Download PHYS 110B - HW #7

PHYS 110B - HW #7
Spring 2004, Solutions by David Pace
Any referenced equations are from Griffiths
Problem statements are paraphrased
[1.] Problem 10.13 from Griffiths
A point charge, q, moves in a loop of radius a. At time t = 0 this charge is at the point
(a, 0) on the x axis; it moves with constant angular acceleration ω. Determine the LiénardWiechert potentials for points on the z axis.
Solution
The general forms of the Liénard-Wiechert potentials for a charge q are,
V (~r, t) =
~ r, t) =
A(~
qc
µo qc~v
~ · ~v )
4π(Rc − R
(1)
Eq. 10.39
~ · ~v )
4πo (Rc − R
=
~v
V (~r, t)
c2
Eq. 10.40
(2)
where my R is equivalent to the lower case script r of Griffiths because making lower case
script r’s appears to be beyond my ability.
The particle is confined to move in a circle. Such a trajectory may be written as,
w(t)
~
= a[cos(ωt) x̂ + sin(ωt) ŷ]
(3)
using the initial condition given in the problem. The location of the particle at the retarded
time is found by replacing t with tr above.
The vector from the source (i.e. the particle at its retarded position) to the observation point
is,
~ = ~r − w(t
R
~ r)
= z ẑ − a[cos(ωtr ) x̂ + sin(ωtr ) ŷ]
The magnitude of this vector, which is needed to solve for the potential in (1), is,
p
~ =
~ ·R
~
R = |R|
R
=
=
=
(4)
(5)
(6)
p
(z ẑ − a[cos(ωtr ) x̂ + sin(ωtr ) ŷ]) · (z ẑ − a[cos(ωtr ) x̂ + sin(ωtr ) ŷ])
q
√
z 2 + a2 (cos2 (ωtr ) + sin2 (ωtr ))
(7)
z 2 + a2
(8)
1
The expressions for the potentials are written in terms of the velocity at the retarded time.
Since w(tr ) represents the position of the particle at the retarded time the velocity is,
~v (tr )
∂
∂
w(t
~ r) =
a[cos(ωtr ) x̂ + sin(ωtr ) ŷ]
∂tr
∂tr
=
(9)
= a[− sin(ωtr )(ω) x̂ + cos(ωtr )(ω) ŷ]
(10)
= aω[− sin(ωtr ) x̂ + cos(ωtr ) ŷ]
(11)
~ · ~v ,
Now compute the quantity R
~ · ~v = (z ẑ − a[cos(ωtr ) x̂ + sin(ωtr ) ŷ]) · (aω[− sin(ωtr ) x̂ + cos(ωtr ) ŷ])
R
(12)
= a2 ω cos(ωtr ) sin(ωtr ) − a2 ω cos(ωtr ) sin(ωtr )
(13)
= 0
(14)
The scalar potential can now be solved for using (1),
qc
√
4πo (c( z 2 + a2 ) − 0)
q
√
=
4πo z 2 + a2
(15)
V (~r, t) =
(16)
The vector potential follows by using (2),
~ r, t) = aω[− sin(ωtr ) x̂ + cos(ωtr ) ŷ] V (~r, t)
A(~
c2
(17)
=
aω[− sin(ωtr ) x̂ + cos(ωtr ) ŷ]
q
√
·
2
c
4πo z 2 + a2
(18)
=
aωq[− sin(ωtr ) x̂ + cos(ωtr ) ŷ]
√
4c2 πo z 2 + a2
(19)
In this problem the particle trajectory is greatly simplified by being constrained to the circular loop. This makes value of R constant and therefore the retarded time is easily found.
