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Transcript
Radiation
Lecture 39: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay
So far we have discussed propagation of electromagnetic waves both in free space and in waveguides
without worrying about how they are produced. In the following two lectures we will discuss the
production of electromagnetic waves. Once produced they carry their energy and momentum and
propagate in free space, may be to be received by an antenna. An antenna is a physical device which
transmits or receives electromagnetic waves.
What causes a source to emit radiation? We have seen that charges at rest creates electrostatic field. A
steady current, on the other hand is a source of magnetic field. Neither of these lead to radiation.
However, it is known that accelerating charges emit radiation.
To discuss this in some detail, we fall back on our workhorse, viz., the Maxwellโ€™s equations. Let us recall
the potential formulation of the electromagnetic field that we had discussed earlier.
Using the expression for Faradayโ€™s law,
๐›ป × ๐ธโƒ— = โˆ’
โƒ—
๐œ•๐ต
๐œ•๐‘ก
We rewrite this in terms of vector potential ๐ด,
๐›ป × (๐ธโƒ— +
๐œ•๐ด
)=0
๐œ•๐‘ก
This enablesus to express the quantity within the bracket as gradient of a scalar function V,
๐ธโƒ— = โˆ’๐›ป๐‘‰ โˆ’
๐œ•๐ด
๐œ•๐‘ก
Using Lorentz gauge condition,
๐›ปโ‹…๐ด+
1 ๐œ•๐‘‰
=0
๐‘ 2 ๐œ•๐‘ก
the Maxwellโ€™s equations can be expressed as a pair of homogeneous wave equations,
๐›ป2๐‘‰ โˆ’
1 ๐œ•2๐‘‰
๐œŒ
=โˆ’
2
2
๐‘ ๐œ•๐‘ก
๐œ–
1
(1)
๐›ป2๐ด โˆ’
1 ๐œ•2๐ด
= โˆ’๐œ‡0 ๐ฝ
๐‘ 2 ๐œ•๐‘ก 2
(2)
Since the structure of both equations is same, we can solve either of them and write the solution to the
remaining equation by similarity. We will solve the equation for the vector potential. The method of
solution depends on defining a โ€œGreenโ€™s functionโ€.
Let us define a time Fourier transform through the equation,
๐ด(๐‘ฅ , ๐‘ก) =
1 โˆž
โˆซ ๐ด(๐‘ฅ , ๐œ”) ๐‘’ ๐‘–๐œ”๐‘ก ๐‘‘๐œ”
2๐œ‹ โˆ’โˆž
the inverse of which is
โˆž
๐ด(๐‘ฅ , ๐œ”) = โˆซ ๐ด(๐‘ฅ , ๐‘ก) ๐‘’ โˆ’๐‘–๐œ”๐‘ก ๐‘‘๐‘ก
โˆ’โˆž
Substituting in Eqn. (2)
๐›ป2๐ด โˆ’
1 ๐œ•2๐ด
1 โˆž
๐œ”2
2
=
โˆซ
+
(๐›ป
) ๐ด(๐‘ฅ , ๐œ”) ๐‘’ ๐‘–๐œ”๐‘ก ๐‘‘๐œ”
๐‘ 2 ๐œ•๐‘ก 2 2๐œ‹ โˆ’โˆž
