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Physics : Electromagnetic Waves: Displacement current and Maxwellâ€™s correction to Ampere's circuital law

**Displacement current**** ****and Maxwellâ€™s correction
to Ampere's circuital law**

In unit 4, we studied
Faradayâ€™s law of electromagnetic induction which states that the change in
magnetic field produces an electric field. Mathematically

where Î¦ is the magnetic
flux and âˆ‚/âˆ‚t is the partial derivative with respect to time. Equation (5.1)
means that the electric field * *is induced along a closed
loop by the* *changing magnetic flux Î¦_{B} in the region encircled
by the loop. Now the question asked by James Clerk Maxwell is â€˜Is converse of
this statement true?â€™ Answer is â€˜yesâ€™. He showed that the change in electric
field also produces magnetic field which is

where Î¦_{B} is
the electric flux. This is known as Maxwellâ€™s law of induction, which explains
that the magnetic field induced along a closed loop by
the changing electric flux Î¦_{E} in the region encircled by that loop.
This in turn, explains the existence of radio waves, gamma rays, infrared rays,
etc.

In order to understand
how the changing electric field produces magnetic field, let us consider a
situation of â€˜charging a parallel plate capacitorâ€™ shown in Figure 5.3 Assume
that the medium in between the capacitor plates is a non-conducting medium.

The electric current
passing through the wire is the conduction current I_{C}. This current
generates magnetic field around the wire (refer Unit 3) connected across the
capacitor. Therefore, when a magnetic needle is kept near the wire, deflection
is observed. In order to compute the strength of magnetic field at a point, we
use Ampere's circuital law (from Unit 3) which states that â€˜the line integral
of the magnetic field around any closed loop is equal to
Âµ_{o} times the net current I threading through the area enclosed by
the loopâ€™. Ampere's law in equation form is

where Âµ_{o} is
the permeability of free space.

To calculate the
magnetic field at a point P near the wire as shown in Figure 5.4, let us draw
an amperian loop (circular loop) which encloses the surface S_{1}
(circular surface). Therefore, using Ampere's circuital law (equation 5.3), we
get

where I_{c} is
the conduction current.

Suppose the same loop is
enclosed by balloon shaped surface S_{2} as shown in Figure 5.5. This
means that the boundaries of two surfaces S_{1} and S_{2} are
same but shape of the enclosing surfaces are different (first surface (S_{1})
is circular in shape and second one is balloon shaped surface (S_{2})).
As the Ampereâ€™s law applied for a given closed loop does not depend on shape of
the enclosing surface, the integrals will give the same answer. But by applying
Ampereâ€™s circuital law (equation 5.3), we get

The right hand side of
equation is zero because the surface S_{2} no where touches the wire
carrying conduction current and further, there is no current in between the
plates of the capacitor (there is a discontinuity). So the magnetic field at a
point P is zero. Hence there is an inconsistency between equation (5.4) and
equation (5.5). J. C.Maxwell resolved this inconsistency as follows:

Due to external source
(battery or cell), the capacitor gets charged up because of current flowing
through the capacitor. This produces an increasing electric field between the
capacitor plates. So, there must be a current associated with the changing
electric field in between the capacitor plates. In other words, the time
varying electric flux (or time varying electric field) existing between the
plates of the capacitor also produces a current known as displacement current.

From Gaussâ€™s law (refer
Unit 1), the electric flux between the plates of the capacitor (Figure 5.6) is

where A is the area of
the plates of capacitor.

The change in electric
flux is

where I_{d} is
known as displacement current. The **displacement current can be defined** **as
the current which comes into play in the region in which the electric field and
the electric flux are changing with time**. In** **other words, whenever
the change in electric field takes place, displacement current is produced.
Maxwell modified Ampere's law as

where I = I_{c}
+ I_{d} which means the total current enclosed by the surface is sum of
conduction current and displacement current. When a constant current is
applied, displacement current I_{d} = 0 and hence I_{c} = I.
Between the plates, the conduction current I_{c} = 0 and hence I_{d}
= I.

**EXAMPLE 5.1**

Consider a parallel
plate capacitor which is maintained at potential of 200 V. If the separation
distance between the plates of the capacitor and area of the plates are 1 and
20 cm^{2}. Calculate the displacement current for the time in Âµs.

**Solution**

Potential difference
between the plates of the capacitor, V = 200 V

The distance between the
plates,

d = 1 mm = 1 Ã— 10^{-3}
m

Area of the plates of
the capacitor,

A = 20 cm^{2} =
20 Ã— 10^{-4} m^{2}

Time is given in
micro-second, Âµs = 10^{-6} s

Displacement current

But electric field, E =
V/d

Therefore,

= 35400 Ã—10^{âˆ’7} = 3 .5 mA

Tags : Electromagnetic Waves , 12th Physics : Electromagnetic Waves

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