A square loop ABCD, carrying a current I₂, is placed near and coplanar with a long straight conductor XY carrying a current I₁, as shown in figure. The net force on the loop will be

$\frac{μ_{0} I_{1} I_{2}}{2 π}$

$\frac{μ_{0} I_{1} I_{2} L}{2 π}$

$\frac{2 μ_{0} I_{1} I_{2} L}{3 π}$

$\frac{2 μ_{0} I_{1} I_{2}}{3 π}$

Solution

$\frac{2 μ_{0} I_{1} I_{2}}{3 π}$

Force on arm AB due to current in conductor XY is

F₁ = $\frac{μ_{0}}{4 π} \frac{2 I_{1} I_{2} L}{(\frac{L}{2})}$

F₁ = $\frac{μ_{0} I_{1} I_{2}}{3 π}$

This force acting towards the wire in the plane of loop.

Force on arm CD due to current in conductor XY is

F₂ = $\frac{μ_{0}}{4 π} \frac{2 I_{1} I_{2} L}{4 π (\frac{3 L}{2})}$

F₂ = $\frac{μ_{0} I_{1} I_{2}}{3 π}$

F₂ is away from the wire the plane loop.

Ner force on the loop is

F = F₁ $-$ F₂

= $\frac{μ_{0} I_{1} I_{2}}{π} - \frac{μ_{0} I_{1} I_{2}}{3 π}$

= $\frac{μ_{0} I_{1} I_{2}}{π} [1 - \frac{1}{3}]$

F = $\frac{2}{3} \frac{μ_{0} I_{1} I_{2}}{π}$

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