JEE Advance - Physics (2018 - Paper 1 Offline - No. 14)
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below, $$[E]$$ and $$[B]$$ stand for dimensions of electric and magnetic fields respectively, while $$\left[ {{\varepsilon _0}} \right]$$ and $$\left[ {{\mu _0}} \right]$$ stand for dimensions of the permittivity and permeability of free space respectively. $$\left[ L \right]$$ and $$\left[ T \right]$$ are dimensions of length and time respectively. All the quantities are given in $$SI$$ units.
The relation between $$\left[ {{\varepsilon _0}} \right]$$ and $$\left[ {{\mu _0}} \right]$$ is
The relation between $$\left[ {{\varepsilon _0}} \right]$$ and $$\left[ {{\mu _0}} \right]$$ is
$$\left[ {{\mu _0}} \right] = \left[ {{\varepsilon _0}} \right]{\left[ L \right]^2}{\left[ T \right]^{ - 2}}$$
$$\left[ {{\mu _0}} \right] = \left[ {{\varepsilon _0}} \right]{\left[ L \right]^{ - 2}}{\left[ T \right]^2}$$
$$\left[ {{\mu _0}} \right] = {\left[ {{\varepsilon _0}} \right]^{ - 1}}{\left[ L \right]^2}{\left[ T \right]^{ - 2}}$$
$$\left[ {{\mu _0}} \right] = {\left[ {{\varepsilon _0}} \right]^{ - 1}}{\left[ L \right]^{ - 2}}{\left[ T \right]^2}$$
Giải thích
$$c = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$$
$${c^2} = {1 \over {{\mu _0}{\varepsilon _0}}}$$
$${\mu _0} = \varepsilon _0^{ - 1}\,.\,{c^{ - 2}}$$
$$[{\mu _0}] = {[{\varepsilon _0}]^{ - 1}}[{L^{ - 2}}{T^2}]$$