Physics - Magnetism Question with Solution | TestHub

The work done in rotating a magnetic dipole in a uniform magnetic field from an initial angle ${\theta }_1$ to a final angle ${\theta }_2$ is given by:

$W=MB(\cos {\theta }_1-\cos {\theta }_2)$

Given that the dipole is initially parallel to the magnetic field, ${\theta }_1=0^{\circ }$. It is turned through an angle ${\theta }_2=30^{\circ }$. Substituting these values:

$W=MB(\cos 0^{\circ }-\cos 30^{\circ })$

$W=MB\left(1-\frac{\sqrt{3}}{2}\right)=MB\left(\frac{2-\sqrt{3}}{2}\right)$

The torque required to hold the magnetic dipole at an angle $\theta$ is given by:

$\tau =MB\sin \theta$

At $\theta =30^{\circ }$, the torque is:

$\tau =MB\sin 30^{\circ }=MB\left(\frac{1}{2}\right)$

From this, we can write $MB=2\tau$. Substituting this into the expression for work done:

$W=2\tau \left(\frac{2-\sqrt{3}}{2}\right)$

$W=\tau (2-\sqrt{3})$

$\tau =\frac{W}{2-\sqrt{3}}$

Answer: $\frac{W}{(2-\sqrt{3})}$