Question

A cylindrical rod magnet has a length of 5.0cm and a diameter of 1.0cm. It has a uniform magnetization of $5.3\times {{10}^{3}}A{{m}^{-1}}$. What is its magnetic dipole moment?
A. $20.8mJ{{T}^{-1}}$
B. $10.8mJ{{T}^{-1}}$
C. $5.8mJ{{T}^{-1}}$
D. $30.8mJ{{T}^{-1}}$

Answer 1

Hint: We have to look for a formula that connects magnetization with magnetic dipole moment of an object. For an object of uniform magnetization, we have the formula in which magnetic dipole moment is given by the product of objects magnetization and its volume. We are given the magnetization of the rod directly and we could find the rod’s volume from the given diameter and length. Also,
$1A{{m}^{2}}=1J{{T}^{-1}}$

Formula used:
Expression for magnetic moment,
$m=M\times V$

Complete step by step answer:
We are given a cylindrical rod magnet and we know its length, diameter and also its magnetization.
We have to find some formula that could find the magnetic dipole moment of the given cylindrical rod with the help of the given quantities in the question.
Magnetic moment represents the magnetic strength and orientation of a magnet. Dimension of Magnetic dipole moment is as that of current times area or energy divided by magnetic flux density. Its SI unit is $A{{m}^{2}}$. It is analogous to the electric dipole moment but magnetic fields lack a ‘charge’ counterpart unlike the electric fields. There exists no magnetic monopoles and hence magnetic dipoles are known to be the fundamental unit that produces magnetic fields. Let us represent magnetic dipole moment by m.
Now, we can look for magnetic dipole moment’s relationship with a magnet’s magnetization. Magnetization gives the density of magnetic dipole moments in a magnetic material. That is, if magnetic dipole gives you the magnetic strength of the whole object, magnetization gives you the net magnetic moment of a particular portion of the object.
So, for an elementary volume $dV$, magnetization (M) is given by,
$M=\dfrac{dm}{dV}$
Where, dm is the elementary magnetic moment. Therefore,
$dm=MdV$ ……………….. (1)
We could integrate equation (1) to get the net magnetic moment.
$m=\iiint{MdV}$
Integration is done for the entire volume hence the triple integral.
For a body with uniform magnetization, magnetic dipole moment is given by,
$m=M\times V$ …………….. (2)
This is exactly the relation required to solve the given question as we are given a cylindrical rod of uniform magnetization.
Magnetization,
$M=5.3\times {{10}^{3}}A{{m}^{-1}}$ ……………………. (3)
Volume of the cylindrical rod, $V=\pi {{r}^{2}}l$
Radius, $r=\dfrac{1cm}{2}=0.5cm=0.5\times {{10}^{-2}}m$
Length, $l=5.0cm=5\times {{10}^{-2}}m$
Now, volume is,
$V=\pi \times {{\left( 0.5\times {{10}^{-2}} \right)}^{2}}\times \left( 5\times {{10}^{-2}} \right){{m}^{3}}$
$V=3.93\times {{10}^{-6}}{{m}^{3}}$ ……………….. (4)
Substituting (3) and (4) in equation (2), we get,
$m=\left( 5.3\times {{10}^{3}} \right)A{{m}^{-1}}\times \left( 3.93\times {{10}^{-6}} \right){{m}^{3}}$
$m=20.8\times {{10}^{-3}}A{{m}^{2}}$
We know, $1A{{m}^{2}}=1J{{T}^{-1}}$
Therefore, we get the magnetic dipole moment of given cylindrical rod magnet as,
$m=20.8\times {{10}^{-3}}J{{T}^{-1}}=20.8mJ{{T}^{-1}}$
Hence, the solution of the given question is option A.

Note:
If the given rod had a non-uniform magnetization, we had to integrate the magnetization over the entire volume using the equation,
 $m=\iiint{MdV}$
And hence find the magnetic dipole moment. We use both $A{{m}^{2}}$ as well as $J{{T}^{-1}}$ units for magnetic dipole moment as it is said to have the dimensions of current times area or energy divided by magnetic flux density.