Download Cavity within a cylinder

Cavity within a cylinder
Submitted by: I.D. 066370016
The problem:
There is a cylindrical cavity of a radius b inside an infinite cylinder. The cylinder’s radius is R and
it is charged with a uniform charge density ρ. The distance between the axis of the cavity and the
axis of the cylinder is a.
Find the electric and the magnetic field as follows:
1. The electric field within the cavity.
2. By using the electric field, find the magnetic field inside the cavity. Consider the cylinder
moving with a velocity v in it’s axis direction.
3. Compare your result to direct method of finding the magnetic field (q.e 45 4 102).
The solution:
1. If we look carefully, and ignoring the cavity, then we notice that we have cylindrical symmetry.
Therefore, we can use the Gauss’ law in order to calculate the electric field which is produced by
the electrical charges in the cylinder and in the cavity (we will consider them negative). According
to the formula
~ = 4πkρ
Div E
(1)
For the entire cylinder we will use a cylindrical Gaussian surface with a radius r and a height h.
Because then the electric field will be perpendicular to the surface and equal at every point on it.
So by using the Gauss’ law we get
I
Z
Z
~
~
Ed~s = Div Edv = 4πk ρdv
(2)
v
s
v
We know that the charge density is uniform and the surface area is 2πrh we get:
I
I
~
E · d~s = Er,cylinder ds = Er,cylinder 2πrh
s
(3)
s
Z
4πk
Z
ρdv = 4πkρ
v
dv = 4πkρπr2 h
(4)
v
Er,cylinder = 2πkρr
(5)
where a − b ≤ r ≤ a + b. And with a vector notation we get that the electric field produced by the
cylinder alone a distance r from the cylinder axis:
~ r,cylinder = 2πkρ~r
E
(6)
1
From the same considerations we can calculate in the same manner the electric field which is
produced by the cavity, because we assign every infinitesimal volume element a charge density −ρ
and say that it is uniform, in order to ”create” the cavity in the first place. Hence the electric field
from the cavity is:
I
I
~
Ed~s = Er,cavity ds = Er,cavity 2π |~r − ~a| h
(7)
Zs
Zs
−ρdv = −4πkρ
4πk
v
dv = −4πkρπ |~r − ~a|2 h
(8)
v
~ r,cavity = −2πkρ(~r − ~a)
E
(9)
By using the superposition idea we can find the total electric field inside the cavity.
~ T OT
E
~ r,cylinder + E
~ r,cavity
= E
(10)
= 2πkρ~r − 2πkρ(~r − ~a)
(11)
= 2πkρ~a
(12)
We can see that the total electric field inside the cavity is constant everywhere inside the cavity
and pointed along the vector ~a.
2. It is given that the entire cylinder is moving along it’s axis with a velocity v. Let us choose a
cylindrical coordinates so that ~v = vẑ. We shall use two system frame that are moving relative to
each other. The Lab system is not moving and we shall call it the S system. The cylinder is in the
rest in another system which is moving with the velocity v, and we shall call it S 0 .
