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 v0 in a constant magnetic field B0 acting perpendicular to the plane of the paper in the inward direction. The current in the resistance is:

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 v0 in a constant magnetic field B0 acting perpendicular to the plane of the paper in the inward direction. The current in the resistance is: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-yx8fgfoqowbw1zew.jpg" /> (A) (2 B0 R v0/r) (B) (B0 R v0/r) (C) (B0 R v0/2 r) (D) 0

Tardigrade Admin · Accepted Answer

As induced emf is Blv for a moving rod, ∴ the arcs produce of emf of 2B0v0R Applying KVL in the loop We get, 2B0v0R=ir i=(2 B0 v0 R/r)

$\frac{2 B _{0} R v _{0}}{r}$

$\frac{B _{0} R v _{0}}{r}$

$\frac{B _{0} R v _{0}}{2 r}$

$0$

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

r2B0​Rv0​​

rB0​Rv0​​

2rB0​Rv0​​

0

As induced emf is Blv for a moving rod, ∴ the arcs produce of emf of 2B0​v0​R Applying KVL in the loop We get, 2B0​v0​R=ir i=r2B0​v0​R​

$\frac{2 B _{0} R v _{0}}{r}$

$\frac{B _{0} R v _{0}}{r}$

$\frac{B _{0} R v _{0}}{2 r}$

$0$

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