Question

A cylindrical rod magnet has a length of $5cm$ and a diameter of $1cm.$ It has a uniform magnetisation of $5.30\times 10^{3}A/m^3.$ What its magnetic dipole moment?

• A. $1\times 10^{-2}J/T$
• B. $2.08\times 10^{-2}J/T$
• C. $3.08\times 10^{-2}J/T$
• D. $1.52\times 10^{-2}J/T$

Answer 1

Answer: (B)

Relation for dipole moment is, $M=I\times V$.
Volume of the cylinder $V=\pi r^{2}l$, where $r$ is the radius and $l$ is the length of the cylinder.
Then dipole moment,
$M=I\pi r^{2}l=\left(5.30\times 10^3\right)\times \frac{22}{7}\times {\left(0.5\times 10^{-2}\right)}^2\left(5\times 10^{-2}\right)$
$=2.08\times 10^{-2}J/T$