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Wires 1 and 2 carrying currents i₁ and i₂ 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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$\frac{μ_{0}}{4 πr} i_{1} i_{2} dlsin θ$

$\frac{μ_{0}}{2 πr} i_{1} i_{2} dlcos θ$

$\frac{μ_{0}}{2 πr} i_{1} i_{2} dlsin θ$

$\frac{μ_{0}}{4 πr} i_{1} i_{2} dl [1 + \cos θ]$

answer is A.

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Detailed Solution

At the location of the element magnetic field produced by wire 1 , $B_{1} = \frac{μ_{0} i_{1} (1 + \cos θ)}{4 π r}$

Therefore magnitude of force experienced by the element , dF = $i_{2} . d l . B_{1}$

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