√
R
z 2 + a2
tr = t −
= t−
(20)
c
c
The final form of the vector potential is,
aωq[− sin ω t −
~ r, t) =
A(~
√
2
x̂ + cos ω t −
√
4c2 πo z 2 + a2
z 2 +a2
c
√
z 2 +a2
c
ŷ]
(21)
[2.] Problem 10.14 from Griffiths
Show that,
V (~r, t)
=
qc
1
p
4πo (c2 t − ~r · ~v )2 + (c2 − v 2 )(r2 − c2 t2 )
Eq. 10.42
(22)
can be written as,
V (~r, t)
=
1
q
q
4πo R 1 − v2 sin2 θ
c2
Eq. 10.44
(23)
~ ≡ ~r − ~v t is the vector from the present location of the particle to the observation
where R
point (do not confuse this symbol with R). Since this relates to the situation of example
10.3 (page 433 in Griffiths), the velocity is a constant. The angle θ is that between the
~ and ~v . Reference figure 10.9 for this geometry. It is also worth noting that at nonvectors R
relativistic velocities where v 2 c2 , the scalar potential in (22) simplifies to the well-known
electrostatic value.
Solution
Noticing how similar (22) and (23) are, the best method for showing their equivalence
seems to be rewriting the square root term in (22). This will require solving for the ~r dependence in terms of the present position of the particle since r does not appear in the final
form.
The velocity is constant in this problem, so the retarded velocity written in the original
solution is equivalent to the velocity at present time. Begin by solving for ~r · ~v using the
~ + ~v t as determined from the problem statement.
fact that ~r = R
~ + ~v t) · ~v
~r · ~v = (R
~ · ~v + v 2 t
= R
(24)
(25)
The magnitude of ~r is,
r
=
q
√
~ + ~v t) · (R
~ + ~v t)
~r · ~r =
(R
q
=
~ · ~v + v 2 t2
R2 + 2tR
~ · ~v + v 2 t2
r2 = R2 + 2tR
(26)
(27)
(28)
where the value of r2 is written because that is what appears in the relevant equation.
Now the square root term in (22) can be simplified. I’ll just take the inside of the square
root here for clarity.
~ · ~v + v 2 t2 − c2 t2 ) (29)
(c2 t − ~r · ~v )2 + (c2 − v 2 )(r2 − c2 t2 ) = (c2 t − ~r · ~v )2 + (c2 − v 2 )(R2 + 2tR
3
~ · ~v + v 2 t2 − c2 t2 )
= c4 t2 − 2c2 t~r · ~v + (~r · ~v )2 + (c2 − v 2 )(R2 + 2tR
(30)
~ · ~v + v 2 t) + (R
~ · ~v + v 2 t)2 + c2 (R2 + 2tR
~ · ~v + v 2 t2 − c2 t2 )
= c4 t2 − 2c2 t(R
~ · ~v + v 2 t2 − c2 t2 )
−v 2 (R2 + 2tR
(31)
2 2
2 2 ~ · ~v + ~ · ~v + ~ · ~v − ~ · ~v )2 + v 4
t2 + c2 R2 + 2c
tR
c2
v 2 t2
2v
tR
= c4
t2 − 2c
tR
2c
v 2 t2 + (R
~
2
v 4
t2 + −
c4
t2 − v 2 R2 − 2tv
R
· ~v − c2
t2 v 2
(32)
~ · ~v )2 + c2 R2 − v 2 R2
= (R
(33)
= R2 v 2 cos2 θ + c2 R2 − v 2 R2
(34)
2
2
2
2
= cR
= cR
v2
v2
2
cos
θ
+
1
−
c2
c2
(35)
v2
v2
2
(1
−
sin
θ)
+
1
−
c2
c2
(36)
v 2 v 2
v 2
2
−
= cR
sin θ + 1 − 2
c2
c2
c
v2
2 2
2
= c R 1 − 2 sin θ
c
2
2
(37)
(38)
Plugging this back into (22) gives,
V (~r, t) =
=
1
qc
q
4πo c2 R2 1 −
v2
c2
(39)
2
sin θ
1
q
q
4πo R 1 − v2 sin2 θ
c2
X
(40)
[3.] Problem 10.17 from Griffiths
Derive the following result,
~
1
qc
R~a
R 2
∂A
2
~
~
(Rc − R · ~v )(−~v +
=
) + (c − v + R · ~a)~v
~ · ~v )3
∂t
4πo (Rc − R
c
c
Eq. 10.63
(41)
by first showing,
∂tr
Rc
=
~
∂t
R · ~u
Solution
4
Eq. 10.71
(42)
From Griffiths,
|~r − w(t
~ r )| = c(t − tr )
~ = ~r − w(t
R
~ r)
(43)
Eq. 10.33
Eq. 10.34
(44)
Using the above as the starting point, write R2 ,
R2
=
~ ·R
~ = c2 (t − tr )2
R
Take the (non-retarded) time derivative of R2 ,
∂
∂ ~ ~
∂
~·
~ +
~ ·R
~
(R · R) = R
R
R
∂t
∂t
∂t
(45)
~·
2R
=
~
∂R
∂t
(46)
The time derivative of the right hand side of (45) is,
∂ 2
c (t − tr )2
∂t
=
∂tr
2c (t − tr ) 1 −
∂t
2
(47)
Equating (46) and (47),
~
∂
R
∂t
r
2
~·
2R
= 2c (t − tr ) 1 −
∂t
∂t
~
∂R
∂tr
~
R·
= cR 1 −
∂t
∂t
(48)
(49)
where the last step makes use of R = c(t − tr ), which follows from the initial equations of
the solution.
