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

Answer:

Magnetic dipole moment is 0.0683 J/T.

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, $M=5.3\times 10^3\ A/m$

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

$M=\dfrac{\mu}{V}$

$\mu=M\times \pi r^2\times l$

Where

r is the radius of rod, r = 0.0075 m

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

$\mu=0.0683\ J/T$

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

Answer 2

Answer:

$0.068317J/T$

Explanation:

Magnetization is defined as the magnetic moment per unit volume means

$Magnetization=\frac{Magnetic\ moment}{Volume}$

So magnetic moment = magnetization × volume

We have given magnetization = $5.3\times 10^3A/m$

Diameter=1.5 cm ,so radius $=\frac{1.5}{2}=0.75cm\ =0.0075m$

Area $A =\pi r^2=3.14\times 0.0075^2=1.766\times 10^{-4}m^2$

Length = 7.3 cm =0.073 m

So volume $=1.766\times10^{-4}\times 0.073=1.289\times 10^{-5}m^3$

Now magnetic moment = magnetization × volume $=5.30\times 10^3\times 1.289\times 10^{-5}=0.068317J/T$