
The value of $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$ is
A) $3 \times {10^6}{\text{m}}{{\text{s}}^{ - 1}}$
B) $3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$
C) $3 \times {10^4}{\text{m}}{{\text{s}}^{ - 1}}$
D) $322{\text{m}}{{\text{s}}^{ - 1}}$
Answer
474.3k+ views
Hint:Electromagnetic waves are considered to be synchronous oscillations of electric field oscillations and magnetic field oscillations. Both the electric field oscillations and the magnetic field oscillations are perpendicular to each other and are perpendicular to the direction of propagation of the electromagnetic wave. In the given quantity, ${\mu _0}$ denotes the permeability of free space and is related to the magnetic field component while ${\varepsilon _0}$ denotes the permittivity of free space and is related to the electric field component.
Complete step by step answer.
Step 1: List the values of ${\mu _0}$ and ${\varepsilon _0}$ to obtain the value of the given term.
The permittivity of a material refers to the capability of that material to allow or transmit electric fields. The permittivity of free space is represented by ${\varepsilon _0}$ and its value is known to be ${\varepsilon _0} = 8 \cdot 85 \times {10^{ - 12}}{{\text{C}}^2}{{\text{N}}^{ - 1}}{{\text{m}}^2}$.
Similarly, the permeability of a material refers to the ability of that material to allow the entry of magnetic field lines in the material. It generally describes the ability of magnetization possessed by the material for the applied magnetic field. The permeability of free space is represented by ${\mu _0}$ and its value is known to be ${\mu _0} = 4\pi \times {10^{ - 7}}{\text{N}}{{\text{A}}^{ - 2}} = 4\pi \times {10^{ - 7}}{\text{N}}{{\text{C}}^{ - 2}}{{\text{s}}^2}$.
We have to determine the value of $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$. So we can substitute the values ${\mu _0} = 4\pi \times {10^{ - 7}}{\text{N}}{{\text{A}}^{ - 2}}$ and ${\varepsilon _0} = 8 \cdot 85 \times {10^{ - 12}}{{\text{C}}^2}{{\text{N}}^{ - 1}}{{\text{m}}^2}$ to obtain the value of the given quantity.
i.e., $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }} = \dfrac{1}{{\sqrt {\left( {4\pi \times {{10}^{ - 7}}{\text{N}}{{\text{C}}^{ - 2}}{{\text{s}}^2}} \right) \times \left( {8 \cdot 85 \times {{10}^{ - 12}}{{\text{C}}^2}{{\text{N}}^{ - 1}}{{\text{m}}^2}} \right)} }} = 3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$
$\therefore $ the value of $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$ is obtained to be $3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$.
So the correct option is B.
Note: Alternate method
The given quantity actually represents the speed of the electromagnetic wave in free space i.e., $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }} = c$.
We know that the speed of electromagnetic wave which is basically light has a speed $c = 3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$ in free space. So the value of the given quantity $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$ will be $3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$ and the correct option will be B.
Complete step by step answer.
Step 1: List the values of ${\mu _0}$ and ${\varepsilon _0}$ to obtain the value of the given term.
The permittivity of a material refers to the capability of that material to allow or transmit electric fields. The permittivity of free space is represented by ${\varepsilon _0}$ and its value is known to be ${\varepsilon _0} = 8 \cdot 85 \times {10^{ - 12}}{{\text{C}}^2}{{\text{N}}^{ - 1}}{{\text{m}}^2}$.
Similarly, the permeability of a material refers to the ability of that material to allow the entry of magnetic field lines in the material. It generally describes the ability of magnetization possessed by the material for the applied magnetic field. The permeability of free space is represented by ${\mu _0}$ and its value is known to be ${\mu _0} = 4\pi \times {10^{ - 7}}{\text{N}}{{\text{A}}^{ - 2}} = 4\pi \times {10^{ - 7}}{\text{N}}{{\text{C}}^{ - 2}}{{\text{s}}^2}$.
We have to determine the value of $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$. So we can substitute the values ${\mu _0} = 4\pi \times {10^{ - 7}}{\text{N}}{{\text{A}}^{ - 2}}$ and ${\varepsilon _0} = 8 \cdot 85 \times {10^{ - 12}}{{\text{C}}^2}{{\text{N}}^{ - 1}}{{\text{m}}^2}$ to obtain the value of the given quantity.
i.e., $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }} = \dfrac{1}{{\sqrt {\left( {4\pi \times {{10}^{ - 7}}{\text{N}}{{\text{C}}^{ - 2}}{{\text{s}}^2}} \right) \times \left( {8 \cdot 85 \times {{10}^{ - 12}}{{\text{C}}^2}{{\text{N}}^{ - 1}}{{\text{m}}^2}} \right)} }} = 3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$
$\therefore $ the value of $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$ is obtained to be $3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$.
So the correct option is B.
Note: Alternate method
The given quantity actually represents the speed of the electromagnetic wave in free space i.e., $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }} = c$.
We know that the speed of electromagnetic wave which is basically light has a speed $c = 3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$ in free space. So the value of the given quantity $\dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$ will be $3 \times {10^8}{\text{m}}{{\text{s}}^{ - 1}}$ and the correct option will be B.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
State the laws of reflection of light

Arrange Water ethanol and phenol in increasing order class 11 chemistry CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

A mixture of o nitrophenol and p nitrophenol can be class 11 chemistry CBSE

How do I convert ms to kmh Give an example class 11 physics CBSE
