Question

A magnet in the form of a cylindrical rod has a length of 5.51 cm and a diameter of 0.865 cm. It has a uniform magnetization of 5.17 x 103 A/m. The magnetic dipole moment?

Answer 1

Magnetic dipole moment will be
`15.51* 10^(-3)J/T`

Step-by-step explanation:

We have given length of the cylinder , that is h = 5.51 cm = 0.051 m

And diameter of the cylinder d = 0.865 cm

So radius
`r=(d)/(2)=(0.865)/(2)=0.4325cm=0.4325* 10^(-2)m`

So volume of cylinder
`V=\pi r^2h=3.14* (0.4325* 10^(-2))^2* 0.051=3* 10^(-6)m^3`

It is given there is uniform magnetization of
`M=5.17* 10^3A/m`

We have to fond the dipole moment

Dipole moment is equal to
`\mu =MV`, here M is magnetization and V is volume

So
`\mu =MV=5.17* 10^3* 3* 10^(-6)=15.51* 10^(-3)J/T`