Question

A magnetic needle of length $l$ and poles of pole strength $\pm m$ is attached to a wooden bar of length $L$ as shown in figure and placed in a uniform magnetic field of intensity $B$. The bar and needle can revolve around point $\text{O}$. Find the magnitude of the moment of force that will cause the rotation of the bar around point $\text{O}$ if the bar makes an angle $\alpha$ with the direction of the lines of force of the magnetic field.

• A. $mBl\sin \alpha$
• B. $mBl\cos \alpha$
• C. $2mBl\sin \alpha$
• D. $2mBl\cos \alpha$

Answer 1

The correct option is A $mBl\sin \alpha$


The moment of forces on the poles of the needle about point $\text{O}$ will be,

${\tau }_1=mB(l+L)\sin \alpha$ (in clockwise direction)

And,
${\tau }_2=mBL\sin \alpha$ (in counter-clockwise direction)

Hence, the net moment of the force acting on the needle is,

$\tau ={\tau }_1-{\tau }_2$

$\tau =mB(l+L)\sin \alpha -mBL\sin \alpha$

$\therefore \tau =mBl\sin \alpha$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(A\right)$ is the correct answer.