A wire is bent in the form of an equilateral triangle PQR of side 10 cm and carries a current of 0.5 A. It is placed in a magnetic field B of magnitude 2.0 T directed perpendicularly to the plane of the loop. Find the forces on the three sides of the triangle.

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Solution

Magnetic force is given by $F = I B l sin θ$
Magnetic field is perpendicular to the length of wire,so $θ = 90^{0}$
Magnetic force pm side $Q R$ ,
$F_{1} = I B l sin 90^{0} = 5 \times 2 \times 0.1 = 1 N$
Magnetic force on side $P R$ ,
$F_{2} = I B l sin 90^{0} = 5 \times 2 \times 0.1 = 1 N$
Magnetic force on side $P Q$ ,
$F_{3} = I B l sin 90^{0} = 5 \times 2 \times 0.1 = 1 N$
Net magnetic force,
$\begin{matrix} F^{'} = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos 120^{0}} = \sqrt{1 + 1 - 1} = 1 N \end{matrix}$
Resultant of $\to F$ and $- \to F_{3}$
$F = 1 - 1 = 0$
$∴$ Net magnetic force of the loop is $Z e r o$ .

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