Answer
Verified
418.8k+ views
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.
Complete step by step answer:
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., \[c=\dfrac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}\]
Additional Information:
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.
Complete step by step answer:
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., \[c=\dfrac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}\]
Additional Information:
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is pollution? How many types of pollution? Define it