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Wires 1 and 2 carrying currents i1 and i...

Wires 1 and 2 carrying currents `i_1 and i_2` respectively are inclined at an angle `theta` to each other. What is the force on a small element `dl` of wire 2 at a distance of r from wire 1 (as shown in the figure) due to the magnetic field of wire 1 ?

A

`(mu_0)/(2pir)i_1i_2dl tan theta`

B

`(mu_0)/(2pir)i_1i_2dl sin theta`

C

`(mu_0)/(2pir)i_1i_2dl cos theta`

D

`(mu_0)/(4pir)i_1i_2dl sin theta`

Text Solution

Verified by Experts

The correct Answer is:
C
B due to current in wire 1 at point P
`B=(mu_0)/(4pi)(i_1)/(r)[cos theta+cos theta]`
`= (mu_0)/(2pi)(i_1cos theta)/(r)`

(Perpendicular to plane of paper `rarr` inward) Force exerted due to this magnet field in current element `i_2dI`
`dF = i_2dI B sin 90^@`
`dF=i_2 dlB sin 90^@ `
`=i_2 dl[(mu_0i_0cos theta)/(2pir)]=(mu_0)/(2pir)i_1i_2dl cos theta`
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