Question

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

The correct option is B $2.08\times {10}^{-2}J/T$

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,
$M=I\pi r^{2}l=(5.30\times {10}^3)\times \frac{22}{7}\times (0.5\times {10}^{-2}{)}^2(5\times {10}^{-2})=2.08\times {10}^{-2}J/T$