
If the relative permeability of a medium is ${\mu _r}$ and its dielectric constant is ${\varepsilon _r}$ then the velocity of light in that medium will be:
A. $\sqrt {\dfrac{{{\mu _r}}}{{{\varepsilon _r}}}} $
B. $\dfrac{c}{{\sqrt {{\varepsilon _r}{\mu _r}} }}$
C. $\sqrt {{\mu _r}{\varepsilon _r}/{\mu _{{\varepsilon _0}}}} $
D. $\sqrt {{\mu _0}{\varepsilon _0}/{\mu _r}{\varepsilon _r}} $
Answer
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Hint:- According to the question, firstly we have to discuss the formulae of velocity of light which is related with relative permeability and dielectric constant. Then, we will change the medium and calculate the velocity of that changed medium.
Complete step by step solution:-
Given that-
Relative permeability of a medium= ${\mu _r}$
Dielectric Constant= ${\varepsilon _r}$
As we know that, the velocity of light is formulated by:
$c = \dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}$
here,
$c$ is the velocity of light
${\varepsilon _0}$ is the dielectric constant for the velocity of light
${\mu _0}$ is the permeability for the velocity of light.
Now, if we change the medium, then the velocity of medium is formulated as:
$v = \dfrac{1}{{\sqrt {\varepsilon \mu } }}$
(here,
$\varepsilon $ is the dielectric constant for the change in medium
$\varepsilon = {\varepsilon _0}{\varepsilon _r}$ or
Dielectric Constant for change in medium=Dielectric constant for velocity of light- dielectric constant
And, $\mu $ is the relative permeability for the change in medium.
$\mu = {\mu _0}{\mu _r}$ .)
$\because v = \dfrac{1}{{\sqrt {{\varepsilon _0}{\varepsilon _r}{\mu _0}{\mu _r}} }}$ $ = \dfrac{c}{{\sqrt {{\varepsilon _r}{\mu _r}} }}$
So, when we change the medium, then the velocity of that particular medium is $\dfrac{c}{{\sqrt {{\varepsilon _r}{\mu _r}} }}$ .
Hence, the correct option is (B.) $\dfrac{c}{{\sqrt {{\varepsilon _r}{\mu _r}} }}$ .
Note:- The dielectric constant and relative permittivity are vital to the activity of capacitors and the assurance of the degrees of capacitance reachable. Permittivity and dielectric steady are two terms that are vital to capacitor innovation. Frequently talk will be known about capacitors with various dielectrics being utilized.
Complete step by step solution:-
Given that-
Relative permeability of a medium= ${\mu _r}$
Dielectric Constant= ${\varepsilon _r}$
As we know that, the velocity of light is formulated by:
$c = \dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}$
here,
$c$ is the velocity of light
${\varepsilon _0}$ is the dielectric constant for the velocity of light
${\mu _0}$ is the permeability for the velocity of light.
Now, if we change the medium, then the velocity of medium is formulated as:
$v = \dfrac{1}{{\sqrt {\varepsilon \mu } }}$
(here,
$\varepsilon $ is the dielectric constant for the change in medium
$\varepsilon = {\varepsilon _0}{\varepsilon _r}$ or
Dielectric Constant for change in medium=Dielectric constant for velocity of light- dielectric constant
And, $\mu $ is the relative permeability for the change in medium.
$\mu = {\mu _0}{\mu _r}$ .)
$\because v = \dfrac{1}{{\sqrt {{\varepsilon _0}{\varepsilon _r}{\mu _0}{\mu _r}} }}$ $ = \dfrac{c}{{\sqrt {{\varepsilon _r}{\mu _r}} }}$
So, when we change the medium, then the velocity of that particular medium is $\dfrac{c}{{\sqrt {{\varepsilon _r}{\mu _r}} }}$ .
Hence, the correct option is (B.) $\dfrac{c}{{\sqrt {{\varepsilon _r}{\mu _r}} }}$ .
Note:- The dielectric constant and relative permittivity are vital to the activity of capacitors and the assurance of the degrees of capacitance reachable. Permittivity and dielectric steady are two terms that are vital to capacitor innovation. Frequently talk will be known about capacitors with various dielectrics being utilized.
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