NAME: KEY
PHGN462: Advanced E&M
Exam 2 - November 18, 2009
1. (30) “Quickies”
a. (7) Sketch a reference frame with a point source located at ~r 0 and field point at ~r and, referring to your sketch, define
“retarded time”.
Solution: See Fig. 10.3 and Eq. 10.18 in Griffiths.
~
b. (7) Write down the general expressions for the retarded potentials, V and A.
Solution: See Eq.10.19 in Griffiths.
c. (7) Explain why the sky is blue.
Solution: See Example 11.1 (page 449) in Griffiths.
d. (9) Consider a proton with rest mass, m ' 1 Gev/c2 , and lab kinetic energy, TL =2 GeV, colliding with another proton
at rest. Use the invariance of the total 4-momentum squared to find the center-of-mass momentum, pCM . Give your answer
in GeV/c units.
Solution: The total 4-momentum in the lab frame is: pµL = (pLp
c, EL + mc2 ), where EL = TL + mc2 = 3 GeV for TL = 2 GeV
√
2
and mc = 1 GeV. Using the Einstein triangle relation: pL c = EL2 − (mc2 )2 = 8 GeV. The total 4-momentum in the CM
2
2 2
2
frame is: pµCM = (0, 2ECM ). Since the square of a four vector is an invariant, we have: (pp
L c) − (EL + mc ) = −(2ECM ) .
√
2
Solving gives: ECM = 2 GeV which, using the Einstein triangle relation, gives pCM = ECM
− (mc2 )2 = 1 GeV/c.
~ r, t) = A0 sin θ cos θ sin ωtr φ̂
2. (35) The vector potential of an oscillating magnetic quadrupole in the radiation zone is found to be A(~
r
where tr = t − r/c.
~ in the radiation zone (to first order in 1/r). Show all work.
a. Find the magnetic field, B,
~ =∇
~ ×A
~ = 1 ∂Aφ r̂ − ∂Aφ θ̂. The first term is of order
Solution: Since there is only a φ̂-component to the vector potential, B
r ∂θ
∂r
2
1/r and can be dropped. The remaining term gives:
ω cos ω(t − r/c) sin ω(t − r/c)
A0 sin θ cos θ ω
~
−
( ) cos ω(t − r/c)θ̂
B = A0 sin θ cos θ (− )
θ̂ →
c
r
r2
r
c
where the 1/r2 term has been dropped.
~ = cB
~ × r̂ to find the Poynting vector, S
~=
b. Use E
~ = cB
~ × r̂ =
Solution: E
cA0 sin θ cos θ ω
( c ) cos ω(t
r
~= c
S
µ0
1 ~
µ0 E
~
× B.
− r/c)(−φ̂). Thus,
A0 sin θ cos θ ω
( ) cos ω(t − r/c)
r
c
2
r̂
c. Integrate the time-averaged Poynting vector through the surface of a large sphere centered on the radiating quadrupole
to find the total power radiated.
R
~ where dA
~ = r2 sin θdθdφr̂,
~ · dA
Solution: The time average: hcos2 ω(t − r/c)i = 1/2. The total power is given by hP i = hSi
giving
Z
Z π
Z 1
A20 ω 2
c A20 ω 2 2 2π
2
2
r
dφ
dθ sin θ sin θ cos θ =
2π
dxx2 (1 − x2 )
(1)
hP i =
2µ0 c2 r2
2µ0 c
0
0
−1
=
4πA20 ω 2
15µ0 c
1
y
I(t)
r'
r-r'
R
ϕ
x
r
3. (35) Consider a semi-circular segment of wire or radius, R, in the upper half x-y plane centered on the origin as shown
above. At t = 0 a counter-clockwise current is turned on which varies as: I(t) = I0 t/τ where τ is a constant.
~
a. Find the vector potential, A(t),
at the origin. Show all work.
Solution: For the field point at the origin we have |~r − ~r 0 | = R (independent of t). Assuming the current turns on at t = 0,
~ giving:
~ < R/c) = 0. For t > R/c, d3 r0 J(~
~ r 0 , tr ) → I(tr )dl,
A(t
Z
µ0 I0 π Rdφ(− sin φx̂ + cos φŷ)(t − R/c)
~
A(0, t) =
4π τ 0
R
Z π
µ0 I0 (t − R/c)
=
dφ(− sin φx̂ + cos φŷ)
4πτ
0
µ0 I0 (t − R/c)
x̂.
