Question

A magnet in the form of a cylindrical rod has a length of $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$and a diameter of$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$. It has a uniform magnetization of$\mathbf{5}\mathbf{.}\mathbf{30}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{A}\mathbf{/}\mathbf{m}$. What is its magnetic dipole moment?

Short answer

The magnetic dipole moment is $2.08\times {10}^{-2}\mathrm{J}/\mathrm{T}$.

Step-by-step solution

Listing the given quantities 

Length of cylinder,$\mathrm{l}=5.0\times {10}^{-2}\mathrm{m}$

Diameter of cylinder,$2\mathrm{r}=1.0\times {10}^{-2}\mathrm{m}$

Magnetization,$\mathrm{M}=5.3\times {10}^3\mathrm{A}/\mathrm{m}$

Understanding the concepts of magnetization

We know that magnetic moment per unit volume is magnetization. We can find the magnetic dipole moment by substituting the given values in the formula for magnetization.

Formula:

$\mathrm{\mu }=\mathrm{M}\cdot {\mathrm{\pi r}}^2\mathrm{l}$

 Step 3: Calculations of the magnetic dipole moment

Magnetization is given by

$\mathrm{M}=\frac{\mathrm{\mu }}{\mathrm{V}}$

Where,$\mathrm{V}$is the volume and $\mathrm{\mu }$is the magnetic dipole moment

The volume of the cylinder is given by$\mathrm{V}={\mathrm{\pi r}}^2\mathrm{l}$

Therefore,

$\begin{array}{c}\mathrm{M}=\frac{\mathrm{\mu }}{\mathrm{V}}\\ =\frac{\mathrm{\mu }}{{\mathrm{\pi r}}^2\mathrm{l}}\end{array}$

Rearranging the above equation, we get,

$\begin{array}{c}\mathrm{\mu }=\mathrm{M}\cdot {\mathrm{\pi r}}^2\mathrm{l}\\ =(5.3\times {10}^3)\times \mathrm{\pi }\times {\left(0.5\right)}^2\times {10}^{-4}\times 5.0\times {10}^{-2}\\ =2.08\times {10}^{-2}\mathrm{J}/\mathrm{T}\end{array}$

The magnetic dipole moment is $2.08\times {10}^{-2}\mathrm{J}/\mathrm{T}$.