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//pi r^(2)B`
B. `mg//pi rB`
C. `mg//2pi r B`
D. `pi rB//mg`

Answer

C

Explanation

The normal to the plane of coil makes `90^(@)` angle with the field direction. Therefore the torque on the coil `tau= BiA= Bi. pi r^(2)`
Since only one edge lifts from the table, the torque required is `= (mg)/(2) r`
`:. Bi pi r^(2) = (mg r)/(2) or i= (mg)/(2pi rB)`