
Dimensions of $\dfrac{1}{{{\mu _o}{\varepsilon _o}}}$, where symbols have their usual meaning?
Answer
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Hint:Let us first get some idea about Dimensions. The dimension of a mathematical space (or object) is specified informally in physics and mathematics as the minimum number of coordinates required to specify some point within it.
Complete step by step answer:
Let us talk about permittivity. The absolute permittivity, also known as permittivity and denoted by the Greek letter $\varepsilon $ (epsilon), is a measure of a dielectric's electric polarizability in electromagnetism. A material with a high permittivity polarises more in response to an applied electric field than one with a low permittivity, allowing it to store more energy.
The relative permittivity ${\varepsilon _r}$, which is the ratio of the absolute permittivity $\varepsilon $ and the vacuum permittivity ${\varepsilon _0}$ is often used to represent permittivity.
${\varepsilon _r} = \dfrac{\varepsilon }{{{\varepsilon _0}}}$
Let's get some idea about permeability. Permeability is the measure of magnetization that a material obtains in response to an applied magnetic field in electromagnetism. The (italicised) Greek letter $\mu $ is commonly used to denote permeability. Oliver Heaviside invented the word in September 1885. Magnetic reluctivity is the reciprocal of permeability.
The permeability constant ${\mu _0}$ also known as the magnetic constant or permeability of free space. ${\varepsilon _0}$ is permittivity of vacuum and its dimension will be [${M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$]. Vacuum Permeability is given by ${\mu _0} = 4\pi \times {10^{ - 7}}N/{A^2}$ and its dimensions will be [$ML{T^{ - 2}}{I^{ - 2}}$]. Therefore, dimensions for $\dfrac{1}{{{\mu _0}{\varepsilon _0}}}$ will be,
$\dfrac{1}{{[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}] \times [ML{T^{ - 2}}{I^{ - 2}}]}} = [{L^2}{T^{ - 2}}]$
Hence, the dimensions of $\dfrac{1}{{{\mu _o}{\varepsilon _o}}}$ is $[{L^2}{T^{ - 2}}]$.
Note:Dimensioning is used to provide a complete and accurate definition of an entity. Only one interpretation is needed to create the part with a full set of dimensions. These rules should be followed while dimensioning. Accuracy: the values must be right.
Complete step by step answer:
Let us talk about permittivity. The absolute permittivity, also known as permittivity and denoted by the Greek letter $\varepsilon $ (epsilon), is a measure of a dielectric's electric polarizability in electromagnetism. A material with a high permittivity polarises more in response to an applied electric field than one with a low permittivity, allowing it to store more energy.
The relative permittivity ${\varepsilon _r}$, which is the ratio of the absolute permittivity $\varepsilon $ and the vacuum permittivity ${\varepsilon _0}$ is often used to represent permittivity.
${\varepsilon _r} = \dfrac{\varepsilon }{{{\varepsilon _0}}}$
Let's get some idea about permeability. Permeability is the measure of magnetization that a material obtains in response to an applied magnetic field in electromagnetism. The (italicised) Greek letter $\mu $ is commonly used to denote permeability. Oliver Heaviside invented the word in September 1885. Magnetic reluctivity is the reciprocal of permeability.
The permeability constant ${\mu _0}$ also known as the magnetic constant or permeability of free space. ${\varepsilon _0}$ is permittivity of vacuum and its dimension will be [${M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$]. Vacuum Permeability is given by ${\mu _0} = 4\pi \times {10^{ - 7}}N/{A^2}$ and its dimensions will be [$ML{T^{ - 2}}{I^{ - 2}}$]. Therefore, dimensions for $\dfrac{1}{{{\mu _0}{\varepsilon _0}}}$ will be,
$\dfrac{1}{{[{M^{ - 1}}{L^{ - 3}}{T^4}{I^2}] \times [ML{T^{ - 2}}{I^{ - 2}}]}} = [{L^2}{T^{ - 2}}]$
Hence, the dimensions of $\dfrac{1}{{{\mu _o}{\varepsilon _o}}}$ is $[{L^2}{T^{ - 2}}]$.
Note:Dimensioning is used to provide a complete and accurate definition of an entity. Only one interpretation is needed to create the part with a full set of dimensions. These rules should be followed while dimensioning. Accuracy: the values must be right.
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