In the figure `AB` is a long straight wire carrying a current of `20 A` and `CDFG` is a rectangular loop of size `20 cm xx 9 cm` carrying a current of `10 A`. The edge `CG` is parallel to `AB`, at a distance of `1 cm` from it. The force exterted on the loop by the magnetic field of the wire is

Options

A. `3.6 xx 10^(-4) N` towards left
B. `3.6 xx 10^(-4) N` towards right
C. `37.2 xx 10^(-4) N` towards right
D. `7.2 xx 10^(-4) N` towards left

Answer

D

Explanation

Force on `CD` is equal and opposite to force on `FG`
`F_(CG) = (mu_(0)i_(1)i_(2))/(2pi d)L = (2 xx 10^(-7) xx 20 xx 10)/(0.01) xx 0.2`
`= 8 xx 10^(-4) N`, towards left
`F_(DF) = (mu_(0)i_(1)i_(2))/(2pi (d + 0.09)) xx L = (2 xx 10^(-7) xx 20 xx 10)/(0.1) xx 0.2`
`= 0.8 xx 10^(-4) N`, towards right
`(F_("loop"))_("net") = F_(CG) - F_(DF) = 7.2 xx 10^(-4)N`, towards left