Displacement Current - Ampere-Maxwell Law

We have read how a magnetic field is produced by a moving charged particle (or what we term as electric current). Now we will study the Maxwell’s law, according to which a changing electric field also produces a magnetic field.

In order to visualize how magnetic field is produced due to a change in the electric field, we consider a capacitor of capacitance C, that is being charged and the current flowing through the capacitor is time dependent and is given by i(t).

Applying the Ampere’s circuital law to the system, we can write,1.1

This equation gives the magnetic field at any point outside a capacitor. In order to find the magnetic field at any point P outside the capacitor, we consider a plane circular loop such that its radius is r and its place is perpendicular to the wire connected to this capacitor. Here, the magnetic field is uniform and its magnitude is the same at all the points in the circular loop under consideration. We can write,1.2

Now, we consider another case in which everything else is the same, only here, there is a pot-shaped surface passing through the interior between the capacitor and the loop. Another such case is taken where a capsule shaped surface is considered which has its rim between the two capacitor plates. Here, the perimeter of each surface is assumed to be the same.

Now, as we apply the Ampere’s Law to both these cases, we notice that, the left hand side of the equation is same as the equation written above, but the right hand side of the equation becomes zero, as no current passes through those surfaces. So, we can say that from case one there is a magnetic field at the point P but from the other two cases, the magnetic field at the point P is zero. This case cannot be explained until we reconsider the Ampere’s Circuital Law. Upon further analysis it was discovered that there was a term missing in the equation representing the Ampere’s circuital law that takes this glitch into consideration. In further analysis done on this system, the electric field was also taken into consideration.

For a capacitor with plate area A and a charge Q, the magnitude of electric field between the plates of the capacitor can be given as,1.3

And as per the Gauss’s law, the flux through the surface is given by,1.4

Now as we change the charge on the capacitor, after charging with a current i = (dQ/dt), we can write,1.5

The above equations are consistent, if and only if,1.6

The above mentioned term was the missing term in the Ampere’s Circuital law. We thus generalize this term and add it to the law, which leads us to the same value of the magnetic field in all situations.

As we know, the current caused due to the flow of charges in a conductor is termed as the conduction current. The current due to the change in the electric field, as defined by the above equation is a term different from conduction current. It is termed as the displacement current.

We can thus write, the total current passing through the capacitor can be given as,sum of conducting current and displacement current

Here, ic is the conduction current and id is the displacement current. The generalized Ampere’s circuital law is given by,1.8

The above equation is termed as the Ampere-Maxwell law.

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Practise This Question

A parallel-plate capacitor is being charged. The displacement current across an area in the region between the plates and parallel to it is equal to