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 $9\text{cm}$ from the axis between the plates.

• A. $1.63\times {10}^{-11}\text{T}$
• B. $3.26\times {10}^{-11}\text{T}$
• C. $4.34\times {10}^{-11}\text{T}$
• D. $2.17\times {10}^{-11}\text{T}$

Answer 1

The correct option is D $2.17\times {10}^{-11}\text{T}$

Given:-
$R=6\text{cm};C=100\text{pF};V=230\text{V}$
$\omega =300\text{rad/s};r=9\text{cm}$

$\text{RMS}$ 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 $\left(9\text{cm}\right)$ is more than the radius of the capacitor $\left(6\text{cm}\right)$, the magnitude field at this point is given by

$B=\frac{{\mu }_0{i}_d}{2\pi r}$

$B=\frac{4\pi \times {10}^{-7}\times 6.9\sqrt{2}\times {10}^{-6}}{2\pi \times 9\times {10}^{-2}}$

$\therefore B=2.17\times {10}^{-11}\text{T}$

Hence, option $\text{D}$ is correct.