Question

An AC rms voltage of $2V$ having a frequency of $50KHz$ is applied to a condenser of the capacity of $10\mu F$. The maximum value of the magnetic field between the plates of the condenser if the radius of a plate is $10cm$ is _____.

Answer 1

Magnetic field $\overrightarrow{B}=\frac{{\mu }_{o}I}{2A}\overrightarrow{r}$

$r=$ radius of plate

$A=$ area of plate

$I=\frac{2}{X_c}=20\times \omega C=20\times 5\times {10}^3\times 10\times {10}^{-6}=1$ $A$

Now, $B=\frac{{\mu }_{o}Ir}{2A}=\frac{4\pi \times {10}^{-7}\times 10\times {10}^{-2}}{2\times pi\times (10\times {10}^{-2}{)}^2}=\frac{2\times {10}^{-7}}{0.1}=2$ $\mu wb/m^2=2\times {10}^{-6}$ $wb/m^2$