๐‘2
= โˆ’๐œ‡0 ๐ฝ(๐‘ฅ , ๐‘ก)
๐œ‡0 โˆž
=โˆ’
โˆซ ๐ฝ(๐‘ฅ , ๐œ”) ๐‘’ ๐‘–๐œ”๐‘ก ๐‘‘๐œ”
2๐œ‹ โˆ’โˆž
Taking inverse Fourier transform for both sides, we get
๐œ”2
(๐›ป 2 + 2 ) ๐ด(๐‘ฅ , ๐œ”) = โˆ’๐œ‡0 ๐ฝ(๐‘ฅ , ๐œ”)
๐‘
๐œ”
๐‘
Defining ๐‘˜ = , we get,
(๐›ป 2 + ๐‘˜ 2 )๐ด(๐‘ฅ , ๐œ”) = โˆ’๐œ‡0 ๐ฝ(๐‘ฅ , ๐œ”)
To obtain the solution of this equation, we define a subsidiary equation for a function known as the
Greenโ€™s function
โƒ—โƒ—โƒ—โ€ฒ ) = โˆ’4๐œ‹๐›ฟ 3 (๐‘ฅ โˆ’ ๐‘ฅ
โƒ—โƒ—โƒ—โ€ฒ )
(๐›ป 2 + ๐‘˜ 2 )๐บ(๐‘ฅ โˆ’ ๐‘ฅ
Thus the Greenโ€™s function satisfies a very similar equation in which the inhomogeneous term of
the original equation has been replaced by a delta function. In terms of the Greenโ€™s function the
formal solution of the equation for the vector potential can be written as a convolution,
๐ด(๐‘ฅ , ๐œ”) =
๐œ‡0
โƒ—โƒ—โƒ—โ€ฒ , ๐œ”)๐‘‘3 ๐‘ฅโ€ฒ
โˆซ ๐บ(๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅ โ€ฒ )๐ฝ(๐‘ฅ
4๐œ‹
To see how it comes about,
2
๐œ‡0
โƒ—โƒ—โƒ—โ€ฒ , ๐œ”)๐‘‘3 ๐‘ฅ โ€ฒ
โˆซ (๐›ป 2 + ๐‘˜ 2 )๐บ(๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅ โ€ฒ )๐ฝ(๐‘ฅ
4๐œ‹
๐œ‡0
โƒ—โƒ—โƒ—โ€ฒ , ๐œ”)๐‘‘ 3 ๐‘ฅ โ€ฒ = โˆ’๐œ‡0 ๐ฝ(๐‘ฅ , ๐œ”)
=
โˆซ โˆ’ 4๐œ‹๐›ฟ 3 (๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅ โ€ฒ )๐ฝ(๐‘ฅ
4๐œ‹
(๐›ป 2 + ๐‘˜ 2 )๐ด(๐‘ฅ, ๐œ”) =
Remember that in the above, the primed coordinate โƒ—โƒ—โƒ—
๐‘ฅโ€ฒ is for the source and the unprimed
quantity refers to the point where the field is being evaluated. The solution must be spherically
symmetric about the source point, i.e. depend on ๐‘… = |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅโ€ฒ|. With this substitution, the
equation for the Greenโ€™s function becomes,
(๐›ป 2 + ๐‘˜ 2 )๐บ(๐‘…) = โˆ’4๐œ‹๐›ฟ 3 (๐‘…)
First consider ๐‘… โ‰  0, for which the equation becomes
(๐›ป 2 + ๐‘˜ 2 )๐บ(๐‘…) = 0
Expressing the Laplacian in spherical coordinates, as there is no dependence on angles,
1 ๐‘‘2
(๐‘…๐บ) + ๐‘˜ 2 ๐บ = 0
๐‘… ๐‘‘๐‘… 2
the solution for which is seen to be
๐‘…๐บ~๐‘’ ±๐‘–๐‘˜๐‘…
so that the Greenโ€™s function, excepting at the origin has the solution as a linear combination of an
outgoing wave and an incoming wave.
๐บ=๐ด
๐‘’ ๐‘–๐‘˜๐‘…
๐‘’ โˆ’๐‘–๐‘˜๐‘…
+๐ต
๐‘…
๐‘…
What about the solution at the origin?
