Question

A thin bar magnet of length L is bent in the middle to form an angle of ${60}^0$. The new length of the magnet is :

• A. $2L$
• B. $L$
• C. $\frac{L}{\sqrt{2}}$
• D. $\frac{L}{2}$

Answer 1

$AD=DC=\frac{L}{2}sin{30}^0$
$\therefore AC=\frac{L}{2}\sin {30}^0+\frac{L}{2}sin{30}^0=Lsin{30}^0=\frac{L}{2}$