Question

A uniform current carrying ring of mass `m` and radius `R` is connected by a massless string as shown in Fig. 1.142. A uniform magnetic field `B_0` exists in the region to keep the ring in horizontal position, then the current in the ring is (l=length of string)
.
A. `(mg)/(piRB_0)`
B. `(mg)/(RB_0)`
C. `(mg)/(3piRB_0)`
D. `(mg)/(piR^2B_0)`

Answer 1

Correct Answer - A
`tau_(mg)` about the left end (from where string is connected)
`=|MxxB|=MBsin90^@`
`or (mgR)=(NiA)B_0=i(piR^2)B_0` ltbgt `or i=(mg)/(piRB_0)`