Let us take a small sphere of radius ๐‘…0 around the origin. Since the argument of the delta function is
included within the sphere, we have,
โˆซ (๐›ป 2 + ๐‘˜ 2 )๐บ(๐‘…)๐‘‘3 ๐‘… = โˆ’4๐œ‹ โˆซ ๐›ฟ 3 (๐‘…)๐‘‘3 ๐‘… = โˆ’4๐œ‹
๐‘…0
๐‘…0
Since ๐บ(๐‘…) is singular only at the origin, we can calculate the integral over it within the sphere, by
replacing it by the expression obtained for it except at the origin,
๐‘…0
๐ด ๐ต
1 2
๐‘…02
โˆซ ๐‘˜ 2 ๐บ(๐‘…)๐‘‘3 ๐‘… = โˆซ ๐‘˜ 2 ( + ) ๐‘‘3 ๐‘… = 4๐œ‹๐‘˜ 2 (๐ด + ๐ต) โˆซ
๐‘… ๐‘‘๐‘… = 4๐œ‹(๐ด + ๐ต)๐‘˜ 2
โ†’0
๐‘… ๐‘…
2
0 ๐‘…
as ๐‘…0 โ†’ 0.
Thus,
3
โˆซ ๐›ป 2 ๐บ(๐‘…)๐‘‘3 ๐‘… = โˆ’4๐œ‹
๐‘…0
As ๐‘…0 โ†’ 0, ๐บ =
๐ด+๐ต
,
๐‘…
so that
๐›ป 2 ๐บ(๐‘…) = (๐ด + ๐ต) ๐›ป 2
1
= โˆ’4๐œ‹(๐ด + ๐ต)๐›ฟ 3 (๐‘…)
๐‘…
Substituting this into the preceding integral, we get
๐ด+๐ต =1
Thus we have,
๐‘’
โƒ—โƒ—โƒ—โƒ—โ€ฒ |
๐‘–๐‘˜|๐‘ฅโˆ’๐‘ฅ
๐‘ฅโ€ฒ|
|๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐บ(๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅโ€ฒ) =
๐‘’
outgoing wave
โƒ—โƒ—โƒ—โƒ—โ€ฒ |
โˆ’๐‘–๐‘˜|๐‘ฅโˆ’๐‘ฅ
๐‘ฅโ€ฒ|
{ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
incoming wave
We will take the outgoing solution and proceed further.
๐ด(๐‘ฅ , ๐œ”) =
๐œ‡0
โƒ—โƒ—โƒ—โ€ฒ , ๐œ”)๐‘‘3 ๐‘ฅ โ€ฒ
โˆซ ๐บ(๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅ โ€ฒ )๐ฝ(๐‘ฅ
4๐œ‹
โƒ—โƒ—โƒ—โƒ—โ€ฒ |
๐‘–๐‘˜|๐‘ฅ โˆ’๐‘ฅ
โˆž
๐œ‡0 ๐‘’
โƒ—โƒ—โƒ—โ€ฒ , ๐‘กโ€ฒ)๐‘’ ๐‘–๐œ”๐‘ก โ€ฒ ๐‘‘๐‘กโ€ฒ๐‘‘3 ๐‘ฅ โ€ฒ
=
โˆซ
โˆซ ๐ฝ(๐‘ฅ
โ€ฒ
4๐œ‹ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅ | โˆ’โˆž
Let us take the inverse Fourier transform of the above to get ๐ด(๐‘ฅ , ๐‘ก). It may, however, be remembered
that ๐‘˜ =
๐œ”
๐‘
has ๐œ” dependence,
๐ด(๐‘ฅ , ๐‘ก) =
โˆž
โˆž
๐‘–๐œ”
1 ๐œ‡0
๐‘‘3 ๐‘ฅ โ€ฒ
โƒ—โƒ—โƒ—โƒ—โ€ฒ
โƒ—โƒ—โƒ—โ€ฒ , ๐‘ก โ€ฒ ) โˆซ ๐‘’ ๐‘–๐œ”๐‘ก โ€ฒ ๐‘’ ๐‘ |๐‘ฅโˆ’๐‘ฅ | ๐‘’ โˆ’๐‘–๐œ”๐‘ก ๐‘‘๐œ”
โˆซ
โˆซ ๐‘‘๐‘ก โ€ฒ ๐ฝ(๐‘ฅ
2๐œ‹ 4๐œ‹ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅ โ€ฒ | โˆ’โˆž
โˆ’โˆž
=
โˆž
1 ๐œ‡0
๐‘‘3 ๐‘ฅ โ€ฒ
๐œ”
โƒ—โƒ—โƒ—โ€ฒ , ๐‘ก โ€ฒ )2๐œ‹๐›ฟ (๐‘ก โ€ฒ โˆ’ ๐‘ก + |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
โˆซ
โˆซ ๐‘‘๐‘ก โ€ฒ ๐ฝ(๐‘ฅ
๐‘ฅ โ€ฒ |)
โ€ฒ
โƒ—โƒ—โƒ—
2๐œ‹ 4๐œ‹ |๐‘ฅ โˆ’ ๐‘ฅ | โˆ’โˆž
๐‘
