Suppose that the electric flux inside a parallel plate capacitor changes at a rate of $7 \times 10^{14} \text{ V} \cdot \text{m/s}$. If the area of the plates is $1 \text{ m}^2$,calculate the magnetic field $B$ at a distance $r = 0.1 \text{ m}$ from the axis of the capacitor. (Given: $\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$,$\mu_0 = 4\pi \times 10^{-7} \text{ T} \cdot \text{m/A}$)

$A$ variable frequency $AC$ source is connected to a capacitor. How will the displacement current change with a decrease in frequency?

The source of a time-varying magnetic field may be:
$(A)$ A permanent magnet
$(B)$ An electric field changing linearly with time
$(C)$ Direct current
$(D)$ A decelerating charged particle
$(E)$ An antenna fed with a digital signal
Choose the correct answer from the options given below:

Seawater at a frequency $f = 9 \times 10^{2} \, Hz$ has permittivity $\varepsilon = 80 \varepsilon_{0}$ and resistivity $\rho = 0.25 \, \Omega m$. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source $V(t) = V_{0} \sin(2 \pi ft)$. Then the conduction current density becomes $10^{x}$ times the displacement current density after time $t = \frac{1}{800} \, s$. The value of $x$ is ......... . $\left(\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \, Nm^{2} C^{-2}\right)$

Displacement current is given by

Write Maxwell's equations.