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)`

A uniform current carrying ring of mass `m` and radius `R` is connecte

1.	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» 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)`.

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