=
๐œ‡0
๐‘‘3 ๐‘ฅ โ€ฒ
๐œ”
โƒ—โƒ—โƒ—โ€ฒ , ๐‘ก โˆ’ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
โˆซ
๐ฝ (๐‘ฅ
๐‘ฅ โ€ฒ |)
โ€ฒ
4๐œ‹ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘
๐‘ฅ|
Parallely, one can write, for the scalar potential
1
๐‘‘3 ๐‘ฅ โ€ฒ
๐œ”
โƒ—โƒ—โƒ—โ€ฒ , ๐‘ก โˆ’ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘‰(๐‘ฅ , ๐‘ก) =
โˆซ
๐œŒ (๐‘ฅ
๐‘ฅ โ€ฒ |)
4๐œ‹๐œ–0 |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘
๐‘ฅโ€ฒ|
4
Note that the potential at time t at the position where it is being calculated is given in terms of current
๐œ”
โƒ—โƒ—โƒ—โ€ฒ |. The potential is influenced by the wave which came
in the source at an โ€œearlier timeโ€ ๐‘ก โˆ’ |๐‘ฅ โˆ’ ๐‘ฅ
๐‘
from the source at a time which is earlier by the time taken by the electromagnetic wave to travel this
distance. This is known as the โ€œretardedโ€ potential. The other solution is the incoming wave solution
which would give to an โ€œadvancedโ€ solution.
Example : Localized oscillation of charge or current
Let us consider a localized source of charge or current, given by
โƒ—โƒ—โƒ—โ€ฒ , ๐‘ก โ€ฒ ) = ๐ฝ(๐‘ฅ
โƒ—โƒ—โƒ—โ€ฒ )๐‘’ โˆ’๐‘–๐œ”๐‘ก
๐ฝ(๐‘ฅ
โ€ฒ
โƒ—โƒ—โƒ—โ€ฒ , ๐‘ก โ€ฒ ) = ๐œŒ(๐‘ฅ
โƒ—โƒ—โƒ—โ€ฒ )๐‘’ โˆ’๐‘–๐œ”๐‘ก โ€ฒ
๐œŒ(๐‘ฅ
๐ด(๐‘ฅ , ๐‘ก) =
=
๐œ‡0
๐‘‘3 ๐‘ฅ โ€ฒ
๐œ”
โƒ—โƒ—โƒ—โ€ฒ , ๐‘ก โˆ’ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
โˆซ
๐ฝ (๐‘ฅ
๐‘ฅ โ€ฒ |)
โ€ฒ
4๐œ‹ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘
๐‘ฅ|
๐œ‡0
๐‘‘3 ๐‘ฅ โ€ฒ
๐œ”
โƒ—โƒ—โƒ—โ€ฒ ) exp (โˆ’๐‘–๐œ” (๐‘ก โˆ’ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
โˆซ
๐ฝ(๐‘ฅ
๐‘ฅ โ€ฒ |))
4๐œ‹ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘
๐‘ฅโ€ฒ|
โƒ—โƒ—โƒ—โƒ—โ€ฒ |
๐‘–๐‘˜|๐‘ฅ โˆ’๐‘ฅ
๐œ‡0 โˆ’๐‘–๐œ”๐‘ก ๐‘’
โƒ—โƒ—โƒ—โ€ฒ )๐‘‘3 ๐‘ฅโ€ฒ
=
๐‘’
โˆซ
๐ฝ(๐‘ฅ
โ€ฒ
โƒ—โƒ—โƒ—
4๐œ‹
|๐‘ฅ โˆ’ ๐‘ฅ |
Thus we have,
โƒ—โƒ—โƒ—โƒ—โ€ฒ |
๐‘–๐‘˜|๐‘ฅ โˆ’๐‘ฅ
๐œ‡0 ๐‘’
โƒ—โƒ—โƒ—โ€ฒ )๐‘‘3 ๐‘ฅโ€ฒ
๐ด(๐‘ฅ ) =
โˆซ
๐ฝ(๐‘ฅ
4๐œ‹ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅโ€ฒ|
and, similarly,
โƒ—โƒ—โƒ—โƒ—โ€ฒ |
๐‘–๐‘˜|๐‘ฅโˆ’๐‘ฅ
1
๐‘’
โƒ—โƒ—โƒ—โ€ฒ )๐‘‘3 ๐‘ฅโ€ฒ
๐‘‰(๐‘ฅ ) =
โˆซ
๐œŒ(๐‘ฅ
4๐œ‹๐œ–0 |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅโ€ฒ|
๐œ”
where ๐‘˜ = ๐‘ .
