Download Magnetic Field

Magnetic Field
Submitted by: I.D. 039120068
The problem:
The coaxial cable is built from two concentric cables. In the internal cable there is a current density
of J~ = J0 ar ẑ and in the external cable there is a current density of J~ = J0 rb ẑ. Find the force of the
magnetic field inside and outside the coaxial cable.
The solution:
In general, we can calculate the magnetic field made by a thick current wire or a cylinder with the
Ampere’s law. Due to the symmetry the field is only in ϕ̂.
I
Z
Z
~
~
~
~
B · dl = 4πκ J · ds = 4πκ2π J(r)rdrϕ̂
(1)
L
Z
~ = 4πκ J(r)rdrϕ̂
B
(2)
r
µ0
(3)
κ =
4π

J0 r

r ≤ R1

a


0
R 1 < r < R2
J(r) =
(4)
J
r
0
 b
R2 ≤ r ≤ R3



0
r > R3 )
In order to calculate the magnetic field in the whole space we will use superposition of the magnetic
field made by the inner cable (B~1 ) and the outer cable (B~2 )
a) for r ≤ R1 :
Z
Z
4πκ r
4πκJ0 r 2
4πκr2 J0
B~1 =
J(r)rdr =
r dr =
ϕ̂
(5)
r 0
ar
3a
0
B~2 = 0
(6)
b) for R1 ≤ r ≤ R2 :
Z R1
Z
Z r
4πκ r
4πκ
J0 2
4πκR13 J0
~
B1 =
J(r)rdr =
r dr +
0 rdr =
ϕ̂
r 0
r
a
3ar
0
R1
B~2 = 0
(7)
(8)
c) for R2 ≤ r ≤ R3 :
Z r
J0 2
4πκR13 J0
r dr +
0 rdr =
ϕ̂
a
3ar
0
0
R1
Z R2
Z
Z r
4πκ r
4πκ
J0 2
4πκJ0 (r3 − R23 )
=
J(r)rdr =
0 rdr +
r dr =
ϕ̂
r 0
r
3br
0
R2 b
3
R1
r3 − R23
= 4πκJ0
+
ϕ̂
3ar
3br
B~1 =
B~2
~ total
B
4πκ
r
Z
r
4πκ
J(r)rdr =
r
Z
R1
1
(9)
(10)
(11)
d) for r ≥ R3 :
B~1 =
B~2 =
~ total =
B
Z r
J0 2
4πκR13 J0
0 rdr =
r dr +
ϕ̂
(12)
a
3ar
0
0
R1
Z R2
Z R3
Z r
Z
J0 2
4πκ r
4πκ
4πκ(R33 − R23 )J0
J(r)rdr =
0 rdr +
0 rdr =
r dr +
ϕ̂
r 0
r
3br
R2 b
0
R3
4πκJ0 R13 R33 − R23
ϕ̂
(13)
+
3r
a
b
4πκ
r
Z
r
4πκ
J(r)rdr =
r
Z
R1
Here is the plot of the magnetic field strength:
2
Electromagnetic Induction
Submitted by: I.D. 41855008
The problem:
~ = B0 (t)ẑ
Inside an infinite cylinder area x2 + y 2 ≤ R2 there is a time-dependant magnetic field B
so that Ḃ = const > 0. Outside the cylinder the magnetic field is zero. We place a conductor rod
l < 2R in x − y plane so the tips of the rod touch the edge of the cylinder.
1. We close an electric circuit by connecting the two tips of the rod with straight conductors to
the origin. What is the EMF in the circuit and the current direction in the rod?
2. Now we close an electric circuit by connecting the two tips of the rod with a conductor along
the short arc. What is the EMF in the circuit and the current direction in the rod?
3. What is the potential between the two tips of the rod when they are not connected?
4. Explain if there is a contradiction between the results of the previous sections.
The solution:
The electmagnetic induction can be found from
I
ΦB = B0 · ds = S · B0
(1)
s
where the area S is constant. According to the Faraday’s law
ε = −Φ̇B = −S Ḃ0
(2)
1. The area is a triangle, therefore,
r
l2
l
S =
R2 −
2
r 4
Ḃ0 l
l2
R2 −
ε = −
2
4
(3)
(4)
1
The current is in clockwise direction in the circuit - from right to left in the rod.
2. In order to calculate the new area we S 0 we calculate the area of the segment M and subtrack
the area of the triangle from the preious section.
M
= R2 arcsin
l
2R
(5)
S0 = M − S
(6)
ε = −Ḃ0 S 0 = −Ḃ0
R2 arcsin
l
l
−
2R 4
r
R2 −
l2
4
!
(7)
The current is in clockwise direction in the circuit - from left to right in the rod.
~ according to:
3. Inside the cylinder there is an electric field E
1
r
~ = −Ḃ0 ẑ
∇×E
!
~ ϕ ∂E
~r
∂rE
ẑ = −Ḃ0 ẑ
−
∂r
∂ϕ
(8)
(9)
Because we do not have charges inside the cylinder so the radial field is zero.
~r = 0
E
(10)
~ϕ
1 ∂rE
ẑ = −Ḃ0 ẑ
r ∂r
~ ϕ = − Ḃ0 r
E
Z 2
ε =
(11)
(12)
Z
Ex dx =
Eϕ cos(ϕ)dx
(13)
where x is a coordinate on the rod measured from its center, ϕ is the angle between Eϕ and the
rod and also the angle between rr̂ and the perpendicular from the center of the circle to the chord.
|y|
Ex
=
Eϕ
r
r
l2
|y| =
R2 −
4
Z
|y|
~ ϕ dx
ε =
E
r
−l/2
Z r
Ḃ0
l2 ~
R2 − dx
ε = −
2
4
cos(ϕ) =
(14)
(15)
(16)
(17)
l/2
l
ε = −Ḃ0
2
r
R2 −
l2
4
(18)
4. We can see that result of the the first and the third sections are equal. The reason is that because
if we calculate the EMF in 1 like the calculation in 3 but we take the entire loop of the circuit:
I
~ = ε
~ ϕ dl
E
(19)
L
where L is the the perimeter of triangle. The integral over this loop is not zero only on the rod l,
because there is no radial component of the field, so that the integrals over the radial sides of the
triangle are zero. However, in the section 2 the integral over the arc is not zero.
2