**CBSE NOTES CLASS 12 PHYSICS **

**CHAPTER 8 ELECTROMAGNETIC WAVES**

**Displacement current – the missing term in Ampere’s circuital law**

An electrical current produces a magnetic field around it. Maxwell showed that for logical consistency, a changing electric field must also produce a magnetic field.

Let us consider the process of charging of a capacitor. According to Ampere’s circuital law,

$$\oint \stackrel{\u20d7}{\mathrm{B}}.\mathrm{d}\stackrel{\u20d7}{\mathrm{l}}={\mathrm{}\mathrm{\mu}}_{\mathrm{o}}\mathrm{i}\left(\mathrm{t}\right)$$

- Let us first find the magnetic field at a point P, outside the parallel plate capacitor, s shown in the diagram.
For this, we consider a plane circular loop of radius r whose plane is perpendicular to the direction of the current-carrying wire, and which is centred symmetrically with respect to the wire. We can see that the magnetic field is directed along the circumference of the circular loop and is the same in magnitude at all points on the loop. If B is the magnitude of the field, the left side of equation becomes B (2πr), i.e.,

B (2πr) = μ

_{o}i(t) - Now let us consider a different surface, which has the same boundary, but different shape This is a pot like surface or shaped like a tiffin box (without the lid). which nowhere touches the current, but has its bottom between the capacitor plates; its mouth is the circular loop.
On applying Ampere’s circuital law to such surfaces with the same perimeter, we find that the

**left hand side**of the equation**has not changed**but the**right hand side**is**zero**(no current passes through the surface) and**not****μ**_{0 }**i (t)**.

This leads to contradiction giving different magnetic field at the same point.

Maxwell pointed out that this is due to the missing term corresponding to electrical field.

Now considering the surface in second case, the electrical field is perpendicular to the surface ‘S’. It has the same magnitude over the area ‘A’ of the capacitor plates, and is zero outside it.

Using Gauss’s law,

$${\mathrm{\u0424}}_{\mathrm{E}}=\left|\mathrm{E}\right|\mathrm{A}=\mathrm{}\frac{\mathrm{q}}{{\mathrm{\varepsilon}}_{\mathrm{o}}}$$

If the charge q on the capacitor plates changes with time, there is a current,

$${i}_{d}=\frac{dq}{dt}$$

Now,

$$\mathrm{}\frac{{\mathrm{d}\mathrm{\u0424}}_{\mathrm{E}}}{\mathrm{d}\mathrm{t}}=\frac{\mathrm{d}}{\mathrm{d}\mathrm{t}}\left(\frac{\mathrm{q}}{{\mathrm{\varepsilon}}_{\mathrm{o}}}\right)$$

$$=\frac{1}{{\mathrm{\varepsilon}}_{\mathrm{o}}}\mathrm{}\frac{\mathrm{d}\mathrm{q}}{\mathrm{d}\mathrm{t}}=\mathrm{}\frac{1}{{\mathrm{\varepsilon}}_{\mathrm{o}}}\mathrm{}{\mathrm{i}}_{\mathrm{d}}$$

$$\Rightarrow \mathrm{}\mathrm{}{\mathrm{i}}_{\mathrm{d}}=\mathrm{}{\mathrm{\varepsilon}}_{\mathrm{o}}\left(\frac{{\mathrm{d}\mathrm{\u0424}}_{\mathrm{E}}}{\mathrm{d}\mathrm{t}}\right)$$

Maxwell pointed out that this is the missing term in Ampere’s circuital law