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

Answer

B

Explanation

The magnetic field inductions at a point P, at a distance d from O in a direction perpendicular to the plane of the wires due to currents through AOB and COD are perpendicular to each other, is:

`B=sqrt(B_(1)^(2)+B_(2)^(2))=[((mu_(0))/(4pi)(2I_(1))/d)^(2)+((mu_(0))/(4pi)(21_(2))/d)^(2)]^(1//2)=(mu_(0))/(2pid)sqrt((I_(1)^(2)+I_(2)^(20))`