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Show that the magnetic field $\mathrm{B}$ at a point in between the plates of a parallel plate capacitor during charging is $\frac{\mu_{0} \varepsilon_{0} \mathrm{r}}{2} \frac{\mathrm{dE}}{\mathrm{dt}}$
Show that the magnetic field $\mathrm{B}$ at a point in between the plates of a parallel plate capacitor during charging is $\frac{\mu_{0} \varepsilon_{0} \mathrm{r}}{2} \frac{\mathrm{dE}}{\mathrm{dt}}$
CBSE | Class 12 | Electromagnetic Waves | NCERT Exemplar | Physics
Show that the magnetic field $\mathrm{B}$ at a point in between the plates of a parallel plate capacitor during charging is $\frac{\mu_{0} \varepsilon_{0} \mathrm{r}}{2} \frac{\mathrm{dE}}{\mathrm{dt}}$

Let $I_{d}$ be the displacement current in the magnetic field region between two parallel plate capacitor plates.

The magnetic field induction at a point between two capacitor plates at a perpendicular distance from the plate axis equals,

$\begin{array}{l}
\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \mathrm{I}_{\mathrm{d}}=\frac{\mu_{0}}{2 \pi \mathrm{r}}\left(\varepsilon_{0} \frac{\mathrm{d} \phi}{\mathrm{dt}}\right) \\
\mathrm{B}=\frac{\mu_{0}}{2 \pi \mathrm{r}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{E} \pi \mathrm{r}^{2}\right) \quad \because \phi_{\mathrm{E}}=\overrightarrow{\mathrm{E}} \overrightarrow{\mathrm{A}} \\
\mathrm{B}=\frac{\mu_{0} \varepsilon_{0} \mathrm{r} \mathrm{dE}}{2 \pi \mathrm{r}} \frac{\mathrm{dt}}{}
\end{array}$

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