Question

Find the magnetization of a BAR magnet of length $10{\text{ cm}}$ and cross section area $4{\text{ c}}{{\text{m}}^2}$ if the magnetic moment is $2{\text{ A}}{{\text{m}}^2}$?

Answer 1

Hint: Magnetic moment: The magnetic moment is the magnetic strength and orientation of a magnet or other object that produces a magnetic field for example loops of electric current, permanent magnets moving elementary particles various molecules and Astronomical objects. Moreover the term magnetic moment refers to a system of magnetic dipole moment, the component of the magnetic moment, that can be represented by an equivalent magnetic dipole: a magnetic North and South pole separated by a very small distance.

Formula Used:
The formula which will uses here is
${M_\tau } = \dfrac{M}{V}$
Here ${M_\tau }$ is the magnetization of a Bar magnet which we have to find.
$M$ is the magnetic moment and formula for the volume is
$\operatorname{Volume} {\text{ = }}\operatorname{Area} {\text{ }} \times {\text{ }}\operatorname{Length} $

Complete step by step solution:
Now here we have the given terms are:
Length of the BAR magnet = $10\,{\text{ }}cm$
Now we have to convert $10cm$ into meters.
So we have
$L = 10cm = 10 \times {10^{ - 2}}meter$
Now
Area of the cross section $ = 4c{m^2}$
Now in $S.I$ system $ = 4 \times {10^{ - 4}}{m^2}$
Given magnetic moment is $2A{m^2}$
Now the formula we have
${M_\tau } = \dfrac{M}{V}$
Firstly we have to find out the volume.
$\operatorname{Volume} {\text{ = }}\operatorname{Area} {\text{ }} \times {\text{ }}\operatorname{Length} $
$V = 4 \times {10^{ - 4}} \times 10 \times {10^{ - 2}}$
Now
$V = 4 \times {10^{ - 5}}{m^3}$
Now putting the value of magnetic moment and volume in the given formula for magnetization.
\[{M_\tau } = \dfrac{{2A{m^2}}}{{4 \times {{10}^{ - 5}}{m^3}}}\]
Now
${M_\tau } = 0.5 \times {10^5}A{m^{ - 1}}$

Now after solving the above we have ${M_\tau } = 50000A{m^{ - 1}}$.

Additional Information:
The magnetic moment of objects are typically measured with devices called magnetometers, though not all magnetometers measure magnetic moment; some are configured to measure magnetic filled instead.

Note: Mathematically magnetic moments can be denoted as
$\tau = M \times B$
Where $\tau = $ tall called torque on the dipole.
$B$ is the external field and \['M'\] is the magnetic moment.
The $S.I$ unit of magnetic moment is $A{m^2}$. Where $A$ is Ampere and $'m'$ is the meter.
Also we can write.
$A.{m^2} = \dfrac{{N.M}}{T} = \dfrac{J}{T}$
$N$ is Newton. $J$ is the Joule . $T$ is tesla
Where as in $C.G.S$ system, there are several different sets of Electromagnetism units,
$1$ stat $Ac{m^2}$ = $3.33564095 \times {10^{ - 14}}A{m^2}$
Stat Ampere = stat $A$