= −
2πτ
b. Consider a field point near the origin at ~r = xx̂ for x R. From the arbitrary current element shown on the figure above
draw the relative position vector (~r − ~r 0 ) and find the retarded time for that element’s contribution to the vector potential
to first order in x/R.
p
Solution: tr = t − |~r − ~r 0 |/c, where |~r − ~r 0 | = (R cos φ − x)2 + (R sin φ)2 . To get this to first order in x/R, expand the
polynomials and factor out R:
q
x
x 1/2
|~r − ~r 0 | =
x2 − 2xR cos φ + R2 cos2 φ + R2 sin2 φ = R 1 − 2 cos φ + ( )2
R
R
1/2
x
≈ R 1 − 2 cos φ
R
where the (x/R)2 term has been dropped. Use Taylor’s expansion to first order, i.e. (1 + )n ≈ 1 + n, to find
1
x
0
|~r − ~r | ≈ R 1 − ( )(2 cos φ) = R − x cos φ
2
R
This gives the retarded time: tr = t − (R − x cos φ)/c.
2
z
-
x
P
-
-
-
-
+
R
r
r-r'
+
r'
β
-
+
+
+
+
+
+
+
+
4. (35) A charged loop of radius, R, lies in the xy-plane and is centered at the origin and carries a charge density, λ(φ, t) =
λ0 sin(φ − β(t)), where β measures the orientation of the loop as shown in the figure. The loop rotates with constant angular
velocity, β = ωt; so the charge density at time, t, is given by λ(φ, t) = λ0 sin(φ − ωt), where φ is measured with respect to
the x-axis and identifies an arbitrary charge element of the loop.
a. Sketch a charge element at some arbitrary angle, φ, on the figure above and find the retarded time for the contribution
that the charge element makes to the retarded potential at a point on the x-axis, ~r = xx̂, to first order in R/x 1.
p
Solution: tr = t − |~r − ~r 0 |/c, where |~r − ~r 0 | = (x − R cos φ)2 + (R sin φ)2 . To get this to first order in R/x, expand the
polynomials and factor out x:
0
|~r − ~r | =
≈
1/2
q
R 2
R
φ
2
2
2
2
x − 2xR cos φ + R cos φ + R sin = x 1 − 2 cos φ + ( )
x
x
1/2
R
x 1 − 2 cos φ
x
where the (R/x)2 term has been dropped. Use Taylor’s expansion to first order, i.e. (1 + )n ≈ 1 + n, to find
1
R
0
|~r − ~r | ≈ x 1 − ( )(2 cos φ) = x − R cos φ
2
x
This gives the retarded time: tR = t − (x − R cos φ)/c where φ is the angle measured with respect to the x-axis.
b. Find sin(φ − ωtR ) to lowest order in R/ −
λ where −
λ = c/ω is the reduced wavelength.
Solution: From part(a) we have: sin(φ − ωtR ) ' sin[φ − ω(t − x/c + R/c cos φ)]. Define tx = t − x/c, and use the trig identity,
sin(a − b) = sin a cos b − sin b cos a, with a = φ − ωtx and b = ωR/c cos φ = Rλ− cos φ, to find:
R
R
cos φ) − sin( − cos φ) cos(φ − ωtx )
−
λ
λ
R
≈ sin(φ − ωtx ) − − cos φ cos(φ − ωtx ),
λ
sin(φ − ωtR ) ≈ sin(φ − ωtx ) cos(
where the approximations ( 1): cos ≈ 1 and sin ≈ have been used.
~ t), to lowest order in 1/x and R/ −
c. Write down an integral expression for the vector potential, A(x,
λ. (For this situation
3 0~ 0
d r J(~r , tR ) → ωRλ(φ, tR )dφ φ̂.)
Solution: Using the indicated substitution, we have
Z
µ0 2π ωRλ0 sin(φ − ωtR )(− sin φx̂ + cos φŷ)dφ
~
A(x, t) =
4π 0
|~r − ~r 0 |
Z 2π
(− sin φx̂ + cos φŷ)[sin(φ − ωtx ) − Rλ− cos φ cos(φ − ωtx )]
µ0 ωλ0 R
≈
dφ
4π
x(1 − R/x cos φ)
0
Z 2π
µ0 ωλ0 R
R
≈
dφ(− sin φx̂ + cos φŷ)[sin(φ − ωtx ) − − cos φ cos(φ − ωtx )](1 + R/x cos φ)
4πx
λ
0
Z 2π
µ0 ωλ0 R
R
≈
dφ(− sin φx̂ + cos φŷ)[sin(φ − ωtx )(1 + R/x cos φ) − − cos φ cos(φ − ωtx )]
4πx
λ
0
where terms second order in 1/x and R/ −
λ have been dropped.
3