Question

If the relative permeability of iron is 2000. It’s absolute permeability in S.I units is
A. \[8\pi \times {10^{ - 4}}\]
B. \[8\pi \times {10^{ - 3}}\]
C. \[\dfrac{{800}}{\pi }\]
D. \[\dfrac{{8\pi \times {{10}^9}}}{\pi }\]

Answer 1

Hint: The absolute permeability of a material is equal to the product of the permeability in free space and the relative permeability of the medium.

Complete step by step solution:
The relation between absolute permeability and relative permeability is as follows:
Absolute permeability is equal to \[4\pi \times {10^{ - 7}}\] times the relative permeability of the material in its respective SI unit.
Therefore, from the above relation we get:
\[\mu = {\mu _0}{\mu _r}\]

Where,
\[\mu \] is the absolute permeability of the material in SI units
\[{\mu _0}\] is equal to \[4\pi \times {10^{ - 7}}\]
\[{\mu _r}\] is the relative permeability of the given material.
Therefore, from the above relation on putting the values in the given formula for absolute permeability we get,
\[\begin{array}{l}
\mu = {\mu _0}{\mu _r}\\
 \Rightarrow \left( {4\pi \times {{10}^{ - 7}}} \right) \times 2000
\end{array}\]

Now after multiplying the values we get,
\[ \Rightarrow 8\pi \times {10^{ - 4}}\]

The above answer is in terms of the absolute permeability of the given material in terms of its SI units.

Additional information: The term absolute permeability \[\mu \] of a material is the product of the permeability of free space \[{\mu _0}\] and the relative permeability of the material \[{\mu _r}\].
The relative permeability of a medium is defined as the force between two poles when it is placed at a fixed distance in vacuum to the force between them at the same distance in that medium.
The relative permeability in its essence is a purely numerical number and therefore has no units. As mentioned before, for air and nonmagnetic materials its values are always unity or equal to 1

Note: Students often do not know the relation between the absolute and relative permeabilities of a medium, so it is extremely hard to solve sums like this for them. Students always need to know the equations with certain quantities with other such quantities.