Question

Find the value of ${\mu _0}{\varepsilon _0}$.Symbols have their usual meanings.

Answer 1

Hint:In order to this question, to find the value of ${\mu _0}{\varepsilon _0}$ , as we know that in the given expression, there are two given different constant that have its fixed value for vacuum, so we will first explain both the constant in the given expression and then with the help of their value, we will find the value of the given expression.

Complete step by step answer:
As we know, \[{\mu _0} = \text{Permeability of vacuum} = 4\pi \times {10^{ - 7}}H\]. Vacuum permeability is the magnetic permeability in a classical vacuum. In all other formulas for magnetic-field output in a vacuum, vacuum permeability is derived from the production of a magnetic field by an electric current or a moving electric charge.

Since the redefinition of SI units in 2019, the vacuum permeability ${\mu _0}$ is no longer a constant (as it was in the previous description of the SI ampere), but must be determined experimentally. whereas,
${\varepsilon _0} = \text{Permeability of vacuum} = 8.85 \times {10^{ - 12}}\,F/m$

The value of the absolute dielectric permittivity of classical vacuum is usually denoted as ${\varepsilon _0}$ (pronounced "epsilon nought" or "epsilon zero"). It's also known as the electric constant, the permittivity of free space, or the distributed capacitance of the vacuum. It is a physical constant that is ideal (or at least close to it). So, now we can find the value of ${\mu _0}{\varepsilon _0}$ :
$\therefore {\mu _0}{\varepsilon _0} = 4\pi \times {10^{ - 7}} \times 8.85 \times {10^{ - 12}} = \dfrac{1}{9} \times {10^{ - 16}}$

Hence, the value of ${\mu _0}{\varepsilon _0}$ is $\dfrac{1}{9} \times {10^{ - 16}}$.

Note:Permittivity is used to measure the obstruction created by a substance during the creation of electric fields. Permeability, on the other hand, is the ability of a substance to allow magnetic lines to pass through it. Permittivity is denoted by $\varepsilon $ and permeability is denoted by $\mu $ .