Two identical conducting wires `AOB and COD` are placed at right angles to each other. The wire `AOB` carries an electric current `I_(1) and COD` carries a current `I_(2)`. The magnetic field on a point lying at a distance `d` from O, in a direction perpendicular to the plane of the wires `AOB and COD`, will be given by

Options

A. `(mu_(0))/(2pid)(I_(1)+ I_(2))`
B. `(mu_(0))/(2pid) (I_(1)^(2) +I_(2)^(2))`
C. `(mu_(0))/(2pid)(O_(1)^(2)+I_(2)^(2))`
D. `(mu_(0))/(2pid) ((I_(1)^(2)+ I_(2)^(2))/(d)) ^(1//2)`

Answer

C

Explanation

The magnetic field at point P due to the current the wire `AOB, B_(1) =(mu_(0)I_(1))/(2pid)" "…(1)`
and the magnetic field at P due to the current in the wire
COD is `B_(2) =(mu_(0)I_(2))/(2pid)" "…(2)`
As the conductor are perpendicular to each other, `B_(1) and B_(2)` will also be perpendicular to each other.
`therefore` Their resultant magnetic field at P is given by
`B=sqrt(B_(1)^(2)+B_(2)^(2))=sqrt(((mu_(0)I_(1))/(2pid))^(2)+((mu_(0)I_(2))/(2pid))^(2))`
`=(mu_(0))/(2pid)sqrt((I_(1)^(2)+I_(2)^(2)))`