Question

In the figure, a three sided frame is pivoted at AC and hang vertically. Its sides are each of the same length and have a linear density (in SI unit) equal to ${100}^{th}$ part of the current (in SI unit) flowing in the frame. which is in a uniform magnetic field of 200 mT directed upwards. The frame will be rotated through an angle $\frac{\pi }{x}$ radian in equilibrium. Find the value of x

Answer 1

Given,

Frame hangs vertically at A and C

Length of each side= same, l

Linear density, $\int =\frac{1}{100}kg{m}^{-3}$; I is current

$\overrightarrow{B}$, magnetic field =200mT(upward)

$\theta$= rotational in equilibrium $=\pi /x$

For equilibrium,

$\sum Forces=0$ (i)

$\sum Moment=0$ (ii)

$|implies$ Magnetic moment+ moment due to weight=0

$IBl\left(l\cos \theta \right)=mgl\sin \theta +2mg\frac{l\sin \theta }{2}$

$⟹\tan \theta =\frac{IBl}{2mg}=\frac{IBl}{2\int lg}=\frac{IB}{2\int g}=1$

From (1)

$⟹\tan \theta =1$

$⟹\theta =\pi /4$ and

$x=4$