Question

A magnet of moment $M$ is bent at its mid point so that angle between the two parts is ${120}^o$, the magnetic moment of the magnet now is

• A. $M$
• B. $M\sqrt{3}$
• C. $\frac{M\sqrt{3}}{2}$
• D. $\frac{M}{2}$

Answer 1

The correct option is C $\frac{M\sqrt{3}}{2}$

Let the initial length be $L$.
Consider an isosceles triangle with the angle between the two equal sides being at angle of ${120}^o$.
Let final length be $a$

Therefore by Cosine rule,
$a^2=(L/2{)}^2+(L/2{)}^2-2\left(L/2{)}^{2}cos\right({120}^0)$
$a^2=3L^2/4$
$a=\sqrt{3}L/2$
$M_{final}=m\sqrt{3}L/2$
$M_{final}=M\sqrt{3}/2$