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Physics-

General

Easy

Question

Wires 1 and 2 carrying currents $i subscript 1 end subscript$ and $i subscript 2 end subscript$ 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 wire1

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

The correct answer is: $fraction numerator mu subscript 0 end subscript over denominator 2 pi r end fraction i subscript 1 end subscript i subscript 2 end subscript d l cos invisible function application theta$

Length of the component dl which is parallel to wire (1) is $d l cos invisible function application theta$ , so force on it.
$F equals fraction numerator mu subscript 0 end subscript over denominator 4 pi end fraction. fraction numerator 2 i subscript 1 end subscript i subscript 2 end subscript over denominator r end fraction left parenthesis d l cos invisible function application theta right parenthesis equals fraction numerator mu subscript 0 end subscript i subscript 1 end subscript i subscript 2 end subscript d l cos invisible function application theta over denominator 2 pi r end fraction$

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