Download Mutual Inductance

Mutual Inductance
Submitted by: I.D. 043423755
The problem:
Two loops of radius a are at a distance b from each other, such that the planes of the loops are
parallel, and perpendicular to the axis connecting them.
1. Assuming b a, what is the mutual inductance of the system?
2. In one loop there is a constant current I and the other loop rotates at the angular velocity
along its diameter. What is the induced EMF in the rotating loop?
The solution:
The method: We assume that there is a changing current in one of the loops and we will calculate
the e.m.f in the second loop and thus extract the inductance coefficient.
~ that is created along the z axis of the ring
1. The magnetic field B
B=
(µ0 I)a2
ẑ
2(a2 + b2 )3/2
(1)
b a - meaning that inside the area of the second ring, the magnetic field is approximately the
same. Using the Taylor expansion.
B=
µ0 Ia2
µ0 Ia2
=
+ O((a/b)2 )
2b3
2b3 (1 + (a/b)2 )3/2
(2)
Let’s calculate the flux
Φ=B·A=
µ0 Ia2 2 µ0 πIa4
πa =
2b3
2b3
(3)
Now we use the Faraday’s law
= −Φ̇ = −
µ0 πa4 ˙
I
2b3
(4)
Finally the mutual inductance coefficient M is the expression multiplying the time derivative of the
current.
M=
µ0 πa4
2b3
(5)
2. The magnetic flux through the loop now changes because the area changes with time
A = πa2 cos(ωt)
(6)
Thus
µ0 πa4 I(t)
cos(ωt)
2b3
µ0 πa4 ˙
= −Φ̇ = −
(I cos(ωt) − ωI sin(ωt))
2b3
Φ = B·A=
1
(7)
(8)
Toroid’s self inductance
Submitted by: I.D. 037706835
The problem:
A toroid shaped inductor with a rectangle profile has an inner radius r1 , an outer radius r2 and a
height a. The toroid has N windings and current I flows through it.
1. What is the magnetic field in the entire area?
a
2. Show that the toroid’s self inductance is L = µ0 N 2 2π
ln ( r2
r1 ).
3. After placing an infinite wire at the center of the toroid (On it’s symmetry axis),
Calculate the mutual inductance between the toroid and the wire in two ways:
• By issuing a current in the infinite wire and calculating the EMF induced in the toroid.
• By issuing a current in the toroid and calculating the EMF induced in the infinite wire.
The solution:
1. The magnetic field cannot be in the z direction because of the symmetry, and cannot be in the
~ ·B
~ = 0. So that the only direction could be ϕ̂.
radial direction since ∇
Using Ampere’s law, we can build numerous closed curves and get the circulation of the magnetic
field, but only a circulation inside the toroid will give us a current through the surface of the circulation. Hence, the only magnetic field in the area will be at direction ϕ̂, inside the toroid (where
r1 < r < r2 , − a2 < z < a2 ):
Inside the toroid:
B · 2πr = µ0 · N · I
~ = µ0 N I ϕ̂
=⇒ B
2πr
(1)
Outside the toroid:
~ =0
B · 2πr = µ0 · 0 =⇒ B
(2)
2. According to the definition of the intuctance Φ = L · I:
Z
Z r2
aN 2 µ0 I
aN 2 µ0 r2
~
ΦB = N · B · d~s =
· dr =
ln · I
2πr
2π
r1
r1
aN 2 µ0 r2
ln
2π
r1
L =
(3)
(4)
3.a. Calculating the flux of the magnetic field from the wire through the toroid, and extracting the
mutual inductance Lwt :
µ0 Iw
ϕ̂
2πr
Z
~ w · d~s = aN µ0 ln r2 · Iw
= N B
2π
r1
aN µ0 r2
=
ln
2π
r1
~ wire =
B
ΦB
Lwt
(5)
(6)
(7)
1
3.b. Using Faraday’s law, and circulating on an infinite rectangle, where one of the sides is on the
infinite wire, and the other three ones are far enough to assume that the electrical field in their area
vanishes, we get that the circulation of the electrical field on the rectangle is equal to the EMF in
the wire, hence:
I
Z
~ · d~l = − ∂
~ · d~a
=
E
B
(8)
∂t
Z r2
aµ0 N It
aµ0 N r2 ˙
∂
∂ aµ0 N It r2
=−
(9)
= −
dr = −
ln
ln · It
∂t r1
2πr
∂t
2π
r1
2π
r1
aN µ0 r2
Ltw =
(10)
ln
2π
r1
And we got Lwt = Ltw as expected.
