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;
\(V=V_0 \sin \omega t\)$\mathrm{}$
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: