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Wires 1 and 2 carrying currents i(1) and...

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 figure) due to the magnetic field of wire 1`?

A

`(mu_(0))/(2 pir) i_(1)i_(2) dl tan theta`

B

`(mu_(0))/(2 pir) i_(1)i_(2) dl sin theta`

C

`(mu_(0))/(2 pir) i_(1)i_(2) dl cos theta`

D

`(mu_(0))/(4 pir) i_(1)i_(2) dl sin theta`

Text Solution

Verified by Experts

The correct Answer is:
C
Magnetic fiweld due to current in weire `1` at point `p` distant `r` from the wire is
`B = (mu_(0))/( 4 pi) ( i_(1))/( r )[ cos theta + cos theta]`
`(mu_(0))/( 4 pi) ( i_(1) cos theta)/( r ) (directed perpendicular to the plane of paper , inwards)`
The force exterted due to this magnetic field on current element `i_(2) dl` is
`dF = i_(2) dl B sin 90^(@)`
:. `dF = i_(2)dl [(mu_(0))/( 4 pi) ( i_(1) cos theta)/( r )] = (mu_(0))/( 2 pi r) i_(1)i_(2) dl cos theta`
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