A cylindrical rod magnet has a length of 5 cm and a diameter of 1 cm. It has a uniform magnetization of `5.30xx10^(3) Amp//m^(3)`. What is it's magnetic dipole moment?

Options

A. `1xx10^(-2)J//T`
B. `2.08xx10^(-2)J//T`
C. `3.08xx10^(-2)J//T`
D. `1.52xx10^(-2)J//T`

Answer

B

Explanation

Relation for dipole moment is, `M=IxxV`. Volume of the cylinder `V=pir^(2)l`, where r is the radius and `l` is the length of the cylinder.
Then dipole moment
`M=Ipir^(2)l=(5.30xx10^(3))xx22/7xx(0.5xx10^(-2))^(2)(5xx10^(-2))`
`=2.08xx10^(-2)J//T`