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1.Units, Dimensions and Measurement
hard
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. $L$ and $T$ are dimensions of length and time respectively. All the quantities are given in $SI$ units.
($1$) The relation between $[E]$ and $[B]$ is
$(A)$ $[\mathrm{E}]=[\mathrm{B}][\mathrm{L}][\mathrm{T}]$
$(B)$ $[\mathrm{E}]=[\mathrm{B}][\mathrm{L}]^{-1}[\mathrm{~T}]$
$(C)$ $[\mathrm{E}]=[\mathrm{B}][\mathrm{L}][\mathrm{T}]^{-1}$
$(D)$ $[\mathrm{E}]=[\mathrm{B}][\mathrm{L}]^{-1}[\mathrm{~T}]^{-1}$
($2$) The relation between $\left[\varepsilon_0\right]$ and $\left[\mu_0\right]$ is
$(A)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][\mathrm{L}]^2[\mathrm{~T}]^{-2}$
$(B)$ $\left[\mu_0\right]=\left[\varepsilon_0\right][\mathrm{L}]^{-2}[\mathrm{~T}]^2$
$(C)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[\mathrm{~L}]^2[\mathrm{~T}]^{-2}$
$(D)$ $\left[\mu_0\right]=\left[\varepsilon_0\right]^{-1}[\mathrm{~L}]^{-2}[\mathrm{~T}]^2$
given the answer question ($1$) and ($2$)
A$B,C$
B$A,B$
C$C,D$
D$B,C,D$
(IIT-2018)
Solution
$\mathrm{F}=\mathrm{qvB}=\mathrm{qE}$
$\mathrm{E}=\mathrm{vB}$
${[\mathrm{E}]=[\mathrm{L}][\mathrm{T}]^{-1}[\mathrm{~B}]}$
$\mu_0 \epsilon_0=\frac{1}{c^2}$
$\mu_0=[\mathrm{L}]^{-2}[\mathrm{~T}]^2\left[\epsilon_0\right]^{-1}$
$\mathrm{E}=\mathrm{vB}$
${[\mathrm{E}]=[\mathrm{L}][\mathrm{T}]^{-1}[\mathrm{~B}]}$
$\mu_0 \epsilon_0=\frac{1}{c^2}$
$\mu_0=[\mathrm{L}]^{-2}[\mathrm{~T}]^2\left[\epsilon_0\right]^{-1}$
Standard 11
Physics
