A thin bar magnet of length $2 L$ is bent at the mid-point so that the angle between them is $60^{\circ} \mathrm{C}$. The new length of the magnet is

• A

  $\sqrt{2} L$
• B

  $\sqrt{3} L$
• C

  $2 L$
• Correct answer
  D

  $L$

Solution

On bending the magnet, the length of the magnet
$\begin{aligned} A C & =A B+B C \\ & =L \sin \left(\frac{\theta}{2}\right)+L \sin \left(\frac{\theta}{2}\right) \\ & =2 L \sin \left(\frac{\theta}{2}\right)=2 L \sin 30^{\circ} \\ & =2 L \times \frac{1}{2}=L\end{aligned}$

Hence, the correct answer is option: 'D'.