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

Answer

a

Explanation

Using the right-hand palm rule we can understand that magnetic fields at a given point due to both the wires are perpendicular to each other.
`B_(1) = mu_(0)/(2 pi d) I_(1)`
`B_(2) = mu_(0)/(2 pi d) I_(2) `
`B = sqrt(B_(1)^(2) + B_(2)^(2)) = mu_(0)/(2 pi d) (I_(1)^(2) + I_(2)^(2))^(1//2) `
Hence option (a) is correct.