Question

A parallel plate capacitor made of circular plates each of radius $R=6\text{cm}$ has a capacitance $C=100\text{pF}.$ The capacitor is connected to a $230\text{V}AC$ supply with an angular frequency of $300\text{rad/s}$. Determine the amplitude of the magnetic field at point $3\text{cm}$ from the axis between the plates.

• A. $3.26\times {10}^{-11}\text{T}$
• B. $1.63\times {10}^{-11}\text{T}$
• C. $4.89\times {10}^{-11}\text{T}$
• D. $8.15\times {10}^{-11}\text{T}$

Answer 1

The correct option is B $1.63\times {10}^{-11}\text{T}$

Given:-
$R=6\text{cm};C=100\text{pF};V=230\text{V}$

$\omega =300\text{rad/s};r=3\text{cm}$

$\text{RMS}$ value of conduction current is

$I_C=\frac{V}{X_C}=V\omega C$

$I_C=230\times 300\times 100\times {10}^{-12}$

$\therefore I_C=6.9\times {10}^{-6}\text{A}$

Since, the magnitude of displacement current is equal to the magnitude of conduction current, we can say that $\text{RMS}$ value of displacement current is

$(I_d{)}_{rms}=6.9\times {10}^{-6}\text{A}$

$\therefore$ The peak value of displacement current is

$I_d=6.9\sqrt{2}\times {10}^{-6}\text{A}$

Since, the given distance $3\text{cm}$ is less than the radius of the capacitor $\left(6\text{cm}\right)$, the magnetic field at that point is given by

$B=\left(\frac{{\mu }_0{I}_d}{2\pi R^2}\right)r$

$B=\frac{4\pi \times {10}^{-7}\times 6.9\sqrt{2}\times {10}^{-6}\times 3\times {10}^{-2}}{2\pi \times (6\times {10}^{-2}{)}^2}$

$\therefore B=1.63\times {10}^{-11}\text{T}$

Hence, option $\left(\text{B}\right)$ is correct.