Question

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. ${\left(\frac{M}{\pi }\right)}^2+({\frac{M}{2})}^2$
• B. $\sqrt{{\left(\frac{M}{\pi }\right)}^2+{\left(\frac{M}{2}\right)}^2}$
• C. $\sqrt{{\left(\frac{M}{\pi }\right)}^2-{\left(\frac{M}{2}\right)}^2}$
• D. $\frac{M}{\pi }+\frac{M}{2}$

Answer 1

Answer: (B)

$\sqrt{{\left(\frac{M}{\pi }\right)}^2+{\left(\frac{M}{2}\right)}^2}$

Given

Bar magnet of moment M is cut into two pieces along axis

Length of magnet is L

Pole strength be m

Solution

Magnetic moment M=m*L

On cutting through axis pole strength get halved.

Piece 1

wire is bent into semi circle of radius r

L=length of magnet=circumference of semi circle

L=$\pi r$

r=$\frac{L}{\pi }$

New Magnetic Moment M'=$\frac{m}{2}\ast 2r$

M'=$mL/\pi$=$M/\pi$

Piece 2

new magnetic moment will be M'=M/2

As both pieces are put perpendicularly

$M_{eff}=\left[\right(\frac{M}{\pi }{)}^2+(\frac{M}{2}{)}^2{]}^{\frac{1}{2}}$

The correct answer is B