In S 0 the parallel electric field is zero, hence the parallel electric field in S is also zero, according to
the equation:
Ez = Ez0 = 0
(13)
In addition, in S 0 there is no magnetic field at all, i.e B 0 = 0. So in S the parallel magnetic field is
also zero, i.e Bz = 0. According to the transformation (from S to S 0 ):
~ × E)
~
~ 0 = γ(B
~⊥ − 1 V
B
⊥
c2
(14)
And the opposite transformation (from S 0 to S):
~ ×E
~ 0)
~ ⊥ = γ(B
~0 + 1 V
B
⊥
c2
~⊥ = γ 1 V
~ ×E
~0
B
c2
Calculating the cross
r̂
0
~
~
V × E = 0
Er
(15)
(16)
~ ×E
~0
product of V
ϕ̂ ẑ 0 v = Er v ϕ̂
0 0 (17)
~ ⊥ is along the ϕ̂ direction, i.e
Which mean that the direction of B
~⊥ = γ 1 V
~ ×E
~ 0 = γ 1 Er v ϕ̂ = γ 1 2πkρav ϕ̂
B
2
c
c2
c2
2
(18)
By using the following relations we can arrive to a simpler solution
1
µ0 ε0
1
k =
4πε0
1
1
γρ
~⊥ = γ
B
2π(
)ρav ϕ̂ = µ0 av ϕ̂
1
2
√
4πε0
2
( µ0 ε0 )
c =
√
(19)
(20)
(21)
3. In order to calculate directly the magnetic field we first have to understand that we have an
infinite cylinder which is uniformly charged and the charge is moving. So we can analog this to an
infinite wire, with a radius R, and a cavity parallel to it’s axis at a distance a and with a radius
b, which has a current. The current density is J~cylinder = J ẑ. Due to the cylindrical symmetry we
can use Ampere’ law to calculate the magnetic field. First, let us conclude, by using the right hand
rule, the direction of the magnetic field. If we shall put our thumb along the z axis (parallel to the
current), we can see that the direction of the magnetic field is pointed along ϕ̂. In addition we shall
indicate the existence of the cavity by a current which is J~cavity = −J~cylinder = −J ẑ.
According to the Maxwell’s equation
~ = 4πκJ~ +
rotB
~
κ ∂E
k ∂t
(22)
~
We know that ∂∂tE = 0 because there is no change in the electric field and we have no EMF. So the
Maxwell equation reduces to
~ = 4πκJ~
rotB
(23)
By using Stokes’s theorem we get:
I
Z
Z
Z
~
~
~
~
B · dl = rotBd~s = 4πκ Jd~s = µ0 J~ · d~s
s
L
s
(24)
s
If we choose a closed circular curved the magnetic field which is created by the cylinder or the cavity
will be tanget to the curve and equal at every point. Also the current density is uniform so we can
consider it as a constant, i.e we can take it out of the integral.
I
I
~ ~l = Bϕ dl = Bϕ 2πr
Bd
(25)
L
L
Z
µ0
s
~ s = µ0 J
Jd~
Z
ds = µ0 Jπr2
(26)
s
Bϕ 2πr = µ0 Jπr2
µ0 Jr
Bϕ =
2
(27)
(28)
And in vector notation Bϕ will be:
~ cylinder = µ0 Jr ϕ̂ = µ0 J ~r × ẑ
B
2
2
(29)
3
By using the same consideration we can find the magnetic field from the cavity:
I
I
~
~
Bdl = Bϕ dl = Bϕ 2π |~r − ~a|
L
(30)
L
Z
~ s = µ0 J
Jd~
µ0
Z
s
ds = −µ0 Jπ |~r − ~a|2
(31)
s
Bϕ 2π |~r − ~a| = −µ0 Jπ |~r − ~a|2
µ0 J |~r − ~a|
Bϕ = −
2
(32)
(33)
And in vector notation Bϕ will be:
~ cavity = − µ0 J |~r − ~a| ϕ̂ = − µ0 J (~r − ~a) × ẑ
B
2
2
(34)
~ cylinder and B
~ cavity to find the total magnetic field in
Using the superposition principle for both B
the cavity, we get:
~ T OT
B
~ cylinder + B
~ cavity
= B
µ0 J
µ0 J
=
~r × ẑ −
(~r − ~a) × ẑ
2
2
µ0 Ja
µ0 J
~a × ẑ =
ϕ̂
=
2
2
(35)
(36)
(37)
In the previus question we have found the magnetic field respect to the lab system.
~ ⊥ = µ0 γρ av ϕ̂
B
2
(38)
In order to get the right answer for the magnetic field we have to ”move” to the cylinder system,
0
0
to the S 0 system. We can do this by presenting γρ as ρ because ρ = γρ. And if we look carefully
0
and the dimensions of the product ρ v we will get J 0 . Eventually the magnetic field will be also
~ 0 = µ0 Ja ϕ̂
B
⊥
2
(39)
So both answers are correct.
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