
The dimension of $\dfrac{{{B^2}}}{{2{\mu _0}}}$, where B is the magnetic field and ${\mu _0}$ is the magnetic permeability of vacuum, is:
(A) $[M{L^{ - 1}}{T^{ - 2}}]$
(B) $[M{L^2}{T^{ - 2}}]$
(C) $[ML{T^{ - 2}}]$
(D) $[M{L^2}{T^{ - 1}}]$
Answer
163.2k+ views
Hint: To solve this question, we should know that every physical quantity can be expressed in terms of their fundamental unit notations called the dimensional formula of a physical quantity, here we will use the dimensional analysis to derive the dimensional formula of $\dfrac{{{B^2}}}{{2{\mu _0}}}$.
Complete answer:
We have been asked for the dimensions of $\dfrac{{{B^2}}}{{2{\mu _0}}}$ where B is the magnetic field and ${\mu _0}$ is the magnetic permeability of vacuum.
Now, since we know from Electromagnetism that the energy density of the magnetic field in an electromagnetic wave is given by the formula $\dfrac{{{B^2}}}{{2{\mu _0}}}$ and energy density is simply the energy per unit volume in space, So given term $\dfrac{{{B^2}}}{{2{\mu _0}}}$ will have the dimensions of Energy density as it’s the formula of energy density hence,
Dimensions of Energy is given by the product of dimensions of Force and displacement so,
Dimension of Displacement is $[L]$
Dimension of force F is $[F] = [ML{T^{ - 2}}]$
So, Dimensions of Energy E will be
$
[E] = [ML{T^{ - 2}}][L] \\
[E] = [M{L^2}{T^{ - 2}}] \\
$
Now, for energy density Divide the dimensions of energy by the dimensions of volume which is $[{L^3}]$ so, we get the dimensions of energy density as
$
= \dfrac{{[M{L^2}{T^{ - 2}}]}}{{[{L^3}]}} \\
= [M{L^{ - 1}}{T^{ - 2}}] \\
$
So, The dimensions of Energy density which is represented by $\dfrac{{{B^2}}}{{2{\mu _0}}}$ is $[M{L^{ - 1}}{T^{ - 2}}]$
Hence, the correct option is (A) $[M{L^{ - 1}}{T^{ - 2}}]$
Note: It should be remembered that there are seven fundamental physical quantities whose units are used to derive any other physical quantity dimensions or units and these seven fundamental physical quantities are Mass, Length, Time, Luminous Intensity, Amount of Substance, Temperature, and Current.
Complete answer:
We have been asked for the dimensions of $\dfrac{{{B^2}}}{{2{\mu _0}}}$ where B is the magnetic field and ${\mu _0}$ is the magnetic permeability of vacuum.
Now, since we know from Electromagnetism that the energy density of the magnetic field in an electromagnetic wave is given by the formula $\dfrac{{{B^2}}}{{2{\mu _0}}}$ and energy density is simply the energy per unit volume in space, So given term $\dfrac{{{B^2}}}{{2{\mu _0}}}$ will have the dimensions of Energy density as it’s the formula of energy density hence,
Dimensions of Energy is given by the product of dimensions of Force and displacement so,
Dimension of Displacement is $[L]$
Dimension of force F is $[F] = [ML{T^{ - 2}}]$
So, Dimensions of Energy E will be
$
[E] = [ML{T^{ - 2}}][L] \\
[E] = [M{L^2}{T^{ - 2}}] \\
$
Now, for energy density Divide the dimensions of energy by the dimensions of volume which is $[{L^3}]$ so, we get the dimensions of energy density as
$
= \dfrac{{[M{L^2}{T^{ - 2}}]}}{{[{L^3}]}} \\
= [M{L^{ - 1}}{T^{ - 2}}] \\
$
So, The dimensions of Energy density which is represented by $\dfrac{{{B^2}}}{{2{\mu _0}}}$ is $[M{L^{ - 1}}{T^{ - 2}}]$
Hence, the correct option is (A) $[M{L^{ - 1}}{T^{ - 2}}]$
Note: It should be remembered that there are seven fundamental physical quantities whose units are used to derive any other physical quantity dimensions or units and these seven fundamental physical quantities are Mass, Length, Time, Luminous Intensity, Amount of Substance, Temperature, and Current.
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