Find the magnetic field B at the center of a rectangular loop of length l and width b, carrying a current i.

Answer

`(2mu_0Isqrt(a^2+b^2))/(piab)`

Explanation

Let `vecB_1, vecB_2, vecB_3` and `vecB_4` be the magnetic field at O due to current in arms 1,2,3 and 4 respectively, figue.
The total magnetic field induction at O due to current through rectangular loop is
`vecB=vecB_1+vecB_2+vecB_3+vecB_4`

where, `B_1=B_3=(mu_0)/(4pi)(I)/((b//2))`
`xx[sin (90^@-alpha)+sin (90^@-alpha)]`
`=(mu_0)/(4pi)(2I)/(b)[cos alpha+cos alpha]`
`=(mu_0)/(4pi)(2I)/(b)xx2 cos alpha=(mu_0)/(pi)I/bxx(a)/(sqrt(a^2+b^2)`
and `B_2=B_4=(mu_0)/(4pi)(I)/((a//2))`
`xx[sin (90^@-beta)+sin (90^@-beta)]`
`=(mu_0)/(4pi)(2I)/(a)[cos beta+cos beta]`
`(mu_0)/(4pi)(2I)/(a)2cos beta=(mu_0)/(pi)I/a(b)/sqrt(a^2+b^2)`
As `vecB_1, vecB_2, vecB_3` and `vecB_4` are acting in the same direction, i.e., perpendicular to the plane of rectangular loop downwards.
`:. B=B_1+B_2+B_3+B_4`
`=(mu_0)/(pi)(I)/(sqrt(a^2+b^2))[a/b+b/a+a/b+b/a]`
`=(mu_0)/(pi)(2I)/(sqrt(a^2+b^2))[a/b+b/a]`
`=(mu_0)/(pi)(2Isqrt(a^2+b^2))/(ab)`