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Two identical conducting wires AOB and C...

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

A

`(mu_(0))/(2 pi d)(I_(1)^(2) + I_(2)^(2))`

B

`(mu_(0))/(2 pi )(I_(1)^(2) + I_(2)^(2))/(d))1//2`

C

`(mu_(0))/(2 pi d)(I_(1)^(2) + I_(2)^(2))1//2`

D

`(mu_(0))/(2 pi d)(I_(1) + I_(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
Clearly, the magnitic fields at a point `P`, equidistant from `AOB and COD` will have directions perpendicular to each other , as they are placed normal to each other.
:. Resultant field , `B = sqrt (B_(1)^(2) + B_(2)^(2))`
But `B_(1) = (mu_(0)I_(1))/(2 pi d) and `B_(2) = (mu_(0)I_(2))/(2 pi d)`
:. `B = sqrt (((mu_(0))/(2 pi d))^(2) (I_(1)^(2) + I_(2)^(2)))`
or, `B = (mu_(0))/(2 pi d) (I_(1)^(2) + I_(2)^(2))1/2`
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Knowledge Check

  • Two identical long conducting wires AOB and Cod are placed at right angles 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. Points P is lying at distance d form O along a direction perpendicular to the plane containing the wires. The magnetic field at the point P will be

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    B
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    C
    `(mu_(0))/(2 pi d)(I_(1)^(2) - I_(2)^(2))`
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  • Two straight long conductors AOB and COD are perpendicular to each other and carry currents I_(1) and I_(2) . The magnitude of the magnetic induction at a point P at a distance d from the point o in a direction perpendicular to the plane ABCD is :

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