Question

The magnetic field at O due to current in the infinite wire forming a loop as a shown in Fig.

A. `(mu_(0)I)/(4pid)(cosphi_(1)+cosphi_(2))`
B. `(mu_(0))/(4pi)(2I)/d(tan theta_(1)+tan theta_(2))`
C. `(mu_(0))/(4pi)(I)/d(sin phi_(1)+sin phi_(2))`
D. `(mu_(0))/(4pi)(I)/d(cos theta_(1)+cos theta_(2))`

Answer 1

Correct Answer - B
`B=(mu_(0)I)/(4pid) (sin theta_(1)+sin theta_(2))`
But `theta_(1)+phi_(1)=90^(@)` or `theta_(1)=90^(@)-phi`
`sin theta_(1)=sin(90^(@)-phi_(1))=cos phi_(1)`
Similarly, `sin theta_(2)=cos phi_(2)`
`B=(mu_(0)I)/(4pid) (cos phi_(1)+cos phi_(2))`