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Two indentical long conducting wires AOB...

Two indentical long conducting wires AOB and COD are placed at right angles to each other, with one above other such that is their common point for the two. The wires carry `I_(1) and I_(2)` currents respectively. A point P is at a height d above the point O, with respect to the plane of the wires. the magnetic field at P is,

A

`(mu_(0))/(2pid) ((I_(1))/(I_(2)))`

B

`(mu_(0))/(2pid)(I_(1)+I_(2))`

C

`(mu_(0))/(2pid)(I_(1)^(2)-I_(2)^(2))`

D

`(mu_(0))/(2pid)(I_(1)^(2)+I_(2)^(2))^(1//2)`

Text Solution

Verified by Experts

Magnetic field at P due to current `I_(1)`.
`B_(1)=(mu_(0))/(4pi) (2I_(1))/(d)`
Magnetic field at P due to current `I_(2)`,
`B_(2)=(mu_(0))/(4pi)(2I_(2))/(d)`
`:.` The resultant magnetic field to AOB and COD,
`B=sqrt(((mu_(0))/(4pi)(2I_(1))/(d))^(2)+((mu_(2))/(4pi)(I_(2))/(d))^(2))`
`:.B=(2mu_(0))/(4pid sqrt(I_(1)^(2)+I_(2)^(2))`
`=(mu_(0))/(2pid) (I_(1)^(2)+I_(2)^(2))^((1)/(2))`
The option C is correct.
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Knowledge Check

  • Two identical long conducting wires AOB and COD are placed at righ 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' form 'O' along a direction perpendicular to the plane containing the wires the magnetic field at the point 'P' will be

    A
    `(mu_0)/(2pid)(I_1/I_2)`
    B
    `(mu_0)/(2pid)(I_1+I_2)`
    C
    `(mu_0)/(2pid)(I_1^2-I_2^2)`
    D
    `(mu_0)/(2pid)(I_1^2+I_2^2)^(1/2)`

  • Two thin long parallel wires separated by a distance b are carrying current I amp each the magnitude of the force per unit length exerted by one wire on the other is:

    A
    `mu_0(i^2//b^2)`
    B
    `mu_0i^2/(2pib)`
    C
    `(mu_0i)/(2pib)`
    D
    `(mu_0I)/(4pib)`

  • Thevresulting magnetic field at the point O due to the current carrying wire shown in the figure

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    B
    points into the page
    C
    is zero
    D
    is the same as due to segment WX along

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