A parallel plate capacitor is made of circular plates each of radius $R=6.0~\text{cm}$ has a capacitance $C=100~\text{pF}.$ The capacitor is connected to a $230~\text V$ AC supply with an (angular) frequency of $300~\text{rad/s}.$ The amplitude of $\vec{B}$ at the point $3~\text{cm}$ from the axis between the plate is:
          
1. $1.12\times 10^{-11}~\text T$
2. $2.01\times 10^{-12}~\text T$
3. $1.63\times 10^{-11}~\text T$
4. $1.01\times 10^{-12}~\text T$

An EM wave radiates outwards from a dipole antenna, with $E_0$, as the amplitude of its electric field vector. The electric field $E_0$, which transports significant energy from the source falls off as:
1. $\dfrac{1}{r^3}$
2. $\dfrac{1}{r^2}$
3. $\dfrac{1}{r}$
4. remains constant

Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor:

Displacement current through a capacitor at a steady state is:
1. infinite
2. zero
3. $\dfrac{\varepsilon_0d\phi_E}{dt}$
4. $\dfrac{\varepsilon_0d\phi_B}{dt}$

A capacitor of capacitance $C$ is connected across an AC source of voltage $V$, given by;
$V=V_0 \sin \omega t$
The displacement current between the plates of the capacitor would then be given by:
1. $I_d=\dfrac{V_0}{\omega C} \sin \omega t$
2. $I_d=V_0 \omega C \sin \omega t$
3. $I_d=V_0 \omega C \cos \omega t$
4. $I_d=\dfrac{V_0}{\omega C} \cos \omega t$

A parallel plate capacitor of capacitance $20~\mu\text{F}$ is being charged by a voltage source whose potential is changing at the rate of $3~\text{V/s}.$ The conduction current through the connecting wires, and the displacement current through the plates of the capacitor would be, respectively: