Download Mutual Inductance

Mutual Inductance
Submitted by: I.D. 043423755
The problem:
Inside an infinite solenoid with n windings per unit length, there is a closed planar loop of area S
which is placed at an angle to the axis of symmetry of the solenoid.
What is the mutual inductance of the system?
The solution:
Lets define a few parameters:
1. n - the density of windings per unit of length
2. θ - the angle that the z axis creates with the solenoid
3. S - the surface of the solenoid
The magnetic field in the solenoid is
~ = µ0 Inẑ
B
(1)
And the flux is
~ · ~s = µ0 Inẑ · S n̂ = µ0 IS cos θ
Φ=B
(2)
Thus the induced e.m.f is
= −Φ̇ = −µ0 nS cos θI˙
(3)
Therefore, the mutual inductance coefficient is
M = µ0 nS cos θ
(4)
1
Self Inductance
Submitted by: I.D. 043423755
The problem:
A long coaxial cable is made out of 2 thin metallic tubes with the radii a, b (b > a).
What is the self inductance per length unit?
The solution:
A. One way of solving the problem:
L=
Φ
I
(1)
For a < r < b
~ = (µ0 )I θ̂
B
2πr
(2)
and zero otherwise. The magnetic field is tangential to the direction θ, so that the area element is
ldr.
dΦ(r) = B(r)ldr
Z b
µ0 Il
µ0
b
Φ =
dr =
Il ln
2π
a
a 2πr
(3)
(4)
The ratio L/l is:
L
Φ
µ0
b
=
=
ln
l
Il
2π a
(5)
B. Second way of solving the problem
Let’s define the density of magnetic energy
u=
1 2
1 µ20 I 2
B =
2µ0
2µ0 4π 2 r2
(6)
We find the total energy stored in the system
1
U = LI 2
2
(7)
And find L from there
Z
Z lZ
U = udV =
0
0
2π
Z
a
b
µ20 I 2
1 µ0 l 2 b
rdrdθdz =
I ln
2µ0 4π 2 r2
2 2π
a
(8)
Hence
L
µ0
b
=
ln
l
2π a
(9)
1
Liquid Filled Capacitor
Submitted by: I.D. 065628786
The problem:
Parallel-plate capacitor built of two round conducting plates with the area A √
is filled with dielectric
liquid with a dielectric constant . The distance between the plates is d d A. Due to a leaking,
the surface of the liquid lowers at a constant rate u.
1. Find the capacitance as a function of t.
2. Find the electric field inside the capacitor.
3. Find the magnetic field between the capacitor’s plates.
The solution:
1. Each part of the capacitor (the part filled with the liquid and the part without the liquid) have
a different dielectric constant, therefore they act as two different capacitors connected in series.
C1 =
C2 =
Cef f
=
0 A
ut
A
d − ut
C1 C2
=
C1 + C2
(1)
(2)
0 A
A
ut · d−ut
0 A
A
ut + d−ut
=
0 A
ut( − 0 ) + 0 d
(3)
2. Let the electrical field at the top part of the capacitor be E0 and the electrical field in the bottom
part of the capacitor be E. The we find the relation between E, E0 from the Gauss’ law (in which
we substitute 0 by ):
E0 A =
EA =
E =
q
0
q
0
E0
(4)
(5)
(6)
1
Then
Z
V
=
E0 =
E =
~ · d~l = E0 ut + 0 E0 (d − ut)
E
V
ut( − 0 ) + 0 d
V 0
ut( − 0 ) + 0 d
(7)
(8)
(9)
3) Magnetic field:
~ ·B
~ =0⇒B
~r = 0
∇
(10)
For the top part of the capacitor:
I
Z
Z
Z
d
d
~
~
J · d~a + µ0 0
E0 · d~a = µo 0
E~0 · d~a
dt
dt
d
V 0 πr2
u( − 0 )πr2
B0φ · 2πr = µo = −µo 0
dt ut( − 0 ) + 0 d
(ut( − 0 ) + 0 d)2
µ0 u( − 0 )0 r
B0φ = −
2 (ut( − 0 ) + 0 d)2
~ · d~l = µ0
B
For the bottom part of the capacitor:
I
Z
~ · d~l = µo d
~ · d~a
B
E
dt
d
V 0 πr2
u( − 0 )0 πr2
Bφ · 2πr = µo = −µo dt ut( − 0 ) + 0 d
(ut( − 0 ) + 0 d)2
µ0 u( − 0 )0 r
Bφ = −
= Boφ
2 (ut( − 0 ) + 0 d)2
Pay attention that the magnetic field is continuous along the capacitor.
2
(11)
(12)
(13)
(14)
(15)
(16)