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# The formula for the velocity of electromagnetic wave in vacuum is given byA.) $c=\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}$B.) $c=\dfrac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}$C.) $c=\sqrt{\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}}}$D.) $c=\sqrt{\dfrac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}}$

Last updated date: 22nd Jul 2024
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Hint: Recall the values of ${{\mu }_{_{0}}}$and ${{\varepsilon }_{0}}$, thereafter by rearranging terms and performing basic mathematical calculations, students will get the required formula easily.

We know that the values of $\dfrac{{{\mu }_{_{0}}}}{4\pi }$and $\dfrac{1}{4\pi {{\varepsilon }_{0}}}$are as follows:
$\dfrac{{{\mu }_{_{0}}}}{4\pi }={{10}^{-7}}H/m$--------(1)
$\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}$--------(2)
On rearranging terms, the value of ${{\mu }_{_{0}}}$and ${{\varepsilon }_{0}}$is given by,
${{\mu }_{_{0}}}=4\pi \times {{10}^{-7}}H/m$--------(3)
${{\varepsilon }_{0}}=4\pi \times 9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}$--------(4)
Now on multiplying equations (3) and (4), we get
$\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=4\pi \times {{10}^{-7}}\times \dfrac{1}{4\pi \times 9\times {{10}^{9}}}$
On solving,
$\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=\dfrac{1}{9\times {{10}^{16}}}$
This may be written as,
$\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=\dfrac{1}{{{\left( 3\times {{10}^{8}} \right)}^{2}}}$
We know that the speed of light in vacuum is, $\text{c=3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{8}}}\text{m/s}$
$\Rightarrow {{\mu }_{_{0}}}{{\varepsilon }_{0}}=\dfrac{1}{{{\left( c \right)}^{2}}}$
$\Rightarrow {{\left( c \right)}^{2}}=\dfrac{1}{{{\mu }_{_{0}}}{{\varepsilon }_{0}}}$
On taking square root,
$\Rightarrow c=\dfrac{1}{\sqrt{{{\mu }_{_{0}}}{{\varepsilon }_{0}}}}$
Hence, the correct option is B, i.e., $c=\dfrac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}$

Here, ${{\varepsilon }_{0}}$ is the permittivity of free space. Basically, this is a mathematical quantity that represents how much electric field is permitted (penetrated) in free space or vacuum.
In this solution, $\dfrac{1}{4\pi {{\varepsilon }_{0}}}=9\times {{10}^{9}}N-{{m}^{2}}/{{C}^{2}}$this number tells us that $9\times {{10}^{9}}$field lines are crossing by a charge in a vacuum but for any medium, this number may change and the number of field lines penetrating is also changed.
Also ${{\mu }_{_{0}}}$is the permeability of free space (vacuum) is a physical constant equal to $1.257\times {{10}^{-6}}H/m$(approximately). Permeability in general is symbolized by µ, and is a constant of proportionality that exists between magnetic flux density and magnetic field strength in a given medium. In certain metals, notably iron and nickel and alloys containing them, µ is greater than ${{\mu }_{_{0}}}$.
Note: Students should try to memorize this kind of very usual relationship. If they do not remember then it is a better option to write basic rememberable values of ${{\varepsilon }_{0}}$ and ${{\mu }_{_{0}}}$ thereafter performing some calculation and rearranging of terms they will get final relation easily.