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`?

Options

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

Answer

C

Explanation

Length of the component `dl` which is parallel to wire (1) is `dl cos theta`, so force on it.
`p = (mu_(0))/(4pi), (2i_(1) i_(2))/(r) (dl cos theta) = (mu_(0) i_(1) i_(2) dl cos theta)/(2pi r)`