Question

A magnet in the form of a cylindrical rod has a length of 5.31 cm and a diameter of 1.34 cm. It has a uniform magnetization of 6.46 × 103A/m. What is its magnetic dipole moment?

Answer 1

Step-by-step explanation:

Given that,

Length of the rod, l = 5.31 cm = 0.0531 m

Diameter of the rod, d = 1.34 cm

Radius of the rod, r = 0.0067 m

Magnetization in the rod,
`M=6.46* 10^3\ A/m`

We need to find the magnetic dipole moment of the rod. The magnetic dipole moment is given by:

`\mu=M* V`

V is the volume of the cylindrical rod

`\mu=M*\pi r^2l`

`\mu=6.46* 10^3* \pi (0.0067)^2* 0.0531`

`\mu=0.049\ J/T`

So, the magnetic dipole moment of the rod is 0.049 J/T. Hence, this is the required solution