Question

A magnet in the form of a cylindrical rod has a length of 7.30 cm and a diameter of 1.5 cm. It has a uniform magnetization of 5.30 x 103 A/m. What is its magnetic dipole moment in units of JIT?

Answer 1

`0.068317J/T`

Step-by-step explanation:

Magnetization is defined as the magnetic moment per unit volume means

`Magnetization=(Magnetic\ moment)/(Volume)`

So magnetic moment = magnetization × volume

We have given magnetization =
`5.3* 10^3A/m`

Diameter=1.5 cm ,so radius
`=(1.5)/(2)=0.75cm\ =0.0075m`

Area
`A =\pi r^2=3.14* 0.0075^2=1.766* 10^(-4)m^2`

Length = 7.3 cm =0.073 m

So volume
`=1.766*10^(-4)* 0.073=1.289* 10^(-5)m^3`

Now magnetic moment = magnetization × volume
`=5.30* 10^3* 1.289* 10^(-5)=0.068317J/T`

Answer 2

Magnetic dipole moment is 0.0683 J/T.

Step-by-step explanation:

It is given that,

Length of the rod, l = 7.3 cm = 0.073 m

Diameter of the cylinder, d = 1.5 cm = 0.015 m

Magnetization,
`M=5.3* 10^3\ A/m`

The dipole moment per unit volume is called the magnetization of a magnet. Mathematically, it is given by :

`M=(\mu)/(V)`

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

Where

r is the radius of rod, r = 0.0075 m

`\mu=5.3* 10^3\ A/m* \pi (0.0075)^2* 0.073\ m`

`\mu=0.0683\ J/T`

So, its magnetic dipole moment is 0.0683 J/T. Hence, this is the required solution.