Sulaimani University - College of Science - Physics Department Solution of third examination of Electromagnetic Theory - Fourth Stage
Derivation of the Lienard-Wiechert Potentials and the Heaviside-Feynman formula for the electromagnetic field of a moving point charge.Descripción completa
Solman AcctgFull description
EMT book U.A Bakshi 8th editionFull description
EMT book U.A Bakshi 8th edition
States Electromagnetic Devices
Electricity and Magnetisme Homework V Solution Roald K. Wangsness, Electromagnetic Fields 2nd Edition, Chapter 19 by Chalis Setyadi 1. Exercise 19-15 : A point dipole m is located at the origin, but it has no special orientatio orientation n with respect to the coordinate coordinate axes. (For (For example,m is not parallel to any of the axes.) Express its potential A at a point r in rectangular coordinates, and find the rectangular components of B . Show that B can be written in the form B(r) =
µ0 [3 (m · ˆr) ˆr − m] 4πr 3
(19 − 55)
and compare with (8-84). Solution : In rectangular coordinate, we can write :
ˆ + my y ˆ + mz zˆ = mx x ˆ + y y ˆ + z z ˆ r = xx
m
r
= |r| = x2 + y 2 + z 2
1/2
(1)
Then inserting equation equation (1) to equation (19-55) we get B(r)
= =
B(x , y , z )
=
µ0 [3 (m · ˆr) ˆr − m] 4πr 3 m µ0 3 (m · r) r − 3 5 4π r r µ0 4π
ˆ + y y ˆ + z zˆ) 3 (mx x + my y + mz z ) (xx 5/2
(x2 + y 2 + z 2 )
−
ˆ + my y ˆ + mz zˆ) ( mx x 3/2
(x2 + y 2 + z 2 )
(2) And we can write Bx
=
µ0 4π
By
=
µ0 4π
Bz
=
µ0 4π
3 (mx x + my y + mz z ) x 5/2
(x2 + y 2 + z 2 )
3 (mx x + my y + mz z ) y (x2 + y 2 + z 2 )
5/2
3 (mx x + my y + mz z ) z 2 5/2
(x2 + y 2 + z )
− − −
mx
(x2 + y 2 + z 2 )
3/2
my
(x2 + y 2 + z 2 )
3/2
mz
(x2 + y 2 + z 2 )3/2
(3)
Then we will show that we can get expressions in equation ( 3) from B = ∇ × A in rectangular coordinat. The first problem problem in this exercise is expressing expressing potential potential vector vector A in rect-
1
angular coordinat. We can can do it using the terms in equation ( 1) µ0 m × r 4π r 3 ˆ + my y ˆ + mz ˆz) × (xx ˆ + y y ˆ + z ˆz) µ0 (mx x 3 2 / 4π (x2 + y 2 + z 2 )
=
A
=
(4)
µ0 (my z − mz y) ˆ x + (mz x − mx z ) ˆ y + ( mx y − my x) ˆ z 3 / 2 4π (x2 + y 2 + z 2 )
=
(5)
Using expression A in equation (4) and vector triple product in equation (1-121) in the book, we can write B =
B
=
r µ0 m × r µ0 ∇ × A = ∇ × = ∇× m× 3 3 4π r 4π r µ0 r r r r ∇ · 3 m − (∇ · m) 3 + 3 · ∇ m − (m · ∇) 3 r r r r 4π
(6)
We know that the second term (∇ · m) r and the third term r · ∇ m of equation (6) are zero (0) because m is constant (not a function of any variable) as it can be seen from equation (1). The fisrt term is also zero because ∇ · r = 0 as we can check it in rectangular coordinat: r
3
r
3
r
3
∇·
r r3
=
=
∂ ∂x
+
∂ ∂z
=
∂ ∂ ∂ ˆ + ˆ ˆz x y + ∂x ∂y ∂z
x
·
2 3/2
(x2 + y 2 + z ) z
ˆ + y y ˆ + z ˆz xx
(x2 + y 2 + z 2 )3/2
+
∂ ∂y
y
(x2 + y 2 + z 2 )3/2
3/2
(x2 + y 2 + z 2 ) 3
(x2 + y 2 + z 2 )3/2
−
3 x2 + y 2 + z 2
(x2 + y 2 + z 2 )5/2
=0
(7)
Then the remaining term of equation (6) is the last term B
=
r µ0 − (m · ∇) 3 4π r
=
µ0 − 4π
∂ ∂ ∂ + mz mx + my ∂x ∂y ∂z
We can solve equation (8) using two methods.
2
ˆ + y y ˆ + z zˆ xx (x2 + y 2 + z 2 )
3/2
(8)
(a) Direct calculation in rectangular coordinate For the x component of B = −
Bx
µ0 4π
µ0 4π
=
mx
∂ ∂ ∂ + my + mz ∂x ∂y ∂z
3x (mx x + my y + mz z )
−
(x2 + y 2 + z 2 )5/2
x
(x2 + y 2 + z 2 ) mx
(x2 + y 2 + z 2 )3/2
3/2
(9)
The y and z component of B can be calculated using the same method By
=
µ0 4π
Bz
=
µ0 4π
3y (mx x + my y + mz z ) (x2 + y 2 + z 2 )5/2 3z (mx x + my y + mz z ) 2 5/2
(x2 + y 2 + z )
− −
my
(x2 + y 2 + z 2 )3/2 mz
(x2 + y 2 + z 2 )
3/2
(10)
Comparing equation (9) and (10) with equation (3), we can see that thats all are the same expression. QED. (b) Using tensor notation We can write equation (8) in tensor notation as B
µ0 (3m · r) r m − 3 5 r r 4π µ0 [(3 m · ˆr) ˆr − m] QED 4πr 3
(11)
2. Exercise 19-10 : A point dipole m1 is located at r1 and another point dipole m2 is at r2 . Either by direct calculation or by transcription of the corresponding results for the electric case, show that the potential energy 3
of m2 in the induction of m1 is given by the dipole-dipole interaction energy µ0 ˆ U DD = (m1 · m2 ) − 3 m1 · R 4πR 3
ˆ m2 · R
(19 − 56)
where R = r 2 − r1 . Similarly, find the force F2 on m2 .
