lecture3.1

Chapter III

Maxwell Equations

Continuity Equation:

To derive the continuity equation we consider a tube of infinitesimal cross-sectional area parallel to flow. If dI is the current flowing then the volume current density is

---(1)

J is actually current per unit area perpendicular to the flow. The current crossing a surface S can be written as

The total charge per unit time leaving volume V is written using Divergence theorem,

---(3)

As the charge is conserved quantity therefore whatever charge is

Flowing out of the volume must be coming at the expense of the charge remaining inside the volume, so we write

---(4) -Ve sign in above Eq. Shows that the outward flow of the charge

Decreases the charge inside.

Faraday’s law of electromagnetic induction: This law states that the Changing magnetic flux can produce the induced emf or induced E.F. In differential form the Faraday’s law is written as

---(1)

Above equation tells us that electric field intensity in a region of time varying magnetic flux density is non-conservative and cannot be

expressed as the gradient of scalar potential.

Using Strokes theorem above Eq. can be written as,

---(2)

Induced Electric Fields: According to faraday’s law, the curl of induced E.F. Is given by

To completely determine the Faraday’s induced E.F. we also need to determine the divergence of this field. As long as E is pure

Faraday’s E.F. (means E is because of change in B), its divergence will be



Electrodynamics Before Maxwell:

So far we studied following equations

---(1)

---(2)

---(3)



However there is one problem with the fourth

equation i.e. Ampere’s Law. Actually Ampere’s

Law is valid only if any E.F. , if Present in problem , is

constant in time. If it is not constant in time then we need to

modify the Ampere law Let us Discuss the issue.

Recall:

We know divergence of curl is always zero.

_{Now we apply above fact to eqn (iii) and eqn (iv)} From eqn (iii), we have

In above eqn left side is zero because divergence of

curl is zero. The right side is also zero because of

eqn(ii). Thus for eqn (iii) everything seems to be fine.



Now let us apply the above to eqn (iv). Taking

divergence of both side of eqn (iv), we have



Now there is trouble with above equation. The left

side must be zero because divergence of curl is always

Zero.

But the right side is not always. For steady current

the divergence of J is zero.

= 0

However when we have time varying current then

divergence is not zero.

.

# 0

Another way to look at the non-validation of Ampere’s Law

for non-steady current:

We consider a charging capacitor as shown in the figure.

When we consider the conduction current I in the figure which charge the capacitor

then this current is present only in wire not in the gap between the plates of the capacitor. Now we consider two surfaces S₁ and

When we consider the surface S₁ then the current enclosed is the conduction current I i.e. I_enc = I. But when we consider the surface S₂ then there is know current passing through this surface and

Modification of Ampere’s Law by Maxwell:



Since we are encountering the problem with the

Right side of eqn. (6), so proceed as follows:

We write continuity equation using differential

form of Gauss law as:

Now in Ampere’s law, the current density J is

combined with the term and

then we will get following eqn

---(8)

modified form of Ampere’s law

Now note that problem of divergence is eliminated.

Because if we take divergence of above eqn

then with the help of eqn (7) the divergence



Note that when we have the case of magneto static

then E will be constant and hence 2

term will be

zero and thus we will again get the Ampere’s law

of magneto static.

Eqn (8) is known as modified form of Ampere’s law.



Note that apart from curing the Ampere’s law, modified

form by Maxwell also says that the changing E.F.

Extra term of Maxwell is known as Displacement current

---(9) Now let us go back to the paradox of capacitor.

The E.F. Between the plates of capacitor

---(10)

Where Q is the charge on the plate of capacitor and A is area of plate. Now between the plates of capacitor, we have

Now integral form of modified form of ampere’s law can be written as

----(12)

Now if we choose flat surface S₁ then E = 0 and I_enc = I.

On the other hand if we choose balloon Shaped surface S₂ then I_enc = 0 in

above eqn but

---(13) .

Thus we will get the same answer for

two surfaces. In 1st case answer comes through conduction