Writing out the left side of (49) gives,
~
~ · ∂ (~r − w(t
~ · ∂R = R
~ r ))
R
∂t
∂t
∂
w(t
~
)
r
~· −
= R
∂t
∂ w(t
~ r ) ∂tr
~
= R· −
∂tr ∂t
∂tr
~
= R · −~v
∂t
Putting (53) back into (49) produces,
∂tr
∂tr
~
R · −~v
= cR 1 −
∂t
∂t
∂tr
∂tr
~
R · −~v
= cR − cR
∂t
∂t
~ · ~v ) ∂tr = cR
(cR − R
∂t
5
(50)
(51)
(52)
(53)
(54)
(55)
(56)
∂tr
cR
=
~ · ~v
∂t
cR − R
(57)
~ · ~u where ~u ≡ cR̂ − ~v (Eq.
and it remains to show that the denominator above is equal to R
10.64).
~ · ~u = R
~ · (cR̂ − ~v )
R
~ · ~v
= cR − R
∴
cR
∂tr
=
~ · ~u
∂t
R
(58)
(59)
(60)
Now for the derivation of equation 10.63. Begin with the equation for the vector potential
as written in (2). Taking the time derivative of this gives,
~
∂A
1 ∂~v
∂V
= 2
V + ~v
∂t
c ∂t
∂t
∂V
1 ∂~v ∂tr
V + ~v
= 2
c ∂tr ∂t
∂t
Replace the scalar potential with its general form, (1),
#
"
~
∂
1 ∂~v ∂tr
∂A
qc
qc
+ ~v
=
2
~
~ · ~u)
∂t
c ∂tr ∂t 4πo (R · ~u)
∂t 4πo (R
(61)
(62)
(63)
where the denominator in the V term has been replaced according to (59) for simplicity. It
~ · ~u term in the
will be easier to carry out the algebra in this form and then replace the R
final expression.
~
∂A
qc
∂~v ∂tr 1
∂ 1
=
+ ~v
~ · ~u
~ · ~u
∂t
4πo c2 ∂tr ∂t R
∂t R
q
∂tr 1
−2 ∂ ~
~
=
~a
+ ~v (−1)(R · ~u)
(R · ~u)
~ · ~u
4πo c
∂t R
∂t
" #
q
cR
1
~v
∂ ~
~a
−
(R · ~u)
=
2
~
~
~
4πo c
R · ~u R · ~u (R · ~u) ∂t
q
∂ ~
=
cR~a − ~v (R · ~u)
~ · ~u)2
∂t
4πo c(R
6
(64)
(65)
(66)
(67)
At this point it is worthwhile to write out the time derivative in (67).
