Q.Show that the magnetic field $B$ at a point in between the plates of a parallel-plate capacitor during charging is $\dfrac{\mu_0\varepsilon_0}{2}r\dfrac{dE}{dt}$ (symbols having usual meaning).

The key idea is that a changing electric field between the capacitor plates acts as a displacement current, which produces a magnetic field just like a conduction current.

Reasoning:

1. Consider a circular parallel-plate capacitor of plate radius $R$, with plates separated by distance $d$. During charging, the electric field $E$ between the plates is uniform and changes with time.
2. By Maxwell’s modification of Ampere’s law, the line integral of $\vec{B}$ around a circular loop of radius $r$ (where $r < R$) centred on the axis is:

Here, $\Phi_E = E \cdot \pi r^2$ is the electric flux through the loop. …

The key insight is that during charging, no conduction current flows between the plates — but the electric field there changes with time. Maxwell realised that a changing electric field produces a magnetic field just as a real current does. This is the displacement current:

where $\Phi_E$ is the electric flux through a surface. For a parallel-plate capacitor, the field between the plates is uniform (ignoring edge effects), so $\Phi_E = E \cdot A$.

Now, to find $B$ at a distance $r$ from the centre (with $r$ less than the plate radius $R$), we apply Ampere-Maxwell law:

Between the plates, $I_c = 0$, so only the displacement current matters.

1. Choose a circular Amperian loop of radius $r$, centred on the axis, lying in a plane parallel to the plates. By symmetry, $B$ is tangential and constant in magnitude along this loop. The left side becomes:

2. Find the displacement current through the loop. The electric flux through the loop is $E \times (\pi r^2)$ (since $E$ is uniform and perpendicular to the loop). So: $I_d = \varepsilon_0 \frac{d}{dt}(E \cdot \pi r^2) = \varepsilon_0 \pi r^2 \frac{dE}{dt}$ …

Method: Applying the Ampere–Maxwell Law with Displacement Current

Use this for any "find/show the magnetic field due to displacement current" problem — typically between capacitor plates, or in any region with a changing electric field but no conduction current.

Steps

Step 1: Confirm there is no conduction current through your chosen surface, only a changing electric flux.

This tells you to use the Ampere–Maxwell form of Ampere's law, $\oint \vec B\cdot d\vec l = \mu_0(I_c+I_d)$, with $I_c=0$ in this region.

Step 2: Choose a closed Amperian loop that respects the field's symmetry.

For a circular parallel-plate capacitor, use a circle of radius $r$ concentric with the axis, so $B$ is tangential and constant in magnitude all along the loop:

Step 3: Compute the displacement current through THAT SAME loop, not through the whole capacitor. …