Baixe wangsness electromagnetic fields solutions - Ch19 e outras Notas de estudo em PDF para Física, somente na Docsity!
Electricity and Magnetisme Homework V Solution Roald K. Wangsness, Electromagnetic Fields 2nd Edition, Chapter 19 by Chalis Setyadi
- Exercise 19-15 : A point dipole m is located at the origin, but it has no special orientation with respect to the coordinate axes. (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) = μ^0 4 πr^3 [3 (m · ˆr) ˆr − m] (19 − 55)
and compare with (8-84). Solution : In rectangular coordinate, we can write : m = mx xˆ + my ˆy + mz zˆ r = xˆx + yyˆ + zˆz r = |r| =
x^2 + y^2 + z^2
Then inserting equation (1) to equation (19-55) we get
B(r) = μ 0 4 πr^3 [3 (m · ˆr) ˆr − m]
= μ^0 4 π
[
3 (m · r) r r^5 − m r^3
]
B(x, y, z) = μ^0 4 π
[
3 (mxx + my y + mz z) (xˆx + yyˆ + zˆz) (x^2 + y^2 + z^2 )^5 /^2
− (mx^ ˆx + my ˆy + mz ˆz) (x^2 + y^2 + z^2 )^3 /^2
]
And we can write
Bx = μ^0 4 π
[
3 (mxx + my y + mz z) x (x^2 + y^2 + z^2 )^5 /^2
− mx (x^2 + y^2 + z^2 )^3 /^2
]
By = μ^0 4 π
[
3 (mxx + my y + mz z) y (x^2 + y^2 + z^2 )^5 /^2
− my (x^2 + y^2 + z^2 )^3 /^2
]
Bz = μ 0 4 π
[
3 (mxx + my y + mz z) z (x^2 + y^2 + z^2 )^5 /^2
mz (x^2 + y^2 + z^2 )^3 /^2
]
Then we will show that we can get expressions in equation (3) from B = ∇ × A in rectangular coordinat.
The first problem in this exercise is expressing potential vector A in rect-
angular coordinat. We can can do it using the terms in equation (1)
A = μ 0 4 π
m × r r^3
μ 0 4 π
(mx ˆx + my yˆ + mz ˆz) × (xˆx + yyˆ + zˆz) (x^2 + y^2 + z^2 )^3 /^2 = μ 0 4 π
(my z − mz y) ˆx + (mz x − mxz) ˆy + (mxy − my x) ˆz (x^2 + y^2 + z^2 )^3 /^2
Using expression A in equation (4) and vector triple product in equation (1-121) in the book, we can write B
B = ∇ × A = ∇ ×
μ 0 4 π
m × r r^3
μ 0 4 π ∇ ×
m × r r^3
= μ^0 4 π
[(
∇ · r r^3
m − (∇ · m) r r^3
( (^) r r^3
m − (m · ∇) r r^3
]
We know that the second term
(∇ · m) (^) rr 3
and the third term
(( (^) r r^3 · ∇
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 ∇ · (^) rr 3 = 0 as we can check it in rectangular coordinat:
∇ · r r^3
∂x xˆ + ∂ ∂y yˆ + ∂ ∂z ˆz
xxˆ + yˆy + zˆz (x^2 + y^2 + z^2 )^3 /^2
∂x
x (x^2 + y^2 + z^2 )^3 /^2
∂y
y (x^2 + y^2 + z^2 )^3 /^2
∂z
z (x^2 + y^2 + z^2 )^3 /^2
(x^2 + y^2 + z^2 )^3 /^2
x^2 + y^2 + z^2
(x^2 + y^2 + z^2 )^5 /^2
Then the remaining term of equation (6) is the last term
B = − μ 0 4 π
[
(m · ∇) r r^3
]
μ 0 4 π
[(
mx
∂x
∂y
∂z
xxˆ + yˆy + zˆz (x^2 + y^2 + z^2 )^3 /^2
]
We can solve equation (8) using two methods.
of m 2 in the induction of m 1 is given by the dipole-dipole interaction energy
U (^) DD′ = μ 0 4 πR^3
[
(m 1 · m 2 ) − 3
m 1 · Rˆ
m 2 · Rˆ
)]
where R = r 2 − r 1. Similarly, find the force F 2 on m 2. Solution : Magnetic field produced by m 1 in position m 2 can be calcu- lated from equation (19-55)
B 1 = μ^0 4 πR^3
[
m 1 · Rˆ
Rˆ − m 1
]
with
R = r 2 − r 1 |R| =
(x 2 − x 1 )^2 + (y 2 − y 1 )^2 + (z 2 − z 1 )^2
x^2 + y^2 + z^2
Then the potentioal energy of m 2 is
U (^) DD′ = −m 2 · B 1 = μ^0 4 πR^3 m 2 ·
[
m 1 − 3
m 1 · Rˆ
Rˆ
]
μ 0 4 πR^3
[
(m 1 · m 2 ) − 3
m 1 · Rˆ
m 2 · Rˆ
)]
and the force on m 2 is
F 2 = −∇U (^) DD′ = ∇ (m 2 · B 1 ) = B 1 × (∇ × m 2 ) + m 2 × (∇ × B 1 ) + (B 1 · ∇) m 2 + (m 2 · ∇) B 1 (15)
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 B 1 is in m 1 but our integral is over m 2 , so there are no source of magnetic field (J = 0) enclosed by the integral ∫ (∇ × B 1 ) · da 2 =
B 1 · ds 2 = μ 0 J = 0 (16)
Then the remaining term for F 2 is
F 2 = ∇ (m 2 · B 1 ) = (m 2 · ∇) B 1
=
m 2 x
∂x 2
∂y 2
∂z 2
μ 0 4 πR^3
[
m 1 · Rˆ
Rˆ − m 1
])
μ 0 4 π
m 2 x
∂x 2
∂y 2
∂z 2
3 (m 1 · R) R R^5
m 1 R^3
There are two methods to solve equation (17).