There are three regions of interest, assuming that the source is confined to a small region of space with
typical dimension d. The distance at which the field is being measured is R= |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅ โ€ฒ | and the
wavelength of the radiation is ๐œ†.
1. Near field for which ๐‘‘ โ‰ช ๐‘… โ‰ช ๐œ†
2. Intermediate zone for which ๐‘‘ โ‰ช ๐‘… โ‰ˆ ๐œ†
3. Radiation zone or Far field with ๐‘‘ โ‰ช ๐œ† โ‰ช ๐‘…
5
Our primary interest will be in the radiation zone, which we will discuss in detail in the next lecture. We
will have very little to say about the intermediate zone and its analysis is very complicated.
Consider the near field for which ๐‘‘ โ‰ช ๐‘… โ‰ช ๐œ†. In this region ๐‘˜๐‘… =
replace ๐‘’
โƒ—โƒ—โƒ—โƒ—โ€ฒ |
๐‘–๐‘˜|๐‘ฅโˆ’๐‘ฅ
2๐œ‹๐‘…
๐œ†
โ‰ช 1 as a result of which we can
โ‰ˆ 1. The expression for the vector potential becomes
๐ด(๐‘ฅ ) =
๐œ‡0
1
โƒ—โƒ—โƒ—โ€ฒ )๐‘‘3 ๐‘ฅโ€ฒ
โˆซ
๐ฝ(๐‘ฅ
4๐œ‹ |๐‘ฅ โˆ’ โƒ—โƒ—โƒ—
๐‘ฅโ€ฒ|
which is the familiar expression for the vector potential in magnetostatics. The only difference from the
magnetostatic case is the time variation of ๐ด(๐‘ฅ , ๐‘ก) which is simply to multiply the expressions obtained
in the static case by ๐‘’ โˆ’๐‘–๐œ”๐‘ก . Thus we can say that the situation is โ€œquasi-staticโ€ and we can borrow all our
results from magnetostatics to discuss this region.
Radiation
Lecture 39: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay
Tutorial Assignment
1. Consider a thought experiment in which a charge q appears abruptly at the origin at time ๐‘ก = ๐‘ก1
and disappears equally abruptly at time ๐‘ก = ๐‘ก2 . Find the retarded potentials for the problem.
Determine the corresponding fields.
2. A long straight wire lying along the z axis carries a current given by
0 (๐‘ก โ‰ค 0)
๐ผ(๐‘ก) = {
๐ผ0 ๐‘ก (๐‘ก > 0)
Determine the retarded potentials and the electric and the magnetic fields.
