Question

If the self-inductance of an air core inductor increases from 0.1mH to 20mH on the introduction of metal core into it. The relative permeability of the core used is
(A) 200
(B) 20
(C) 2
(D) 1

Answer 1

Permeability is the measure of magnetization that a material can obtain in response to the magnetic field applied on it. It is given by the symbol, ${\mu _{}}$ . Similarly relative permeability which is given by ${\mu _T}$ ,is nothing but the ratio of the permeability of any specific medium/substance to that of free space, which is given by ${\mu _0}$ .
${\mu _T} = \dfrac{\mu }{{{\mu _0}}}$
Formulas used: We will be using the formula for self-inductance of a substance,
${L_0} = \dfrac{{{\mu _0}{N^2}A}}{l}$ where ${L_0}$ is the self-inductance of a material, ${\mu _0}$ is the relative permeability in air / free space, $N$ is the number of turns in the solenoid, $A$ is the area of cross-section of the solenoid, and $l$ is the length of the solenoid.
We would also be using the formula, ${\mu _T} = \dfrac{\mu }{{{\mu _0}}}$ where ${\mu _T}$ is relative permeability, $\mu$ is permeability of a substance in a medium , and ${\mu _0}$ is permeability of a substance in free space .

Complete Step by Step answer:
We know that the self induction is the induction of voltage in a current carrying wire when the current in the wire itself is changing. And mathematically this can be given by, ${L_0} = \dfrac{{{\mu _0}{N^2}A}}{l}$ .
According to our question, we can see that the self-inductance changes when a metal core is introduced into the field of a solenoid.Thus we have two different self-inductances,
${L_0} = \dfrac{{{\mu _0}{N^2}A}}{l} = 0.1mH$ and $L = \dfrac{{\mu {N^2}A}}{l} = 20mH$ . Since we know all other factors of the solenoid remain unchanged, the only factor that reflects the change in self-inductance is the permeability of the media. Thus, dividing both the inductances we get,
$\dfrac{L}{{{L_0}}} = \dfrac{{20}}{{0.1}} = \dfrac{\mu }{{{\mu _0}}}$
Thus the relative permeability of the metal core given by, ${\mu _T} = \dfrac{\mu }{{{\mu _0}}}$ is equivalent to the ratio of the self-inductances, that is, ${\mu _T} = \dfrac{L}{{{L_0}}} = \dfrac{\mu }{{{\mu _0}}}$ .
$\Rightarrow {\mu _T} = \dfrac{\mu }{{{\mu _0}}} = \dfrac{{20}}{{0.1}}$
$\Rightarrow {\mu _T} = \dfrac{\mu }{{{\mu _0}}} = 20 \times 10 = 200$
Thus the relative permeability of the metallic core introduced in the solenoid is ${\mu _T} = 200$

Hence the correct answer is option A.

Note: The physical quantity permeability is measured in Henry(H). But relative permeability does not have units because it is a ratio of similar physical quantities in different media.