A current loop `ABCD` is held fixed on the plane of the paper as shown in figure. The arcs `BC( radius = b) and DA ( radius = a)` of the loop are joined by two straight wires `AB and CD` at the origin `O` is 30^(@)`. Another straight thin wire with steady current `I_(1)` flowing out of the plane of the paper is kept at the origin .

Due to the process of the current `I_(1)` at the origin:

Options

A. the force on AB and DC are zero
B. the force on AD and BC are zero
C. the magnitude of the net force on the loop is given by `(I_(1)I)/(4 pi) mu_(0) [2 (b - a) + pi/3 (a + b)]`
D. the magnitude of the net force on the loop is given by `(mu_(0)II)/(24 ab) (b - a)`.

Answer

b

Explanation

Magnetic field due to outward current `I_(1)` at the origin are concentric circles and hence parallel to current in branches AD and BC. So no magnetic force will act on these but the same magnetic field is perpendicular to AB and CD, thus, it will apply magnetic force on AB and CD. Although force on AB and CD will be equal and opposite, the net force on the complete system will be zero. Therefore, option (b) is correct.