Question

A cylindrical rod magnet has a length of $5\, cm$ and a diameter of $1\, cm .$ 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,
$M=I \pi r^{2} l=\left(5.30 \times 10^{3}\right) \times \frac{22}{7} \times\left(0.5 \times 10^{-2}\right)^{2}\left(5 \times 10^{-2}\right)$
$=2.08 \times 10^{-2} J / T$