Question

A square loop PQRS carrying a current of 6.0 A is placed near a long wire carrying 10 A as shown in figure (35-E10). (a) Show that the magnetic force acting on the part PQ is equal and opposite to that on the part Rs. (b) Find the magnetic force on the square loop.

Answer 1

Here =$i_1=6A,i_2=10A,$

$F_{PQ}=\frac{{\mu }_0{i}_1{i}_2}{2\pi }{\int }_1^3\frac{dx}{x}$

$=2\times {10}^{-7}\times 6\times 10[logx{]}_1^3$

$=120\times {10}^{-7}log3N$

$F_{RS}=\frac{{\mu }_0{i}_1{i}_2}{2\pi }{\int }_3^1\frac{dx}{x}$

$=120\times {10}^{-7}log\left(\frac{1}{3}\right)$

$=-120\times {10}^{-7}log3N$

Both forces are same in magnitude but opposite in direction.