A circular loop of radius R carrying a current I is placed in a uniform magnetic field B perpendicular to the loop. The force on the loop is

Options

A. `irB_0`
B. `2pi irB_0`
C. zero
D. `pi irB_0`

Answer

c

Explanation

The magnetic field is perpendicular to the plane of the
paper. Let us consider two diametrically opposite elements.
By Fleming's left hand rule on element AB, the direction of
force will be leftward and the magnitude will be
`df=IdlB sin 90^@=IdlB`

On element CD, the direction of force will be towards right on
the plane of the paper and the magnitude will be dF=IdlB.
These two forces will cancel on.
Similarly, all forces acting on the dimetrically opposite
elements will cancel out in pair. The net force on the loop
will be zero.