Q. A bar magnet of moment M is cut into two identical pieces along the length. One piece is bent in the form of a semi circle. If two pieces are perpendicular to each other, then resultant magnetic moment is

(a) $\displaystyle (\frac{M}{\pi})^2 + (\frac{M}{2})^2 $

(b) $ \displaystyle \sqrt{\frac{M}{\pi})^2 + (\frac{M}{2})^2 }$

(c) $ \displaystyle \sqrt{\frac{M}{\pi})^2 – (\frac{M}{2})^2 }$

(d) $ \displaystyle \frac{M}{\pi} + \frac{M}{2} $

**Click to See Answer : **

$\displaystyle M_1 = m \times 2R $

$ \displaystyle M_1 = m \times 2 \times \frac{L}{\pi} $

$ \displaystyle M_1 = \frac{M}{\pi} $

For other Part ,

$ \displaystyle M_2 = \frac{M}{2} $

Since two magnetic moment are in perpendicular direction ;

Hence net magnetic moment is

$ \displaystyle M_R = \sqrt{M_1^2 + M_2^2} $

$ \displaystyle M_R = \sqrt{(\frac{M}{\pi})^2 + (\frac{M}{2})^2} $