Answer:
inside plates
[tex]B = (2.4 \times 10^{-6})r(e^{-500 t})[/tex]
Outside plates
[tex]B = \frac{0.24 \times 10^{-7} e^{-500 t}}{r}[/tex]
Explanation:
As capacitor is connected with the battery and resistance then we have
[tex]\frac{q}{C} + iR = V[/tex]
so we have
[tex]q = CV(1 - e^{-t/RC})[/tex]
now the current in the circuit is given as
[tex]i = \frac{dq}{dt}[/tex]
[tex]i = \frac{V}{R}e^{-t/RC}[/tex]
we have
[tex]R = 100 ohm[/tex]
V = 12 Volts
[tex]C = 20 \mu F[/tex]
now we have
[tex]i = \frac{12}{100} e^{-t/(100\times 20\times 10^{-6})}[/tex]
[tex]i = 0.12 e^{-500t}[/tex]
now to find the magnetic field at a distance "r" from the axis inside the plates we know
[tex]B = \frac{\mu_0 i r}{2\pi R^2}[/tex]
so we have
[tex]B = \frac{2\times 10^{-7} (0.12 e^{-500 t}) r}{0.10^2}[/tex]
[tex]B = (2.4 \times 10^{-6})r(e^{-500 t})[/tex]
Now for outside the plates we have
[tex]B = \frac{\mu_0 i}{2\pi r}[/tex]
[tex]B = \frac{2\times 10^{-7} (0.12 e^{-500 t})}{r}[/tex]
[tex]B = \frac{0.24 \times 10^{-7} e^{-500 t}}{r}[/tex]
