Question

An iron rod of volume \[{10^{ - 3}}{\text{ }}{{\text{m}}^3}\]and relative permeability \[1000\] is placed as core in a solenoid with \[10{\text{ turns/cm}}\]. If a current of \[0.5\]A is passed through the solenoid, then the magnetic moment of the rod will be:
A. \[0.5 \times {10^2}{\text{ A}}{{\text{m}}^2}\]
B. \[50 \times {10^2}{\text{ A}}{{\text{m}}^2}\]
C. \[5 \times {10^2}{\text{ A}}{{\text{m}}^2}\]
D. \[500 \times {10^2}{\text{ A}}{{\text{m}}^2}\]

Answer 1

Hint: In this question, we need to find the magnetic moment of the rod. Magnetic moment depends on the number of turns of an object, current, area and the relative permeability.

Formula used:
The magnetic moment of an iron rod is given by
\[M = {\mu _r}niV\]
\[M\] is the magnetic moment of a rod, \[\mu \] is the relative permeability, \[n\] is the number of turns per unit length, \[i\] is the current, \[V\] is the volume of a rod.

Complete step by step solution:
The magnetic moment of an iron rod is \[M = {\mu _r}niV\].
Here, we need to convert the number of terms per centimeter into the number of turns per meter.
\[n = 10{\text{ turns/cm}}\\
{\text{n}} = 1000{\text{ turns/m}}\\ \]
Thus, we get
\[M = 1000 \times 1000 \times 0.5 \times {10^{ - 3}}\]
By simplifying, we get
\[M = 500{\text{ A}}{{\text{m}}^2}\]
That is \[M = 5 \times {10^2}{\text{ A}}{{\text{m}}^2}\]

Therefore, the option (B) is correct.

Additional Information: The magnetic moment is a unit of measurement that indicates the magnetic strength and direction of a magnet or other item that generates a magnetic field. A magnetic moment is a characteristic of a magnet that interacts with an applied field to produce a physical moment. The magnetic moment's orientation is from the magnet's south pole to its north pole. The magnetic field generated by the magnet is proportionate to its magnetic moment.

Note: This example can also be solved using another way.
\[I = \left( {\mu - 1} \right)ni\\
\Rightarrow I = \left( {1000 - 1} \right)1000 \times 0.5\\
\Rightarrow I = 4.99 \times {10^5}{\text{ A}}{{\text{m}}^{ - 1}}\\ \]
\[M = IV\\
\Rightarrow M = 4.9 \times {10^5} \times {10^{ - 3}}\\
\Rightarrow M = 4.9 \times {10^2}\\ \]
\[M \approx 5 \times {10^2}{\text{ A}}{{\text{m}}^2}\]
In this way, we can find the magnetic moment of a rod.