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. `(mgl)/(piR^2B_0)`

Answer 1

Correct Answer - a
Torque due to magnetic field `tau_(mag)=MB_0=ipiR^2B_0....(i)`
Torque due to weight about the point where string is connected
`tau_("weight")=mgR....(ii)`
If ring remains horizontal, then `tau_(mag)=tau_("weight")`
`IpiR^2B_0=mgRimpliesI=(mg)/(piRB_0)`.