Question

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.

Answer 1

Magnetic force is given by $F=IBl\sin \theta$

Magnetic field is perpendicular to the length of wire,so $\theta ={90}^0$

Magnetic force pm side $QR$,

$F_1=IBl\sin {90}^0=5\times 2\times 0.1=1N$

Magnetic force on side $PR$,

$F_2=IBl\sin {90}^0=5\times 2\times 0.1=1N$

Magnetic force on side $PQ$,

$F_3=IBl\sin {90}^0=5\times 2\times 0.1=1N$

Net magnetic force,

$\begin{array}{c}{F}^{\prime }=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos {120}^0}\\ =\sqrt{1+1-1}=1N\end{array}$

Resultant of $\overrightarrow{F}$ and $\overrightarrow{F_3}$

$F=1-1=0$

$\therefore$ Net magnetic force of the loop is $Zero$.