Question

A cylindrical rod magnet has a length of 5.0 cm and a diameter of 1.0 cm. It has a uniform magnetization of 5.3*${10}^3$ A/m. What is its magnetic dipole moment?

• A. 20.8$mJ{T}^{-1}$
• B. 10.8$mJ{T}^{-1}$
• C. 5.8$mJ{T}^{-1}$
• D. 30.8$mJ{T}^{-1}$

Answer 1

The correct option is A 20.8$mJ{T}^{-1}$

$M=I\times V$ where volume of cylinder $v=\pi r^{2}l$

$M=I\pi r^{2}l=5.3\times {10}^3\times \frac{2}{2}7\times (0.5\times {10}^{-2}{)}^2\times (5\times {10}^{-2})$

$2.08\times {10}^{-2}J/T=20.8mJ{T}^{-1}$