Question

A metallic ring (radius $R$ ) of negligible resistance has a resistance $r$ connected across its diameter as shown in the figure. It is moving with velocity $v_{0}$ in a constant magnetic field $B_{0}$ acting perpendicular to the plane of the paper in the inward direction. The current in the resistance is:

• A. $\frac{2 B_{0} R v_{0}}{r}$
• B. $\frac{B_{0} R v_{0}}{r}$
• C. $\frac{B_{0} R v_{0}}{2 r}$
• D. $0$

Answer 1

Answer: (A)

As induced emf is $Blv$ for a moving rod,
$\therefore \, \, $ the arcs produce of emf of $2B_{0}v_{0}R$
Applying KVL in the loop
We get,
$2B_{0}v_{0}R=ir$
$i=\frac{2 B_{0} v_{0} R}{r}$