(a) Direct calculation in Cartesian coordinat The x component of F 2 are
F 2 x = μ 0 4 π
m 2 x
∂x
∂y
∂z
3 x (m 1 xx + m 1 y y + m 1 z z) (x^2 + y^2 + z^2 )^5 /^2
m 1 x (x^2 + y^2 + z^2 )^3 /^2
= μ^0 4 π m 2 x
3 (2xm 1 x + ym 1 y + zm 1 z ) (x^2 + y^2 + z^2 )^5 /^2 − 15 x
(^2) (xm 1 x + ym 1 y + zm 1 z ) (x^2 + y^2 + z^2 )^7 /^2
- 3 xm^1 x (x^2 + y^2 + z^2 )^5 /^2
3 xm 1 y (x^2 + y^2 + z^2 )^5 /^2 − 15 xy^ (xm^1 x^ +^ ym^1 y^ +^ zm^1 z^ ) (x^2 + y^2 + z^2 )^7 /^2
- 3 ym^1 x (x^2 + y^2 + z^2 )^5 /^2
3 xm 1 z (x^2 + y^2 + z^2 )^5 /^2 − 15 xz^ (xm^1 x^ +^ ym^1 y^ +^ zm^1 z^ ) (x^2 + y^2 + z^2 )^7 /^2
- 3 zm^1 x (x^2 + y^2 + z^2 )^5 /^2
μ 0 4 π
3 m 1 x(xm 2 x + ym 2 y + zm 2 z ) (x^2 + y^2 + z^2 )^5 /^2
3 m 2 x(xm 1 x + ym 1 y + zm 1 z ) (x^2 + y^2 + z^2 )^5 /^2
μ 0 4 π
3 x(m 1 xm 2 x + m 1 y m 2 y + m 1 z m 2 z ) (x^2 + y^2 + z^2 )^5 /^2
μ 0 4 π
15 x(xm 2 x + ym 2 y + zm 2 z )(xm 1 x + ym 1 y + zm 1 z ) (x^2 + y^2 + z^2 )^7 /^2
Using the same method, y and z components of F 2 are
F 2 y = μ 0 4 π
3 m 1 y (xm 2 x + ym 2 y + zm 2 z ) (x^2 + y^2 + z^2 )^5 /^2
3 m 2 y (xm 1 x + ym 1 y + zm 1 z ) (x^2 + y^2 + z^2 )^5 /^2
μ 0 4 π
3 y(m 1 xm 2 x + m 1 y m 2 y + m 1 z m 2 z ) (x^2 + y^2 + z^2 )^5 /^2
μ 0 4 π
15 y(xm 2 x + ym 2 y + zm 2 z )(xm 1 x + ym 1 y + zm 1 z ) (x^2 + y^2 + z^2 )^7 /^2
F 2 z = μ 0 4 π
3 m 1 z (xm 2 x + ym 2 y + zm 2 z ) (x^2 + y^2 + z^2 )^5 /^2
3 m 2 z (xm 1 x + ym 1 y + zm 1 z ) (x^2 + y^2 + z^2 )^5 /^2
μ 0 4 π
3 z(m 1 xm 2 x + m 1 y m 2 y + m 1 z m 2 z ) (x^2 + y^2 + z^2 )^5 /^2
− μ^0 4 π
15 z(xm 2 x + ym 2 y + zm 2 z )(xm 1 x + ym 1 y + zm 1 z ) (x^2 + y^2 + z^2 )^7 /^2
Then
F 2 = F 2 x xˆ + F 2 y ˆy + F 2 z ˆz
3 μ 0 4 πR^6
(m 1 · m 2 ) Rˆ + (m 1 · Rˆ)m 2 + (m 2 · Rˆ)m 1 − 5(m 1 · Rˆ)(m 2 · Rˆ) Rˆ
(b) Using tensor notatation
Then the force on m is
F = −∇U = −∇
μ 0 mIa^2 2 (a^2 + z^2 )^3 /^2
= 3 μ^0 mIa
(^2) z 2 (a^2 + z^2 )^5 /^2
ˆz (26)