Solution : Magnetic field produced by m1 in position m2 can be calcu-
The first and third term of equation ( 15) are zero because m 2 is contant. The second term is also zero because the source of radiation B1 is in m1 but our integral is over m 2 , so there are no source of magnetic field ( J = 0) enclosed by the integral
(∇ × B1 ) · da2 =
B1 · ds2 = µ 0 J = 0
(16)
Then the remaining term for F2 is F2
= ∇ (m2 · B1 ) = (m2 · ∇) B1 = =
∂ ∂ ∂ + m2y + m2z m2x ∂x 2 ∂y 2 ∂z 2
µ0 4π
m2x
µ0 ˆ ˆ 3 m1 · R R − m1 4πR 3
∂ ∂ ∂ + m2y + m2z ∂x 2 ∂y 2 ∂z 2
There are two methods to solve equation (17). 4
3 (m1 · R) R 5
R
−
m1 R3
(17)
(a) Direct calculation in Cartesian coordinat The x component of F2 are =
F 2x
= + + = + −
3x (m1x x + m1y y + m1z z ) m1x − 2 2 2 2 5/2 (x + y + z ) (x + y 2 + z 2 )3/2 µ0 3 (2xm1x + ym 1y + zm 1z ) 15 x2 (xm1x + ym 1y + zm 1z ) 3xm1x m2x − + 2 2 2 2 5/2 2 2 2 7/2 4π (x + y + z ) (x + y + z ) (x + y 2 + z 2 )5/2 µ0 3xm1y 15 xy (xm1x + ym 1y + zm 1z ) 3ym 1x + 2 m2y − 2 2 2 5/2 2 2 2 7/2 4π (x + y + z ) (x + y + z ) (x + y 2 + z 2 )5/2 µ0 3xm1z 15 xz (xm1x + ym 1y + zm 1z ) 3zm 1x − + 2 m2z 2 2 2 5/2 2 2 2 7/2 4π (x + y + z ) (x + y + z ) (x + y 2 + z 2 )5/2 µ0 3m1x (xm2x + ym 2y + zm 2z ) 3 m2x (xm1x + ym 1y + zm 1z ) + 4π (x2 + y 2 + z 2 )5/2 (x2 + y 2 + z 2 )5/2 µ0 3x(m1x m2x + m1y m2y + m1z m2z ) 4π (x2 + y 2 + z 2 )5/2 µ0 15x(xm2x + ym 2y + zm 2z )(xm1x + ym 1y + zm 1z ) (18) 4π (x2 + y 2 + z 2 )7/2 µ0 4π
∂ ∂ ∂ m2x + m2y + m2z ∂x ∂y ∂z
Using the same method, y and z components of F2 are F 2y
F 2z
=
µ0 4π
+
µ0 4π
−
µ0 4π
=
µ0 4π
+
µ0 4π
−
µ0 4π
3m1y (xm2x + ym 2y + zm 2z ) 3 m2y (xm1x + ym 1y + zm 1z ) + (x2 + y 2 + z 2 )5/2 (x2 + y 2 + z 2 )5/2 3y(m1x m2x + m1y m2y + m1z m2z ) (x2 + y 2 + z 2 )5/2 15y (xm2x + ym 2y + zm 2z )(xm1x + ym 1y + zm 1z ) (x2 + y 2 + z 2 )7/2 3m1z (xm2x + ym 2y + zm 2z ) 3 m2z (xm1x + ym 1y + zm 1z ) + (x2 + y 2 + z 2 )5/2 (x2 + y 2 + z 2 )5/2 3z (m1x m2x + m1y m2y + m1z m2z ) (x2 + y 2 + z 2 )5/2 15z (xm2x + ym 2y + zm 2z )(xm1x + ym 1y + zm 1z ) (19) (x2 + y 2 + z 2 )7/2
Then F2
ˆ + F 2y y ˆ + F 2z ˆz = F 2x x 3µ0 ˆ + (m1 · R ˆ )m2 + ( m2 · R ˆ )m1 − 5(m1 · R ˆ )(m2 · R ˆ )R ˆ = (m1 · m2 )R 4πR 6 (20)
µ0 R R R R 4π 3µ0 ˆ + (m1 · R ˆ )m2 + ( m2 · R ˆ )m1 − 5(m1 · R ˆ )(m2 · R ˆ )R ˆ (m1 · m2 )R 4πR 6
3. Exercise 19-12 : A circular ring of radius a lies in the xy plane with the origin at its center. It carries a current I circulating clockwise when viewed from a point on the positive z axis. A point dipole m = m zˆ is on the z axis. Find the z component of the force on m . Solution : Magnetic field produced by I in the dipole m position is µ0 Ids × R (22) B = 4π C R3
with R
ˆ + y y ˆ ) = z ˆz − (a cos φx ˆ + a sin φy ˆ) = z ˆz − (xx 1/2
= a2 + z 2 ˆ + dy y ˆ = −a sin φdφx ˆ + a cos φdφy ˆ ds = dxx ˆ + z sin φy ˆ + azˆ) dφ ds × R = a (z cos φx
R
(23)
and because the current I circulates clockwise, it has negative sign, so B
