Physics - Magnetism Question with Solution | TestHub

Given, $\theta =60^{\circ }=\frac{\pi }{3}\mathrm{rad}$ The new magnetic length of the dipole will be, $I^{\prime }=r$ Now, $M=m\cdot I$ and $M^{\prime }=m\cdot I^{\prime }$ $\therefore \frac{\mathrm{M}}{{\mathrm{M}}^{\prime }}=\frac{\mathrm{I}}{{\mathrm{I}}^{\prime }}\dots (1)$ Now, angle in radians $=\frac{\text{ arc length }}{\text{ radius }}$ $\therefore \frac{\pi }{3}=\frac{1}{r}\Rightarrow 1=\frac{\pi r}{3}$ In the equilateral triangle, marked by dotted lines, $I^{\prime }=r$ $\therefore \frac{1}{I^{\prime }}=\frac{\pi }{3}\dots (2)$ From and, we get, $\begin{array}{rl}& \frac{M}{M^{\prime }}=\frac{\pi }{3}\\ & \therefore M^{\prime }=\frac{3M}{\pi }\end{array}$ Hence, $(C)$ is the correct answer. Alternative method: Using the short cut formula about bending of a bar magnet, $M=\frac{2M\sin \left(\frac{\theta }{2}\right)}{\theta }$ $=\frac{2M\sin 30^{\circ }}{\frac{\pi }{3}}=\frac{3M}{\pi }$ Hence, (C) is the correct answer.