right)$ is correct.