Question

A cylindrical magnetic rod has a length of $5{{ cm}}$ and a diameter of $1{{ cm}}$. It has a uniform magnetization of $5.3 \times {10^3}{{ A/m}}$. What is its magnetic dipole moment?

Answer 1

Hint: First, we must understand what is meant by a magnetic dipole moment and how it is calculated. Then we can continue to learn how it changes with the variation of other factors. Finally, we will use this concept to figure out our desired answer. In this problem, we will try to think about how the magnet's magnetic dipole moment is related to its dimension and magnetization per unit volume.

Complete step by step solution:
The magnetic moment or magnetic dipole moment of an object is defined as the torque that an object can experience in a magnetic field produced by the magnet. The unit of magnetic dipole moment is ${{A}}{{.}}{{{m}}^2}$ or ${{Joule/Tesla}}$. Therefore, the magnetic dipole moment is also given by the total force applied on a particle to displace it by a unit length in a unit magnetic flux density. In other forms, the magnetic moment is given by the product of magnetization per unit volume of the magnet's magnet and volume.
In the simplest form, magnetic dipole moment can be written as-
$\mu = M \times V$ ……….$(1)$
where,
$M$ is the magnetization per unit volume
$V$ is the volume of the magnet
Hence, the magnetic dipole moment of an object is directly proportional to the volume of that object.
From the equation $(1)$, we have-
$\mu = M \times \pi {r^2}l$ ……….$(2)$
where,
$r$ is the radius of the cylindrical magnet
$l$ is the length of the cylindrical magnet
Therefore, we put
$\Rightarrow M = 5.3 \times {10^3}{{A/m}}$, $r = 0.5 \times {10^{ - 2}}{{m}}$, and $l = 5 \times {10^{ - 2}}{{m}}$ in the equation $(2)$-
$\Rightarrow \mu = \left( {5.3 \times {{10}^3}} \right) \times \pi \times {\left( {0.5 \times {{10}^{ - 2}}} \right)^2} \times \left( {5 \times {{10}^{ - 2}}} \right)$
Putting $\pi = \dfrac{{22}}{7}$ we get,
$\Rightarrow \mu = 20.82 \times {10^{ - 3}}{{A}}{{.}}{{{m}}^2}$
We switch the unit from ${{A}}{{.}}{{{m}}^{ - 2}}$ to ${{J}}{{.}}{{{T}}^{ - 1}}$-
$\Rightarrow \mu = 20.82 \times {10^{ - 3}}{{J}}{{.}}{{{T}}^{ - 1}}$
$\because {{1J = 1}}{{{0}}^3}{{mJ}}$-
$\Rightarrow \mu = 20.82{{ mJ}}{{.}}{{{T}}^{ - 1}}$

Therefore, the required magnetic dipole moment of the given rod magnet is $20.82{{ mJ}}{{.}}{{{T}}^{ - 1}}$.

Note: Magnetic moments can be found in various physical systems such as permanent magnets, electric current-carrying loop (i.e., electromagnets), moving elementary particles (e.g., electrons). The magnetic dipole moment of an object can be measured with magnetometers.