# Displacement currents

My book gives a really vague picture of the displacement current. Could someone explain how it was actually arrived at? My book does something like this: It assumes a loop just before the capacitor plate to apply Ampere's law and calculate the magnetic field at a certain point P. That's all good. Next it takes a closed surface (pot-shaped) such that the point P is included to calculate the magnetic field. My doubt is this: Doesn't Ampere's law consider a loop? How can one apply it to a surface? Is my book wrong?

Edit

Method 1= (a) Method 2= (b)

My book says calculated by method 1, there's a finite magnetic field at P. Calculated by method 2, there's no magnetic field. But how? Isn't there a current which still passes through the surface in method 2? I guess my perceptions of loop and surface are messed up.

• Related/duplicate physics.stackexchange.com/q/240839/104696 Mar 10 '17 at 6:19
• @Farcher I read your answer but I still couldn't make much sense of it. Maybe I've been taught Amperes law incompletely. What confuses me in my text book is this: 'Calculated one way the magnetic field is finite...calculated another way the electric field is zero'. What I fail to understand is how is the magnetic field calculated in the second manner zero? A time varying current still threads through the second choice of surface, doesn't it? Mar 10 '17 at 6:33
• There is no current of the movement of charges type between the plates of the capacitor. So with the butterfly net arrangement the right hand side of the amperes law equation ($\mu_oI$)is zero whereas the integral along a loop on the left hand side has a finite value. This would mean that the law does not work unless you postulate that there is another type of current to be considered on the right hand side and that is exactly what Maxwell did. He called it the displacement current. Mar 10 '17 at 6:54
• @Farcher But that butteryfly net does enclose a bit of the wires connecting the capacitor plate to the source right? There is a current in those connecting wires. So...? Mar 10 '17 at 7:14
• The current does not just have to enter into the butterfly net it actually has to pass through the butterfly net. I have written out an answer for you Mar 10 '17 at 7:33

Remember that the area can be that of any surface as long the closed loop defines its edge as shown by the butterfly net diagram below.
Also one must remember that it is the current passing through the butterfly net which must be considered not just the current entering the butterfly net.

So now look at example (a) and example (b) which have the same loop and hence the same integral on the left hand side of ampere's law $B2\pi r$.

In example (a) the butterfly net has been collapsed and is the blue shaded area.
For the right hand side of ampere's law you have to evaluate the total current passing through the butterfly net which is the blue area.

In example (a) it is easy - the current is $I$ and so you get that $B2\pi r = \mu_o I \Rightarrow B= \dfrac{\mu_oI}{2 \pi r}$.

Now look at example (b).
The current passing through the butterfly net (blue area) is zero so $B2\pi r = 0 \Rightarrow B= 0$.

You are evaluating the same magnetic field $B$ at a distance $r$ from the wire on the left hand side of the capacitor so how can you have two different values?

The answer is that you cannot and that is why Maxwell said that there was another type of current passing between the capacitor plates which he called the displacement current.

When you apply amperes law with the displacement current you find that example (b) also predicts that $B= \dfrac{\mu_oI}{2 \pi r}$.

Update in response to a comment from @samjoe

Here is a diagram of a capacitor which is charging with and amperian loop shown in blue and the amperian surface shown in pink.

The area vector is in the same direction as the electric field $\vec E$ and so the positive direction around the loop is anticlockwise looking from the top - blue arrow.

$\displaystyle \oint_{\rm loop} \vec B \cdot d\vec l = \mu_o I_{\rm surface}+ \mu_o\epsilon_o \dfrac {\partial}{\partial t}\left ( \int_{\rm surface} \vec E \cdot d\vec S\right)$

Left hand side
$\displaystyle \oint_{\rm loop} \vec B \cdot d\vec l = 2 \pi r B$

Right hand side
$\mu_o I_{\rm surface} = 0$

For a parallel plate capacitor $E = \dfrac \sigma \epsilon_o$ where $\sigma$ is the surface charge density which is equal to $\dfrac{Q}{\pi R^2}$

$\Rightarrow E = \dfrac{Q}{\epsilon_o \pi R^2} \Rightarrow \displaystyle \int_{\rm surface} \vec E \cdot d\vec S = \dfrac{Q}{\epsilon_o \pi R^2} \pi r^2 = \dfrac{Q r^2}{\epsilon_o R^2}$

$\displaystyle \mu_o\epsilon_o \dfrac {\partial}{\partial t}\left ( \int_{\rm surface} \vec E \cdot d\vec S\right) = \dfrac{\mu_o I r^2}{R^2}$ because $\dfrac{\partial Q}{\partial t}=I$

Equating the left hand side and the right hand side gives a value for the magnetic field at a distance $r$ from the central axis of the capacitor

$B = \dfrac{\mu_oIr}{2\pi R^2}$ for $0\le r\le R$

and with $r=R$ this gives the familiar $B = \dfrac{\mu_oI}{2\pi R}$

• Nice answer! But I have one question that how can we show that magnetic field inside capacitor plates at some distance r from axis, will be same as outside the plates? Outside, B is given as $B = \dfrac{\mu_o i}{2\pi r}$, but what about inside? Mar 13 '17 at 8:31
• @samjoe I have updated my answer in response to your question. Mar 13 '17 at 9:49
• Thank you for the answer, basically we ahve to apply the ampere's corrected law! Mar 13 '17 at 16:33
• Thank you for the answer, basically we ahve to apply the ampere's corrected law! Mar 13 '17 at 16:34

This is a typical setup to introduce displacement currents.

(Image credit: Haliday and Resnick)

The key in Ampere's law is how to calculate the current enclosed by the Amperian loop. This is done by considering an open surface bounded by the loop, and adding the currents that "puncture" this surface.

Ampere's law does NOT prescribe how this surface should be drawn, only that it should be bounded by the Amperian loop. Thus, both surfaces in your example are technically valid and they both bound the loop, and thus should give the same $\vec B$ field.

There is technically no current that "punctures" second "bottle-shaped" surface since the two plates of the capacitor are not connected by any conducting wire. What "punctures" this second surface is some electric flux, associated with the buildup of the electric field between the plates when charging the capacitor. Hence, to make sure the second case gives the same as the first surface (which is punctured by a conduction current), one writes $$\oint_C \vec H\cdot d\vec \ell = \int_S \frac{\partial \vec D}{\partial t}\cdot d\vec S\, .$$ Since $\vert\vec D\vert= \frac{Q}{A}$ where $Q$ is the charge on the cap, and $A$ is the area of the cap, $\frac{\partial \vec D}{\partial t}\cdot d\vec S=\frac{\partial \vec Q}{\partial t}$ is actually a "current" due to the change with time of the electric displacement; in this specific example this displacement current in fact must be equal in magnitude to the conduction current feeding the capacitor.

This "capacitor argument" is pedagogical because the field $\vec D$ between the plates of a capacitor is constant. A similar argument "in free space" can be made by considering a sufficient small volume of space over which the $\vec D$ would be likewise constant in magnitude, and for which one would ignore the capacitor setup of this problem (with feeding wire). The general argument simply shows that a changing flux of $\vec D$ must induce a magnetic field.

• Still don't get it :( Mar 10 '17 at 6:54