NEET-XII-Physics
40: Electromagnetic Waves
- #1Show that the dimensions of the displacement current
ε0dφEdtare that of an electric current.Ans : Displacement current,
`` {I}_{\,\mathrm{\,D\,}}=\frac{{\in }_{0}d{\phi }_{\,\mathrm{\,e\,}}}{dt}``
Electric flux,
`` {\varphi }_{\,\mathrm{\,e\,}}=EA``
`` \left[{\varphi }_{e}\right]\mathit{=}\left[E\right]\left[A\right]``
`` =\left[\frac{\mathit{1}}{\mathit{4}\pi {\mathit{\in }}_{\mathit{0}}}\frac{q}{{r}^{\mathit{2}}}\right]\left[A\right]``
`` \,\mathrm{\,Also\,},\left[{\mathit{\in }}_{0}\right]=\left[{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{-3}{\,\mathrm{\,T\,}}^{4}{\,\mathrm{\,A\,}}^{2}\right]``
`` \Rightarrow \left[{\varphi }_{\,\mathrm{\,e\,}}\right]=\left[{\,\mathrm{\,M\,}}^{1}{\,\mathrm{\,L\,}}^{3}{\,\mathrm{\,T\,}}^{-4}{\,\mathrm{\,A\,}}^{-2}\right]\left[\,\mathrm{\,AT\,}\right]\left[{\,\mathrm{\,L\,}}^{-2}\right]\left[{\,\mathrm{\,L\,}}^{2}\right]``
`` =\left[{\,\mathrm{\,ML\,}}^{3}{\,\mathrm{\,T\,}}^{-3}{\,\mathrm{\,A\,}}^{-1}\right]``
`` ``
Displacement current,
`` \left[{I}_{\,\mathrm{\,D\,}}\right]=\left[{\in }_{0}\right]\left[{\varphi }_{\,\mathrm{\,e\,}}\right]\left[{\,\mathrm{\,T\,}}^{-1}\right]``
`` \left[{I}_{\,\mathrm{\,D\,}}\right]=\left[{\,\mathrm{\,M\,}}^{-1}{\,\mathrm{\,L\,}}^{-3}{\,\mathrm{\,T\,}}^{4}{\,\mathrm{\,A\,}}^{2}\right]\left[{\,\mathrm{\,ML\,}}^{3}{\,\mathrm{\,T\,}}^{-3}{\,\mathrm{\,A\,}}^{-1}\right]\left[{\,\mathrm{\,T\,}}^{-1}\right]``
`` \left[{I}_{\,\mathrm{\,D\,}}\right]=\left[\,\mathrm{\,A\,}\right]``
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