Solutions to Tutorial Assignments
๐‘ฅ
1. Since the charge is static only retarded scalar potential is generated. For time ๐‘ก < ๐‘ก1 + ๐‘ the
information that a charge has appeared at the origin has not reached the position x. Thus the
potential is zero for such time. Likewise, though the charge has disappeared at time ๐‘ก2 , this
๐‘ฅ
information would not reach the position x till a later time ๐‘ก2 + ๐‘ . Thus the potential is
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1
๐‘‰(๐‘ฅ ) = 4๐œ‹๐œ–
๐‘ž
4๐œ‹๐œ–0 ๐‘Ÿ 2
๐‘ž
0๐‘ฅ
๐‘ฅ
for ๐‘ก1 < ๐‘ก โˆ’ ๐‘ < ๐‘ก2 and zero at other times. The corresponding field is ๐ธโƒ— =
๐‘Ÿฬ‚ in the same time interval and is zero at other times.
2. The wire being charge neutral, the scalar potential is zero. The vector potential, because of
cylindrical symmetry depends only on the distance s of the point of observation P from the wire,
and is given by
๐œ‡0
๐ผ(๐‘ก โ€ฒ )
๐ด(๐‘ , ๐‘ก) =
๐‘งฬ‚ โˆซ
๐‘‘๐‘ง
4๐œ‹
๐‘Ÿ
r
z
O
P
s
Consider an element dz at a distance z form the origin (see figure). Information about the
๐‘Ÿ
current element would reach the point P at a time ๐‘ =
โˆš๐‘ง 2 +๐‘ 2
๐‘
later than the time the current
was actually established. Thus the current, as seen by the point P is given by
๐ผ(๐‘ก โ€ฒ ) = {
๐‘Ÿ
๐‘
where ๐‘ก โ€ฒ = ๐‘ก โˆ’ = ๐‘ก โˆ’
โˆš๐‘ง 2 +๐‘ 2
.
๐‘
0 (๐‘ก โ€ฒ โ‰ค 0)
๐ผ0 ๐‘ก โ€ฒ (๐‘ก โ€ฒ > 0)
Now let us look at the range of integration. For an element dz
located at a distance z from the origin to influence the field at the point P, its distance from P
should be less than ct. Thus, โˆš๐‘ง 2 + ๐‘  2 < ๐‘๐‘ก . Hence the potential is given by
โˆš๐‘
๐œ‡0
๐ด(๐‘ , ๐‘ก) = 2
๐ผ0 ๐‘งฬ‚ โˆซ
4๐œ‹
0
2 ๐‘ก 2 โˆ’๐‘ 2
โˆš๐‘ง 2 +๐‘ 2
๐‘
๐‘‘๐‘ง
โˆš๐‘ง 2 + ๐‘  2
๐‘กโˆ’
(the factor of 2 in front is because only the upper half of the wire is considered in the
integration). The integral is straightforward and gives,
๐ด(๐‘ , ๐‘ก) =
๐œ‡0
๐‘๐‘ก + โˆš๐‘ 2 ๐‘ก 2 โˆ’ ๐‘  2
โˆš๐‘ 2 ๐‘ก 2 โˆ’ ๐‘  2
๐ผ0 ๐‘งฬ‚ [๐‘ก ln (
]
)โˆ’
2๐œ‹
๐‘ 
๐‘
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The electric field is given by
๐ธโƒ— = โˆ’
๐œ•๐ด
๐œ‡0
๐‘๐‘ก + โˆš๐‘ 2 ๐‘ก 2 โˆ’ ๐‘  2
= โˆ’ ๐ผ0 ๐‘งฬ‚ ln (
)
๐œ•๐‘ก
2๐œ‹
๐‘ 
and the magnetic field is given by
โƒ— =โˆ‡×๐ด=
๐ต
๐œ•๐ด๐‘ง
๐œ‡0
๐œ‘ฬ‚ =
๐ผ โˆš๐‘ 2 ๐‘ก 2 โˆ’ ๐‘  2 ฬ‚
๐œ‘
๐œ•๐‘ 
2๐œ‹๐‘ ๐‘ 0
Radiation
Lecture 39: Electromagnetic Theory
Professor D. K. Ghosh, Physics Department, I.I.T., Bombay
Self Assessment Questions
1. An long straight wire lies along the z direction. At time ๐‘ก = 0, a constant current ๐ผ0 is
switched on. Calculate the vector potential and the electric and the magnetic fields.
