A rectangular loop of conductor of length a and breadtlı b carrying current / is shown in the figure. The magnetic field at the centre o of the loop is

Options

A. `(mu_(0)(a+b))/(2pisqrt(a^(2)+b^(2)))`
B. `(mu_(0)Iab)/(2pisqrt(a^(2)+b^(2)))`
C. `(mu_(0)(a+b))/(pisqrt(a^(2)+b^(2)))`
D. `(2mu_(0)(a^(2)+b^(2)))/(piab)`

Answer

D

Explanation

The total megnetic field at O due to current I thought rectangular loop is B. Let `B^(1),B^(2),B^(3) and B^(4)` be the magnetic field at O due to current in arms 1,2,3 and 4 respestively.
`:.B=B^(1)+B^(2)+B^(3) + B^(4)`
where `(B^(1))=(B^(2))`
`B^(1)=B^(3)=(mu_(0)I)/(4pi(a"/"2))[sin(90^(@)-alpha)+sin(90^(@)-alpha)]`
`=(mu_(0)I)/(4pi(a"/"2))(cosalpha+cosalpha)=(mu_(0)Icosalpha)/(pia)`
`(B_(2))=(B_(4))=(mu_(0))/(4pi(a"/"2))[sin(90^(@)-beta)+sin(90^(@)-beta)]`
`=(mu_(0))/(4pi(a"/"2))(cosbeta+cosbeta)=(mu_(0)Icosbeta)/(pib)`
As `B^(1),B^(2),B^(3) and B^(4)` are acting in the same direction perpendicular to the plane of rectangular loop downwards.
`:.B=B^(1)+B^(2)+B^(3) + B^(4)or B=(2mu_(0)I)/(pi)[(cosalpha)/(a)(xosbeta)/(b)]`
Puting`cosalpha=(b)/(sqrt(a^(2)+b^(2)))andcosbeta=(a)/(sqrt(a^(2)+b^(2)))`
we get , `B=(2mu_(0)sqrt(a^(2)+b^(2)))/(piab)`