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Show that the magnetic field B at a point in between the plates of a parallel-plate capacitor during charging is  \left ( \frac{\varepsilon _{0}\mu r}{2} \right )\frac{dE}{dt}.

Answers (1)

Here, I_{d} is the displacement current between the two plates of a parallel plate capacitor.

r = distance between the two plates

Therefore,

B=\mu _{0}*\frac{2I_{d}}{4\pi r}

      =\frac{(\mu _{0}I_{d})}{2\pi r}

      =\frac{\mu _{0}}{2\pi r} \varepsilon _{0}\left ( \frac{d\phi _{E}}{dt} \right )\; \; \; \; \; \; \; \; \left [ as, I_{d}=\varepsilon _{0} \left ( \frac{d\phi _{E}}{dt} \right )\right ]                     

     =\left ( \frac{\mu _{0}\varepsilon _{0}}{2\pi r} \right ) \frac{d(E\pi r^{2})}{dt}\; \; \; \; \; \; \; \; \; \; \; \; \left [ as,\phi _{E}=E\pi r^{2} \right ]

     =\left ( \frac{\mu _{0}\varepsilon _{0}r}{2} \right )\left ( \frac{dE}{dt} \right )

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