Question

As shown in the figure, three sided frame is pivoted at $\text{P}$ and $\text{Q}$ and hangs vertically. Its sides are of same length and have a linear mass density of $\sqrt{3}{\text{kgm}}^{-1}$. A current of $10\sqrt{3}\text{A}$ is sent through the frame, which is placed in a uniform magnetic field of $2T$ directed upwards as shown. Then the angle (in degree) through which the frame will be deflected in equilibrium is (Take $g=10{\text{ms}}^{-2}$)

Answer 1

If we look at the position of frame having an angle $\theta$ in equilibrium, the magnetic forces acting on the two parallel sides will not yield any torque.

Torque due to magnetic force on wire frame about $\text{PQ}$ is (due to horizontal portion only)

${\tau }_m=\left(IlB\right)\times l\cos \theta$


Now considering that mass per unit length ${}^{\prime }{\lambda }^{\prime }$ , the torque due to gravitational force about $\text{PQ}$ on the wire frame (taking into account all three sections)

${\tau }_g=\left(\lambda lg\right)l\sin \theta +2\times \left[\right(\lambda l\left)g\frac{l\sin \theta }{2}\right]$

${\tau }_g=2\lambda l^{2}g\sin \theta$

Now at the equilibrium both torques must balance,

${\tau }_B={\tau }_g$

$BI{l}^2\cos \theta =2\lambda l^{2}g\sin \theta$

$\tan \theta =\frac{BI{l}^2}{2\lambda l^{2}g}=\frac{BI}{2\lambda g}$

Substituting the data given in the question,

$\tan \theta =\frac{2\times 10\sqrt{3}}{2\times \sqrt{3}\times 10}=1$

$\therefore \theta ={45}^{\circ }$

Accepted answers : $45,45.0,45.00$