Wires 1 and 2 carrying currents $i_{1} and i_{2}$ respectively are inclined at an angle $θ$ to each other. What is the force on a small element dl of wire 2 at a distance r from wire 1 due to magnetic field of wire l

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\frac{μ_{0}}{2 π r} i_{1} i_{2} d l \sin θ

\frac{μ_{0}}{4 π r} i_{1} i_{2} d l \sin θ

\frac{μ_{0}}{4 π r} i_{1} i_{2} d l [1 + \cos θ]

\frac{μ_{0}}{2 π r} i_{1} i_{2} d l \cos θ

answer is C.

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Detailed Solution

Length of the component dl which is parallel to wire (i) is

l = d l \cos θ

Force acting on it is

$F = \frac{μ_{0}}{2 π} = \frac{i_{1} i_{2} l}{r} = \frac{μ_{0}}{2 π r} i_{1} i_{2} (d l \cos θ)$

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