# z + − y

advertisement ```PHGN462: Advanced E&amp;M
Group Quiz 6 – November 2, 2009
NAMES: KEY
z
−
−
−
x
+
−
+
+
−
y
+
β
−
+
−
+
+
A loop of radius, R, is centered at the origin and lies in the xy-plane. The loop consists of two oppositely-charged half loops
of uniform linear charge density, &plusmn;λ0 . Let β(t) be the angle with respect to the x-axis to one of the joints where the half-loops
are joined (as shown). Initially, β(t = 0) = 0 and dβ
dt (t = 0) = 0; so at this time the loop is oriented with the positive side
centered on the positive y-axis, that is,
(
+λ0 , 0 ≤ θ &lt; π
λ(θ, t = 0) =
−λ0 , π ≤ θ &lt; 2π.
where θ identifies an element of the loop. The loop rotates in the +ẑ direction with an angular velocity which increases
~ is:
linearly in time according to, ω(t) = dβ/dt = αt. The general formula for the retarded vector potential, A,
~ r, t) = &micro;0
A(~
4π
Z ~ 0
J(~r , tR )d3 r0
,
|~
%R |
where %
~R = ~r − ~r 0 (tR ) with the retarded time given by tR = t − |~
%R |/c.
a. Through what angle will the loop have rotated after time, t?
Solution: Integrating the angular acceleration twice gives, β(t) = 21 αt2 (just like constant linear acceleration).
b. What is the retarded time for a√
point on the z-axis?
Solution: tR = t − |~r − ~r0 |/c = t − R2 + z 2 /c, shift is independent of time.
~ r 0 , tR ) → Rω(tR )λ(θ, tR )dθ φ̂, where λ(θ, tR ) is the linear charge density at time,tR ,
c. For this case, we can write d3 r0 J(~
similar to the t = 0 case shown above. Find λ(θ, tR ).
Solution: Using the result from part b, the linear charge density at arbitrary time, t, is
(
2
2
+λ0 αt2 ≤ θ &lt; π + αt2
λ(θ, t) =
2
2
−λ0 π + αt2 &lt; θ ≤ 2π + αt2
The charge density at the retarded time is this expression evaluated at the retarded time, part b.
~ t), for a point on the z-axis.
d. Find the vector potential, A(z,
Solution: Using the result from part c and setting ω(tR ) = αtR , the vector potential is
~ t)
A(z,
=
=
Z
2π
R(αtR )λ(θ, tR )
√
(cos θŷ − sin θx̂)
R2 + z 2
0
αt2
Z
Z 2π+ αt22R R
&micro;0 RαtR λ0 π+ 2
√
dθ −
dθ (cos θŷ − sin θx̂)
αt2
αt2
R
4π R2 + z 2
π+ 2R
2
&micro;0
4π
dθ
&micro;0 RαtR λ0
αt2
αt2
= − √
(cos( R )x̂ + sin( R )ŷ)
2
2
2
2
π R +z
where tR = t −
√
R2 + z 2 /c and φ̂ = − sin θx̂ + cos θŷ was used.
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