Question

A copper wire with cross-sectional area $S=2.5m{m}^2$ bent to make three sides of a square can turn about a horizontal axis $O{O}^{\prime }$ (Fig.). The wire is located in uniform vertical magnetic field. Find the magnetic induction if on passing a current $I=16A$ through the wire the latter deflects by an angle $\theta ={20}^{\circ }$.

Answer 1

The Ampere forces on the sides $OP$ and $O^{\prime }{P}^{\prime }$ are directed along the same line, in opposite directions and have equal values, hence the net force as well as the net torque of these forces about the axis $O{O}^{\prime }$ is zero. The Ampere-force on the segment $P{P}^{\prime }$ is effective and is deflecting in nature.
In equilibrium (in the dotted position) the deflecting torque must be equal to the restoring torque, developed due to the weight of the shape.
Let, the length of each side be $l$ and $\rho$ be the density of the material then,
$ilB\left(l\cos \theta \right)=\left(Sl\rho \right)g\frac{l}{2}\sin \theta +\left(Sl\rho \right)g\frac{l}{2}\sin \theta +\left(Sl\rho \right)gl\sin \theta$
or, $i{l}^{2}B\cos \theta =2S\rho g{l}^2\sin \theta$
Hence, $B=\frac{2S\rho g}{i}\tan \theta$