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Question: Find the dimensional formula for \({\mu _ \circ }\) and \({\varepsilon _ \circ }\) :...

Find the dimensional formula for μ{\mu _ \circ } and ε{\varepsilon _ \circ } :

Explanation

Solution

First let us see what μ{\mu _ \circ } and ε{\varepsilon _ \circ } is:
ε{\varepsilon _ \circ } stands for permittivity of free space. μ{\mu _ \circ } stands for vacuum permeability in free space.
It reflects the power of vacuum to permit the use of electrical fields. It is also related to the energy contained in the electrical field and to the capacitance.

Complete step by step answer:
The permittivity of free space ε{\varepsilon _ \circ } , is a physical constant often used in electromagnetism.
The permittivity of free space is given mathematically as:
ε=1μc28.8542×1012Fm1{\varepsilon _ \circ } = \dfrac{1}{{{\mu _ \circ }{c^2}}} \approx 8.8542 \times {10^{ - 12}}F{m^{ - 1}}
Where μ{\mu _ \circ } is the permeability of free space and cc is the speed of light.
An electrical field, EE , in an area of space, has field energy associated with it, which is energy density.
Energyvolume=εE22\dfrac{{Energy}}{{volume}} = \dfrac{{{\varepsilon _ \circ }{E^2}}}{2}

The energy stored in a capacitor is also given by:
energy=εA2dV2energy = \frac{{{\varepsilon _ \circ }A}} {{2d}}{V^2}

Where
AA is the area of the plates,
dd is the distance between the plates
VV is the voltage between the plates
The permittivity of free space can also be used to detect the Coulomb power. The constant shows how intense the force is between two charges separated by a distance:
F=14πεq1q2r2F = \dfrac{1} {{4\pi {\varepsilon _ \circ }}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}

Where
FF is the Coulomb force
q1{q_1} and q2{q_2} are two charges
rr is the separation between the charges.
k=14πεk = \dfrac{1}{{4\pi {\varepsilon _ \circ }}}

Thus, the dimension of ε{\varepsilon _ \circ } is:
ε=[k1] =[ML3T4A2]1 =[M1L3T4A2]  {\varepsilon _ \circ } = \left[ {{k^{ - 1}}} \right] \\\ = {\left[ {M{L^3}{T^{ - 4}}{A^{ - 2}}} \right]^{ - 1}} \\\ = \left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right] \\\

The permeability of free space μ{\mu _ \circ } is a physical constant often used in electromagnetism. It is defined to have an exact value of 4π×107NA24\pi \times {10^{ - 7}}\,N{A^{ - 2}} . It is linked to the energy contained in a magnetic field.
It is related to the speed of light as:
c=1μεc = \dfrac{1}{{\sqrt {{\mu _ \circ }{\varepsilon _ \circ }} }}

Where
cc is the speed of light
ε{\varepsilon _ \circ } is the permittivity of free space
Magnetic field BB in the space area has field energy correlated with it. The density of energy is defined as:
energy=B22μenergy = \dfrac{{{B^2}}}{{2{\mu _ \circ }}}

It also gives the usage of magnetic force. The constant shows how intense the force is between two electrical currents divided by a distance:
F=I1I22πμrF = \dfrac{{{I_1}{I_2}}}{{2\pi {\mu _ \circ }r}}

Where
FF is the magnetic force
I1{I_1} and I2{I_2} are two currents
rr is the separation between the wires
So, the dimension of μ{\mu _ \circ } is given as:
μ=[c2ε]1 =[MLT2A2]  {\mu _ \circ } = {\left[ {{c^2}{\varepsilon _ \circ }} \right]^{ - 1}} \\\ = \left[ {ML{T^{ - 2}}{A^2}} \right] \\\ .

Note:
Although the permeability and permittivity may appear the same, they are different.
Permittivity measures the obstruction created by the material in the creation of an electrical field, while
The permeability is the ability of the material to allow magnetic lines to conduct through it.