∂
∂ ~
~ · ~v )
(R · ~u) =
(cR − R
∂t
∂t
= c
~
∂R ∂ R
~ · ∂~v
−
· ~v − R
∂t
∂t
∂t
The first term is,
∂R
∂t
=
(68)
∂
c(t − tr )
∂t
=
∂tr
c 1−
∂t
(69)
(70)
The second term is solved in (53), and the third term is solved in equations (61) through
(65). The result is,
∂tr
∂tr
∂ ~
∂tr
2
~
(R · ~u) = c 1 −
− −~v
· ~v − R · ~a
(71)
∂t
∂t
∂t
∂t
∂tr
∂tr
~ · ~a ∂tr
+ v2
−R
∂t
∂t
∂t
~ · ~a ∂tr
= c2 + v 2 − c2 − R
∂t
cR 2
2
2
~ · ~a
= c + v −c −R
~ · ~u
R
= c2 − c2
Plug the result of (74) back into (67).
cR ~
∂A
q
2
2
2
~
cR~a − ~v c + v − c − R · ~a
=
~ · ~u)2
~ · ~u
∂t
4πo c(R
R
"
!#
~ · ~u
q
cR
R
~ · ~u)~a − ~v
~ · ~a
=
(R
c2 + v 2 − c2 − R
2
~
~
cR
4πo c(R · ~u) R · ~u
qR
c~
v
2
2
~ · ~u)~a − (R
~ · ~u) − v ~v + c ~v + (R
~ · ~a)~v
=
(R
~ · ~u)3
R
4πo (R
qR
c~
v
2
2
~ · ~u)(~a − ) + (c − v + R
~ · ~a)~v
=
(R
~ · ~u)3
R
4πo (R
c qR
R~a
R 2
2
~
~
=
(R · ~u)(
− ~v ) + (c − v + R · ~a)~v
~ · ~u)3 R
c
c
4πo (R
qc
R~
a
R
2
2
~ · ~u)(
~ · ~a)~v
=
(R
− ~v ) + (c − v + R
~ · ~u)3
c
c
4πo (R
~ · ~u terms and rearranging gives,
Finally, replacing the R
~
∂A
1
qc
R~a
R 2
2
~
~
=
) + (c − v + R · ~a)~v
(Rc − R · ~v )(−~v +
~ · ~v )3
∂t
4πo (Rc − R
c
c
which matches (41) as intended.
7
(72)
(73)
(74)
(75)
(76)
(77)
(78)
(79)
(80)
(81)
[4.] Problem 10.19 from Griffiths
(a) Use
~ r, t)
E(~
1 − v 2 /c2
R̂
q
3/2
4πo 1 − v 2 sin2 θ/c2
R2
=
Eq. 10.68
(82)
to solve for the electric field a distance d away from an infinite wire that is carrying a
uniform line charge λ that moves down this wire with constant speed v.
(b) Use
1
~
(~v × E)
c2
to solve for the magnetic field due to this wire.
~
B
Eq. 10.69
=
(83)
Solution
Let the wire lie along the x axis. This problem quickly reduces to a form similar to chapter
2 in Griffiths. Notice that for the linear charge density the q in the expression for the electric
field may be taken in infinitesimal pieces, dq = λ dx.
The electric field is,
~ r, t)
E(~
=
1 − v 2 /c2
4πo
Z
λ dx
1 − v 2 sin2 θ/c2
3/2
R̂
R2
(84)
where the λ is also a constant and may be taken outside the integral. It remains to set the
limits of this integral and perform the mathematical steps, all of which have been employed
previously.
From the geometry of the problem we know that the x component of the electric field will
be zero. Choosing any location along the wire, the symmetry will cause the x components
of the electric field due to that part of the wire to the left to cancel those x components due
to that part on the right. This geometry is similar to figure 10.9 in Griffiths, but with the
distance from the line to the observation point set at a length d.
The y component will be all that remains of the electric field and this is recorded through
R̂ = sin θ ŷ. Furthermore, sin θ = d/R so we can make the replacement 1/R2 = sin2 θ/d2 .
