Question

The quantity \sqrt{\mu_{0} \varepsilon_{0}} represents
(a) speed of sound
(b) speed of light in vacuum
(c) speed of e.m.w.
(d) inverse of speed of light in vacuum

Answer 1

Answer:

Electromagnetic waves are created by the vibration of an electric charge. This vibration creates a wave which has both an electric and a magnetic component. An electromagnetic wave transports its energy through a vacuum at a speed of 3.00 x 108 m/s (a speed value commonly represented by the symbol c).

Answer 2

The quantity $\sqrt {\frac{1}{{{\mu _0}{\varepsilon _0}}}}$ represents the speed of the light in vacuum. Therefore, the correct answer is option (b) speed of light in vacuum.

Explanation:

• From electrostatics,  the Coulomb's constant is given by

        $\frac{1}{{4\pi {\varepsilon _0}}} = 9.0 \times {10^9}{{N{m^2}} \mathord{\left/ {\vphantom {{N{m^2}} {{C^2}}}} \right. \kern-\nulldelimiterspace} {{C^2}}}$      ... (i)

       Here, ${\varepsilon _0}$ is the permittivity of the free space.

• From magnetostatics, the constant from the Biot-Savart law is

        $\frac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}N \cdot {A^2}$                   ... (ii)

        Here, ${{\mu _0}}$ is the permeability of free space.

• Now, dividing the equation (i) by equation (ii), we get,

       $\[\begin{gathered} \frac{{\frac{1}{{4\pi {\varepsilon _0}}}}}{{\frac{{{\mu _0}}}{{4\pi }}}} = \frac{{9.0 \times {{10}^9}{{N{m^2}} \mathord{\left/ {\vphantom {{N{m^2}} {{C^2}}}} \right. \kern-\nulldelimiterspace} {{C^2}}}}}{{{{10}^{ - 7}}N{A^2}}} \hfill \\ \frac{1}{{{\mu _0}{\varepsilon _0}}} = 9.0 \times {10^{16}}{{{m^2}} \mathord{\left/ {\vphantom {{{m^2}} {{s^2}}}} \right. \kern-\nulldelimiterspace} {{s^2}}} \hfill \\ \end{gathered} \]$

• Taking the square root of the above equation, we have,

        $\begin{gathered} \sqrt {\frac{1}{{{\mu _0}{\varepsilon _0}}}} = \sqrt {9.0 \times {{10}^{16}}{{{m^2}} \mathord{\left/ {\vphantom {{{m^2}} {{s^2}}}} \right. \kern-\nulldelimiterspace} {{s^2}}}} \hfill \\ \sqrt {\frac{1}{{{\mu _0}{\varepsilon _0}}}} = 3.0 \times {10^8}{m \mathord{\left/ {\vphantom {m s}} \right. \kern-\nulldelimiterspace} s} \hfill \\ \end{gathered}$

• This value is equal to the speed of light in vacuum. Therefore, option (b) is correct.

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