A square loop PQRS carrying a current of 6.0 A is placed near a long wire carrying 10 A as shown in figure . (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. `

Explanation

Here `= i_1 = 6A, i_2= 10A,`
`F_(pQ) = (mu_oi_1i_2)/(2pi) int_1^3(dx)/(x)`
`=2xx10^(-7)xx6xx10[logx]1^3`
`=120xx10^(-7) log 3N`
`F_(RS) = (mu_oi_1i_2)/(2pi) int_3^1(dx)/(x)`
`=120xx10^(-7)log ((1)/(3))`
`= -120 10^(-7) log 3 N`
Both forces are same in magnitude but
opposite in direction.