A thin bar magnet of length 2 L is bent at the mid-point so that the angle between them is 60°. The new length of the magnet is

$\frac{L}{2 \sqrt{3}}$

$\frac{\sqrt{3} L}{2}$

L

$\frac{2 L}{3}$

Solution

On bending the magnet, the angle between two parts of the magnet is expressed in the diagram below

Now, the minimum distance between the ends P and R

= L sin 30^o + L sin30^o

= 2L sin30^o

= 2L × $\frac{1}{2}$ ....( sin30^o = $\frac{1}{2}$ )

= L

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