# How to compute the field lines of an induced magnetic field inside a capacitor?

Consider a capacitor with a varying voltage applied to it. As the voltage changes over time, the electrical field $$\vec{E}$$ inside the plates does too.

Assumption We assume that the direction of $$\vec{E}$$ is the same inside the capacitor, hence ignoring border effects.

According to Ampere-Maxwell (Maxwell's 4th equation):

$$\nabla \times \vec{B} = \mu_0 \vec{j} + \frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}$$

A magnetic field $$\vec{B}$$ is generated. How do I find out the direction of the magnetic field? Basically, how do I find out the field lines of $$\vec{B}$$ inside the capacitor?

Ultimately, how can we find the equations of the field lines?

## 2 Answers

Assuming that the approximation that the electric field is constant in space but changes with time is reasonable, then since $$\vec j = 0$$ between the plates, a possible solution for the $$\vec \nabla \times \vec B$$ equation which also satisfies $$\vec \nabla \cdot \vec B = 0$$ is $$\vec B = \frac{1}{2c^2} \vec r \times \frac{\partial \vec E}{\partial t}$$. To see what other solutions can be, imagine another solution $$\vec B_2$$, that satisfies both of these equations. The difference $$\vec \Delta = \vec B-\vec B_2$$ satisfies $$\vec \nabla \times \vec \Delta = 0$$, so it can be written as the gradient of a scalar field. Since the divergence of $$\vec \Delta$$ is also zero, this scalar field satisfies Laplaces equation. Adding the gradient of any solution to Laplaces equation to $$\vec B$$ will give another solution. You need to add boundary conditions, i.e. deal with the shape of the capacitor plates and the charging current, to set these boundary conditions and get a unique solution.

A concrete example is given by problem 6.14 in J.D. Jackson's Classical Electrodynamics, third edition, where he asks for the fields in a capacitor with circular plates in the quasi-static approximation.

Using the integral form of the Ampére-Maxwell law without the current, that is zero across the plates:

$$\oint_{\partial \Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \varepsilon_0 \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{S}$$

we can see that around the borders of the plates, the LHS is maximum, because the integral of RHS includes all the area. The B-vectors are pointing clockwise if viewed from positive to negative side when $$\mathbf E$$ is increasing. For regions closer to the center, the direction is the same, but the magnitude of $$\mathbf B$$ is smaller, due to the smaller enclosed area.