Question

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

Answer 1

Correct Answer - 3
Length of the combonent `dl` which is parallel to wire `(1)` is `dl cos theta`, so force on it
`F=(mu_(0))/(2pi)(i_(1)i_(2))/(r)(dl cos theta)`