The integral is now,
Z
sin2 θ
dx
λ(1 − v 2 /c2 )
~
(sin
θŷ)
(85)
E(~r, t) =
3/2
4πo
d2
1 − v 2 sin2 θ/c2
It appears as though converting the integral in terms of the angle θ is the best method. This
is once again accomplished from geometrical concerns,
tan θ =
d
−x
(86)
x = −d tan−1 θ
=
dx = −d(− csc2 θ) dθ
=
d
dθ
sin2 θ
8
−d cot θ
(87)
(88)
(89)
The integral becomes,
2 2
~ r, t) = λ(1 − v /c )
E(~
4πo
Z
λ(1 − v 2 /c2 )
=
4πo
Z
sin2 θ d
dθ
3/2 (sin θŷ) 2
d sin2 θ
1 − v 2 sin2 θ/c2
1
π
0
sin θ
d 1 − v 2 sin2 θ/c2
(90)
(91)
3/2 dθ ŷ
where the limits follow from the nature of the infinite wire. When the charge density is all
the way to the left, the angle is zero. As the charge density gets all the way to the right this
angle goes to π.
This integral may be solved with the following substitution,
Let z = cos θ
sin2 θ = 1 − z 2
dz = − sin θ dθ
(92)
Making these substitutions leads to,
2 2
~ r, t) = λ(1 − v /c )
E(~
4πo d
−1
Z
−dz
1−
1
v2
c2
3/2 ŷ
(1 − z 2 )
(93)
where the limits have changed according to the substitution.
This problem is being solved in Cartesian coordinates though it clearly has cylindrical
symmetry. We use this to our advantage by claiming that the vector nature of our solution
is simply known from geometry. I keep the ŷ direction here because that will facilitate the
cross product necessary for finding the magnetic field. Since the linear charge density is
uniform it is known that the electric field lines will be directed along the radial vector.
2 2
~ r, t) = − λ(1 − v /c )
E(~
4πo d
−1
Z
1
dz
1−
v2
c2
+
v 2 2 3/2
z
2
c
ŷ
(94)
Here is an integral solution,
Z
dx
(a + cx2 )3/2
1
x
√
a a + cx2
=
(95)
This brings the solution to,
"
#−1
2 2
λ(1
−
v
/c
)
1
z
~ r, t) = −
E(~
ŷ
2 ·
1/2
2
2
4πo d
1 − vc2
1 − vc2 + vc2 z 2
1
"
#−1
λ
z
ŷ
= −
2
4πo d 1 − v + v2 z 2 1/2
2
2
c
c
1
"
#
1
λ
1
−
= −
−
ŷ
2
2 1/2
2
2 1/2
4πo d
1− v + v
1− v + v
c2
= −
=
λ
(−2) ŷ
4πo d
c2
c2
(96)
(97)
(98)
c2
(99)
λ
ŷ
2πo d
(100)
9
This solution is time-independent. It also matches the solution we would have found for a
uniformly charged wire in the electrostatic case (hence the claim that this problem follows
those of chapter 2 in the text).
(b) The magnetic field is found using this solution for the electric field. Since the velocity
is also a constant we have,
~
B
1
λ
ŷ
vx̂ ×
2
c
2πo d
=
=
λv
ẑ
2πo dc2
(101)
This can be written in a more familiar form noting c2 = 1/µo o .
~ = µo λv ẑ
B
2πd
(102)
which is again the common static result.
The magnetic field still forms loops around the wire. The z direction is not unique in
this case and as we rotate around the wire the direction of the magnetic field will rotate
accordingly.
[5.] Problem 10.20 from Griffiths
Using the setup of problem 10.13 find the fields at the center. Using your solution for this
magnetic field, derive the field at the center of a loop carrying a steady current I. Compare
this solution to that of example 5.6 in the text.
Solution
We solved for the potentials in the setup of problem 10.13 at the beginning of this problem
set. It is tempting to simply apply the established relations between the fields and these
potentials to quickly get an answer for this problem. It is not this simple, however, as
section 10.3.2 in the text shows. The differentiation due to the retarded terms is non-trivial
and the expressions we must use for the fields are,
~ r, t) =
E(~
i
R h 2
q
2
~
(c − v )~u + R × (~u × ~a)
~ · ~u)3
4πo (R
~ r, t)
~ r, t) = 1 R̂ × E(~
B(~
c
Eq. 10.65
Eq. 10.66
(103)
(104)
We can still use some of the results from problem 10.13, and these results are simpler now
because we are concerned with the center of the loop (i.e. z = 0). The plan is to determine
the electric field first. Relevant terms needed for (103) that can be taken from 10.13 at the
center of the loop are,
~ = −w(t
R
~ r)
R = a
=
−a[cos(ωtr ) x̂ + sin(ωtr ) ŷ]
From (8)
From (5)
(105)
(106)
10
~a =
∂
~v (tr )
∂tr
velocity given in (11)
(107)
= −aω 2 (cos(ωtr ) x̂ + sin(ωtr ) ŷ)
(108)
= −ω 2 w(t
~ r)
(109)
where we must be careful not to confuse a (the loop radius) with ~a (the acceleration at the
retarded time).