Another method is using the differential version of Faraday’s law, taking under consideration that
along the wire (at the area of the toroid, since only there a magnetic flux changes) there is an
electrical field in the ẑ direction (cylindrical coordinates) which has no dependence on ϕ since the
problem is symmetrical:
~ = µ0 N I · ϕ̂
B
2πr
~
~ = − ∂ E · ϕ̂ = − ∂ B
∇×E
∂r
∂t
Z
∂
~ =
~ · dr
E
B
∂t
Z
Z a Z r2
2
aN µ0
r2
∂
µ0 N I
~
=
E · d~z =
drdz =
ln ( )I˙
∂t − a r1 2πr
2π
r1
~ = E(r,z) ẑ
E
(11)
(12)
(13)
(14)
2
Ltw =
aN µ0 r2
ln
2π
r1
(15)
2
Induction
Submitted by: I.D. 303689129
The problem:
A surface of an isolating cylinder, with a radius R and A length h, is charged with a charge Q.
The charge is distributed uniformly on the surface of the cylinder. A mass m is wrapped around
the center of the cylinder. Due to the force of gravity the mass is falling down, which rotates the
cylinder around its axis. Find the acceleration of the mass ignoring the cylinder’s moment of inertia.
The solution:
Let the cylinder stand on the XY (ρϕ) plane and rotate around the Z axis in the direction of ϕ.
We can assume that an induced electromagnetic force may cause a torque along the rotation axis.
Therefore, the torque equation would be:
X
~ × m~a = R
~ × m~g + R
~ × F~q
~τ = R
(1)
Now we shall evaluate the induced electromagnetic force.
The charge density on the cylinder surface is given by:
σ=
Q
2πRh
(2)
The charge is rotating, therefore the current density will be:
J~ = σ~v = σ(ω × R)
(3)
and the total current on the cylinder is:
Z
I = J~ · ds = σ k(ω × R)k h
(4)
h
Using the Ampere’s law and taking the same considerations as with a coil, the magnetic field inside
of the cylinder is:
I
~ · d~l =4πKIin ⇒ B ẑ = 4πKI ẑ = 4πKσωRẑ
B
(5)
h
And the total flux through the cylinder is:
Φ = 4πKσωRẑ · πR2 ẑ
(6)
Now, using the Faraday’s Law we can calculate the electric field along the cylinder’s surface:
I
∂ 4π 2 KσωR3
Eϕ̂ dl = −Φ̇ ⇒ 2πRE ϕ̂ = −
= −4π 2 Kσ ω̇R3
∂t
2πR
1
(7)
E ϕ̂ = −2πKσR2 ω̇ ϕ̂
(8)
The total elctromagnetic force can be calulated by integrating the electric field on each charge on
the cylinder’s surface:
Z
−Q2 KRω̇
F~q = E ϕ̂dq = −2QπKσR2 ω̇ ϕ̂ =
ϕ̂
(9)
h
S
ω̇ =
a
R
(10)
−Q2 Ka
F~q =
ϕ̂
h
(11)
Now we can insert equation (11) into equation (1) and evaluate the total torque on the cylinder.
X
~τ = Rmaẑ = Rmgẑ + R
−Q2 Ka
ẑ
h
(12)
Therefore the acceleration is:
~a =
m~g h
mh + Q2 K
(13)
2