Figure shows a square loop of edge a made of a uniform wire. A current i enters the loop at the point A and leaves it at the point C. Find the magnetic field at the point P which is on the perpendicular bisector of AB at a distance a/4 from it. `

Explanation

B at P due to AD `=(mu_o)/(4pi) . (i)/(2) . (4)/(a^2). a[(((a)/(2)))/(sqrt(((a)/(2))^2 + ((a)/(4))^2)) + (((a)/(2)))/(sqrt(((a)/(2))^2 + ((3a)/(4))^2))] along`
`=(mu_oi)/(2pia)[(((a)/(2)))/(sqrt(((a)/(2))^2 + ((a)/(4))^2)) + (((a)/(2)))/(sqrt(((a)/(2))^2 + ((3a)/(4))^2))]` along`
`B` at `P` due to AC `=(mu_o)/(4pi).(i)/(2). (16)/(9a^2.a 2.[(((3a)/(4)))/(sqrt(((3a)/(4))^2 + ((a)/(2))^2))`
`(4mu_oi)/(9pia) [(((3a)/(4)))/(sqrt(((a)/(4))^2 +((3a)/(2))^2))]along`
`B at P due to AB = (mu_oi)/(4pi).(i)/(2) . (16)/(9a^2). a. 2 [(((a)/(4)))/sqrt(((a)/(4))^2 + ((a)/(2))^2 along`
`B` at `P` due to `BC = (mu_o)/(4pi).(i)/(2). (4)/(a^2).a. [(((a)/(2)))/ sqrt(((a)/(2))^2 + ((a)/(2))^2) +(((a)/(2)))/sqrt(((a)/(2))^2 + ((3a)/(4))^2)] along`
`=(um_oi)/(2pia)[(((a)/(2)))/sqrt(((a)/(2))^2 +((a)/(2))^2) + (((a)/(2)))/(sqrt(((a)/(2))^2 +((3a)/(4))^2))]` along ox
so, `net `B = (4mu_oi)/(pia)[(((a)/(4)))/sqrt(((a)/(2))^2 +((a)/(4))^2)] - (4mu_oi)/(9pia) [(((3a)/(4)))/(sqrt(((a)/(2))^2 +((3a)/(4))^2))]`
`=(4mu_oi)/(pia).(1)/(4)[(1)/sqrt((1)/(4) +(1)/(16)]-(4mu_oi)/(9pia).(3)/(4)[(1)/(sqrt((1)/(4)) +(9)/(16))]`
`=(4mu_oi)/(pia)[(4)/sqrt5]-(mu_oi)/(3pia)[(4)/(sqrt13)] [(4)/(sqrt13)] along`
`=(4mu_oi)/(2pia)[(1)/sqrt5- (1)/(3sqrt13)]along =(2pi_oi)/(pia)[(1)/(sqrt5)-(1)/(3sqrt13)]along `.