Two identical long conductin wires AOB and COD are placed at right angle to each other with one above other such that O is their common point for the two. The wires carry `I_(1)"and"I_(2)` currents, respectively . Point P is lying at distance d from O along a direction perpendicular to the plane containing the wires. the magnetic field at the point P will be

Options

A. `(mu_(0))/(2pia)(I_(1)+I_(2))`
B. `(mu_(0))/(2pia)(I_(1)-I_(2))`
C. `(mu_(0))/(2pia)(I_(1)^(2)+I_(2)^(2))^(1//2)`
D. `(mu_(0))/(2pia) (I_(1)I_(2))/(I_(1)+I_(2))`

Answer

C

Explanation

The point P is lying symmetrical with respect to the two long conductors. Field at P due to these currents are mutually perpendicular.
`B_(1)=B_(AOB)=(mu_(0)I_(1))/(2pia), B_(2)=B_(COD)=(mu_(0)I_(2))/(2pa)`

`:.B=sqrt(((m_(0))/(2pia))^(2)(i_(1)^(2)+I_(2)^(2)))impliesB=(mu_(0))/(2pia)(I_(1)^(2)+I_(2)^(2))^(1//2)`