Question

The SI unit of magnetic permeability is –
\[\begin{align}
& \text{A) A}{{\text{m}}^{-1}} \\
& \text{B) A}{{\text{m}}^{-2}} \\
& \text{C) H}{{\text{m}}^{-2}} \\
& \text{D) H}{{\text{m}}^{-1}} \\
\end{align}\]

Answer 1

Hint: We need to understand the usage of the constant magnetic permeability so that we can relate it to the physical parameter involved, then compare the units and find the SI unit which is the required solution in this case.

Complete answer:
The magnetic permeability is a characteristic feature of a medium. It is the ability of the medium or material to allow the maximum utilisation of the available magnetic field lines.
The magnetic permeability appears as the proportionality constant in the Biot-Savart’s law. According to the law, the magnetic field at a point is proportional to the current flowing through the conductor, the length of the conductor and inversely proportional to the square of the distance between the conductor and the point.
i.e.,
\[\begin{align}
& B=\dfrac{\mu }{4\pi }\dfrac{Idl\sin \theta }{{{r}^{2}}} \\
& \therefore \mu =\dfrac{4\pi {{r}^{2}}B}{Idl\sin \theta } \\
\end{align}\]
Where, \[\mu \] is the magnetic permeability, B is the magnetic field, I is the current and dl is the element of length of the conductor.
We can use the physical parameter involved to get the magnetic permeability after removing the constants involved. We can write the magnetic permeability in terms of the SI units as of the given parameter as –
\[\begin{align}
& \mu =\dfrac{4\pi {{r}^{2}}B}{Idl\sin \theta } \\
& \Rightarrow [\mu ]=\dfrac{{{m}^{2}}T}{Am} \\
& \therefore [\mu ]=\dfrac{Tm}{A} \\
\end{align}\]
Now, let us find how the unit of magnetic field Tesla, ‘T’ can be further broken in terms of the current unit Ampere ‘A’. From the Lorentz’ force we know that the magnetic field B can be written as –
\[\begin{align}
& F=qvB \\
& \Rightarrow B=\dfrac{F}{qv} \\
& \Rightarrow [B]=\dfrac{N}{Cm{{s}^{-1}}} \\
& \text{but, }C{{s}^{-1}}=A \\
& \Rightarrow [B]=\dfrac{N}{Am} \\
& \therefore [B]=\dfrac{J}{A{{m}^{2}}} \\
\end{align}\]
Now, we can need to convert the Joules and the Ampere into the units of Henry. The Henry is basic Si units can be defined as the work done per square current flow in an inductor. i.e.,
\[H=\dfrac{J}{{{A}^{2}}}\]
We can apply this relation in the magnetic field units to get –
\[\begin{align}
& H=\dfrac{J}{{{A}^{2}}} \\
& [B]=\dfrac{J}{A{{m}^{2}}} \\
& \therefore [B]=\dfrac{HA}{{{m}^{2}}} \\
\end{align}\]
Now, we can substitute this in the equation for the magnetic permeability to get its unit as –
\[\begin{align}
& [B]=T=\dfrac{HA}{{{m}^{2}}} \\
& [\mu ]=\dfrac{Tm}{A} \\
& \Rightarrow [\mu ]=\dfrac{\dfrac{HA}{{{m}^{2}}}m}{A} \\
& \therefore [\mu ]=\dfrac{H}{m}=H{{m}^{-1}} \\
\end{align}\]
The SI unit of magnetic permeability is \[H{{m}^{-1}}\].

The correct answer is option D.

Note:
The magnetic permeability is the characteristic property of a medium in which the magnetic fields are produced. The air has a comparatively low permeability because of which we use iron cores in the inductors for better trapping of the magnetic fields.