JEE Advance - Physics (2022 - Paper 2 Online - No. 4)
In a particular system of units, a physical quantity can be expressed in terms of the electric charge $e$, electron mass $m_{e}$, Planck's constant $h$, and Coulomb's constant $k=\frac{1}{4 \pi \epsilon_{0}}$, where $\epsilon_{0}$ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is $[B]=[e]^{\alpha}\left[m_{e}\right]^{\beta}[h]^{\gamma}[k]^{\delta}$. The value of $\alpha+\beta+\gamma+\delta$ is _______.
Answer
4
Explanation
$$
\begin{aligned}
& \mathrm{B}=\mathrm{e}^\alpha\left(\mathrm{m}_{\mathrm{e}}\right)^\beta \mathrm{h}^\gamma \mathrm{k}^\delta \\\\
& {[\mathrm{~B}]=\left[\mathrm{e}^\alpha\right]\left[\mathrm{m}_{\mathrm{e}}\right]^\beta[\mathrm{h}]^\gamma\left[\mathrm{k}^\delta\right]} \\\\
& {\left[\mathrm{M}^1 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]=[\mathrm{AT}]^\alpha[\mathrm{m}]^\beta\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^\gamma\left[\mathrm{ML}^3 \mathrm{~A}^{-2} \mathrm{~T}^{-4}\right]^\delta} \\\\
& \mathrm{M}^{\mathrm{l}} \mathrm{T}^{-2} \mathrm{~A}^{-1}=\mathrm{m}^{\beta+\gamma+\delta} \mathrm{L}^{2 \mathrm{r}+3 \delta} \mathrm{T}^{\alpha-\gamma-4 \delta} \mathrm{A}^{\alpha-2 \delta} \\\\
& \text {Compare }: \beta+\gamma+\delta=1 ; 2 \gamma+3 \delta=0, \alpha-\gamma-4 \delta=-2, \alpha-2 \delta=-1 \\\\
& \text {On solving } \alpha=3, \beta=2, \gamma=-3, \delta=2 \\\\
& \alpha+\beta+\gamma+\delta=4
\end{aligned}
$$
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