2. A conducting wire is bent into a loop of two semicircles, one of radius ๐‘…1 and the other
๐‘…2 , (< ๐‘…1 ) connected by straight segments. The loop carries a current ๐ผ(๐‘ก) = ๐ผ0 ๐‘ก, which is
directed anticlockwise on the shorter semicircle and clockwise on the other one.. Calculate
the vector potential at the centre of the loop. Calculate the electric field at the centre. Can
you calculate the magnetic field from the expression of the vector potential?
๐‘…1
๐‘…2
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Solutions to Self Assessment Questions
1. The problem is very similar to Problem 2 of the Tutorial. The vector potential is given by
๐ด(๐‘ , ๐‘ก) =
๐œ‡0
๐‘๐‘ก + โˆš๐‘ 2 ๐‘ก 2 โˆ’ ๐‘  2
๐ผ0 ๐‘งฬ‚ ln (
)
2๐œ‹
๐‘ 
2.
y
๐‘…1
๐‘…2
O
x
Since the current is switched on at t=0, the (retarded) vector potential at the centre O is given
by
๐‘Ÿ
๐œ‡0 ๐ผ๐‘œ (๐‘ก โˆ’ ๐‘ )
๐œ‡0 ๐ผ๐‘œ
1
๐œ‡0 ๐ผ๐‘œ
๐ด=
โˆซ
๐‘‘๐‘™ =
๐‘กโˆซ ๐‘‘๐‘™โƒ— โˆ’
โˆซ ๐‘‘๐‘™โƒ—
4๐œ‹
๐‘Ÿ
4๐œ‹
๐‘Ÿ
4๐œ‹๐‘
Since โˆฎ ๐‘‘๐‘™ = 0, the second integral vanishes. We are left with ๐ด =
๐œ‡0 ๐ผ๐‘œ
1
๐‘กโˆซ ๐‘Ÿ ๐‘‘๐‘™โƒ— .
4๐œ‹
Note that the
integral is a line integral over the loop, which is directed differently at different points of the
loop. Let us take the loop to be in the x-y plane with directions as shown in the figure. Recall
that the direction of ๐‘‘๐‘™โƒ— is along the direction of the current. The contribution from the two
straight sections give โˆ’
๐œ‡0 ๐ผ๐‘œ
๐‘…
๐‘ก ln ๐‘…2 ๐‘–ฬ‚,
4๐œ‹
1
the negative sign is due to the fact that the current in both
the sections are directed along the negative x direction.
We now have to perform the integration over the two semicircles. Consider the integration
over the bigger semicircle. We can resolve ๐‘‘๐‘™ = ๐‘…1 ๐‘‘๐œƒ(cos ๐œƒ๐‘–ฬ‚ + sin ๐œƒ๐‘—ฬ‚) . The denominator of
the integral also has ๐‘…1 so that the integral becomes independent of ๐‘…1 (In a similar way, the
integral of the smaller semicircle will be independent of ๐‘…2 ). The y-component of the integral
would vanish by symmetrically placed elements on either side of the y-axis, leaving us with
๐œ‡0 ๐ผ0 ๐‘ก
๐‘–ฬ‚
4๐œ‹
as the value of the integral. The integral over the smaller semicircle is done in a similar
way but its direction would be opposite because of the direction of the current. Thus the
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contribution from the two semicircles would cancel and the vector potential at the centre would
be given by โˆ’
๐œ‡0 ๐ผ๐‘œ
๐‘…
๐‘ก ln ๐‘…2 ๐‘–ฬ‚
4๐œ‹
1
. The electric field is trivial and is given by
๐ธโƒ— = โˆ’
๐œ•๐ด ๐œ‡0
๐‘…1
=
๐ผ0 ln ๐‘–ฬ‚
๐œ•๐‘ก 4๐œ‹
๐‘…2
Since the vector potential has been calculated only at a single point, it is not possible to
calculate its curl and hence the magnetic field cannot be determined by the above calculation.
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