A circular loop of mass m and radius r in X-Y plane of a horizontal table as shown in figure. A uniform magnetic field B is applied parallel to X-axis. The current I in the loop, so that its one edge just lifts from the table is

Options

A. `(mg)/(pirB)`
B. `(mg)/(2pirB)`
C. `(mg)/(pir^2B)`
D. `(2pirB)/(mg)`

Explanation

Here, area of the coil `A=pir^2`. The direction of `vecA` is normal to the plane of coil, making an angle `90^@` with the field direction. Therefore, the torque on the coil is `tau=BIA=BIpir^2`
Since only one edge of circular coil lifts from the table, the torque required is `=(mg)/(2)r`
`:. BIpir^2=(mgr)/(2)` or `I=(mg)/(2pirB)`