Question

A wire of cross-sectional are A forms three sides of a square and is free to rotate about axis OO'. If the structure is deflected by an angle $\theta$ from the vertical when current i is passed through it in a magnetic field B acting vertically upwards and density of the wire is $\rho$, then the value of $\theta$ is given by

• A. $\frac{2A\rho g}{iB}=cot\theta$
• B. $\frac{2A\rho g}{iB}=tan\theta$
• C. $\frac{A\rho g}{iB}=sin\theta$
• D. $\frac{A\rho g}{2iB}=cos\theta$

Answer 1

Answer: (A)

$\frac{2A\rho g}{iB}=cot\theta$

The torque on the current loop due to the magnetic field

${\tau }_B=i{a}^{2}Bsin(90-\theta )=i{a}^{2}Bcos\left(\theta \right)$

Torque due to gravity,

${\tau }_g={\tau }_{O{A}^{\prime }}+{\tau }_{A^{\prime }{B}^{\prime }}+{\tau }_{O^{\prime }{B}^{\prime }}=mg\frac{a}{2}sin\left(\theta \right)+mgasin\left(\theta \right)+mg\frac{a}{2}sin\left(\theta \right)$

In equilibrium, torque due to magnetic force is balanced by torque due to gravity.

$i{a}^{2}B\cos \theta =2mga\sin \theta$

$cot\theta =\frac{2mg}{iaB}=\frac{2Aa\rho g}{iaB}=\frac{2A\rho g}{iB}$