[NOTES/EM-10007]-Time Varying Fields ,Ampere's Law

We explain how Maxwell's addition of a displacement current in the fourth equation \begin{equation} \nabla \times \vec{B} = \mu_0 \vec{j} + \epsilon_0\mu_0 \pp[\vec{E}]{t}. \end{equation} resolves problem with the Ampere's law.

1  Ampere's law --- A problem without displacement current

Let us assume that a capacitor is being charged by connecting to a battery. Let the area of the plates, distance between the plates and the capacitance be denoted, respectively, by \(A,d\) and \(C\). These parameters are related by \begin{equation} C = \frac{\epsilon_0 A}{d}\end{equation} The key $K$ is closed at time $t$. Then the charge on capacitor plates at time $t$ is $q(t)$ $$ I = \frac{dq}{dt} $$ Consider two loops $\Gamma_1,\Gamma_2$. The current produces magnetic field $\vec{B}$ and Amperes Law $$ \int_{\Gamma_1}\vec{B}\cdot\overrightarrow{dl} = \mu_0~~ \text{current enclosed} $$ without any modification, takes the form \begin{equation} \int \vec{B}\cdot\overrightarrow{dl} = \begin{cases} \mu_0 I & \text{for}~ \Gamma_1\\ \mu_0 I & \text{for}~ \Gamma_2\end{cases} \end{equation} However, if the loops $\Gamma_1$ and $\Gamma_2$ are moved toward capacitor plates. The value of the line integral is non zero, and constant, until the loops reach the plates. As soon as a loop crosses the plates, and encloses the region between plates, the value of the current and hence the value of the line integral suddenly drops to zero. So for example, without any changes in the $4^\text{th}$ equation, \begin{equation} \int\vec{B}\overrightarrow{dl} = \begin{cases} \mu_0 I & \text{ for }\Gamma_1 \text{ and } \Gamma_2 \\ 0 & \text{for}~\Gamma^{'}_3,\Gamma^{'}_4 \text{and } \Gamma_5 \end{cases} \end{equation}

2. Ampere's law --- Problem resolved

So what happens to the charge conservation and to the Ampere's law \underline{with} the extra term? by Maxwell. We will now verify that, that addition of the displacement current in the right hand side removes the anomalies mentioned above. \paragraph*{Charging a capacitor} With the displacement current the Ampere's law has an extra term and gets modified to \begin{align} \int\vec{B}\overrightarrow{dl} = \iint \mu_0 \vec{j} \cdot \overrightarrow{dS} + \mu_0\epsilon_0 \iint \pp[\vec{E}]{t} \cdot \overrightarrow{dS}. \end{align} We will now show that there is no discontinuity in the right hand side as the amperean loop is moved from \(\Gamma_1\) to \(\Gamma_2\) through positions \(\Gamma_1{'}, \Gamma_3, \Gamma_2{'}\) successively. For loops such as \(\Gamma_2,\Gamma_3, \Gamma_4\), see \Figref{Amp}, the first term is zero. Only the displacement current contributes to Ampere's law. The potential difference between the plates \(= q(t)/C\)
The electric field between the plates \(E=\dfrac{q(t)}{d C}\)
The displacement current is \[(\mu_0\epsilon_0) \pp[E]{t}= (\mu_0\epsilon_0) \frac{1}{d C }\dd[q]{t} = \frac{\mu_0 \epsilon_0}{d C}\, I(t). \] \begin{align*} \int_{\Gamma_3} \vec{B}\cdot\overrightarrow{dl} =& \int_{S_3} \nabla\times\vec{B}\cdot\overrightarrow{dl}\\ =& \int \left(\mu_0\vec{j}+ \mu_0\epsilon_0 \frac{\partial\vec{E}}{\partial t}\right)\cdot\hat{n}\cdot dS\\ =& \mu_0\epsilon_0 \int \frac{\partial\vec{E}}{\partial t}~\hat{n}\cdot dS = \mu_0\epsilon_0 \pp[E]{t} \times A \\ =&\dfrac{\mu_0\epsilon_0}{d C A} I(t)\qquad \mbox{{$\because C =\frac{\epsilon_0 A}{d} $}} \\ =& \mu_0I(t) \end{align*}

1. Sec 18-1 Maxwell's Equations Sec 18-2 How the New Term WorksR. P. Feynman, Robert B. Leighton and Mathew Sands Lectures on Physics, vol-II, B.I. Publications (1964)
2. Sec 7.3.1 Electrodynamics Before MaxwellSec 7.3.2 How Maxwell Fixed Ampere's LawSec 7.3.3  Maxwell's Equations David Griffiths, Introduction to Electrodynamics, 3rd EEE edn, Prentice Hall of India Pvt Ltd New Delhi, (2002).