Wires 1 and 2 carrying currents `I_(1) and I_(2)` respectively 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 fig. due to the magnetic field of wire 1 ?

Options

A. `(mu_(0))/(2 pi r) I_(1), I_(2) dl tan theta`
B. `(mu_(0))/(2pi r) I_(1)I_(2) dl sin theta`
C. `(mu_(0))/(2 pi r) I_(1)I_(2) dl cos theta`
D. `(mu_(0))/( 4 pi r) I_(1) I_(2) dl sin theta`

Answer

C

Explanation

Magnetic field on dl element due to current `I_(1)` is
`B = (mu_(0))/(2 pi) (I_(1))/(r)` . . . (1)
Unit force acting on dl element
`(F)/(dl) = I_(2) B sin (90^(@) - theta)` . . .(2)
Thus, `(F)/(dl) = (mu_(0))/(2pi) (I_(1)I_(2))/(r) cos theta`
`F = (mu_(0) I_(1) I_(2) cos theta)/( 2 pi r)` Per unit length