We also need to solve for ~u ≡ cR̂ − ~v . The R̂ term is,
R̂
=
~
R
R
=
−(cos(ωtr ) x̂ + sin(ωtr ) ŷ)
(110)
Leading to,
~u = −c(cos(ωtr ) x̂ + sin(ωtr ) ŷ) − aω[− sin(ωtr ) x̂ + cos(ωtr ) ŷ]
= x̂ (−c cos(ωtr ) + aω sin(ωtr )) + ŷ (−c sin(ωtr ) − aω cos(ωtr ))
(111)
(112)
~ u = (−a[cos(ωtr )x̂+sin(ωtr )ŷ])·(x̂ (−c cos(ωtr )+aω sin(ωtr ))+ŷ (−c sin(ωtr )−aω cos(ωtr )))
R·~
= (ac cos2 (ωtr ) − a2 ω cos(ωtr ) sin(ωtr )) + (ac sin2 (ωtr ) + aω cos(ωtr ) sin(ωtr )) (113)
= ac(cos2 (ωtr ) + sin2 (ωtr )) + a2 ω(cos(ωtr ) sin(ωtr ) − cos(ωtr ) sin(ωtr ))
(114)
= ac
(115)
For the triple product use the following vector identity (the general form is in the front
cover of Griffiths),
~ × (~u × ~a) = (R
~ · ~a)~u − (R
~ · ~u)~a
R
(116)
The last term to solve for is,
~ · ~a = (−a[cos(ωtr ) x̂ + sin(ωtr ) ŷ]) · (−ω 2 w(t
R
~ r ))
(117)
= −w(t
~ r ) · (−ω 2 w(t
~ r ))
(118)
= ω 2 w2
(119)
= ω 2 R2
(120)
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Therefore,
~ × (~u × ~a) = ω 2 R2~u − ac~a
R
(121)
All of the terms are known so we return to the general expression for the electric field.
q
a 2
~ r, t) =
E(~
(c − v 2 )(x̂ (−c cos(ωtr ) + aω sin(ωtr ))
3
4πo (ac)
(122)
+ŷ (−c sin(ωtr ) − aω cos(ωtr ))) + ω 2 R2~u − ac~a
The final result is,
~ =
E
2 2
q
(a ω − c2 ) cos(ωtr ) + acω sin(ωtr ) x̂
2
2
4πo a c
+ (a2 ω 2 − c2 ) sin(ωtr ) − acω cos(ωtr ) ŷ
(123)
The magnetic field is found from,
~
B
=
1
(R̂x Ey − R̂y Ex )ẑ
c
(124)
Algebra and substitutions...
(125)
= −
=
q
1
2
2
−acω
cos
(ωt
)
−
acω
sin
(ωt
)
ẑ
r
r
4πo a2 c3
qω
ẑ
4πo ac2
(126)
(127)
To convert (127) to an expression for the field due to a loop of steady current we need to
define I = λv. In this case λ is the charge density and must be λ = q/2πa by definition since
the charge is entirely contained within this loop. The velocity is known from the radius of
the loop and the angular velocity at which it is spinning, v = aω.
q = λ(2πa)
λ =
=
q =
I
v
(129)
I
aω
(130)
2πI
ω
(131)
Replacing this into the previous solution for the magnetic field gives,
ω
2πI
~ =
B
ẑ
4πo ac2
ω
=
I
ẑ
2o ac2
12
(128)
(132)
(133)
Once again employing the relation c2 = 1/µo o ,
~
B
=
µo I
ẑ
2a
(134)
and this is equivalent to the solution provided by equation 5.38 in Griffiths’ example 5.6.
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