Question

A bar magnet of magnetic moment $80A{m}^2$ is cut into $16$ equal parts (both parallel to length as well as breadth). The magnetic moment of each part is

• A. $15A{m}^2$
• B. $10A{m}^2$
• C. $20A{m}^2$
• D. $5A{m}^2$

Answer 1

The correct option is D $5A{m}^2$

For each part, Magnetic length $=L/4$
Pole strength $=m/4$
$\therefore$ Magnetic moment is reduced by a factor of 16 for each part.
$\therefore M_{new}=80/16=5A{m}^2$