In the shown diagram, all the wires are made up of same material, lenth of all section is same. But area of cross section of AD and DC is double that of AB and BC. Find magnetic field at point 'O' due to the squar. The current through AB and BC is I:-

Options

A. `zero`
B. `sqrt(2mu_(0)i)/(piL)o+`
C. `sqrt(2mu_(0)i)/(piL)o.`
D. `(2sqrt(2mu_(0)i))/(piL)ox`

Answer

C

Explanation

`B_(1)=(mu_(0)i)/(4pi(L)/(2))[2sin45]xx2`
`=sqrt(2mu_(0)i)/(2pil)xx2=sqrt(2mu_(0)i)/(pil)ox`
`B_(2)=(mu_(0)2i)/(2pil)[2sin45]xx2o.`
`(2sqrt(2mu_(0)I))/(pil)`
`B_("net")=B_(2)-B_(1)`