Download Examples of magnetic field calculations and applications 1 Example

Examples of magnetic field calculations and
applications
Lecture 12
1
Example of a magnetic moment calculation
We consider the vector potential and magnetic field due to the magnetic moment created
by a rotating surface charge, σ, on a cylinder. The geometry is shown in Figure 1. The
magnetic moment, I(area), of a small loop at the position, z, as shown in the figure is ;
σ(Rdθ dz)
(πR2 )ẑ = πσR3 ω dz ẑ
dt
dm
~ =
The vector potential is then;
~ × ~r = µ0 [πσR3 ω ẑ × ~r ] dz
~ = µ0 dm
dA
4π
4π
r3
r3
~
This is integrated using ~r = α
~ − ~z to get A.
~ =
A
L/2
R
~
dA
−L/2
Choose α
~ to lie in the (x, y) plane along the x̂ direction. Then ẑ × ~r = αŷ
L/2
R dm
1
~ × ~r = [ µ0 [πσR3 ωα] R dz
~ = µ0
]ŷ
A
2
4π
4π
r3
[α + z 2 ]3/2
−L/2
The vector potential above has been calculated in the magnetic moment approximation,
(ie in the first non-zero order expansion of the vector potential in a power series of the loop
radius divided by the distance to the field point.)
2
The field and action of a Quadrupole lens
The quadrupole field is illustrated by the magnetic field shown in Figure 2 and given by the
equations;
Bz = gx
Bx = gz
By = 0
1
z
ω
P
r
L/2
α
dm
R
σ
x
y
−L/2
Figure 1: The vector potential of a rotating, cylindrical charge distribution
z
N
S
x
S
N
Figure 2: A quadrupole magnet used to focus charged particles
2
The field is generated by wire windings that create the magnetic poles shown in the figure
and parallel to the equipotential curves perpendicular to the field lines. We suppose the
length of the field into the page is L, and the field lines are shown in the figure. The field
strength increases linearly with distance from the axis.
B 2 = Bx2 + Bz2 = g 2 r 2
~ into the page, the Lorentz force converges a
For positive particles moving with velocity V
particle beam horizontally, and diverges it vertically. Check the force direction and note
the further away from the axis, the stronger the force. A magnetic lens is created if two
quadupole fields are placed in line and rotated by 90 deg with respect to each other. Then
with a proper choice of spacing, and field strength can provide focusing of a parallel beam
of particles to a point some distance behind the two magnets.
3
Power and the magnetic field
~ +V
~ × B].
~ This force causes the charge to move
The Lorentz force on a charge is F~ = q[E
in a direction perpendicular to the field and velocity, ˆl. Then we determine the power due
to this movement.
~ · F~ = q V
~ ·E
~
P = V
~ × (V
~ × B),
~ vanishes, so the
As indicated, the force term involving the magnetic field, V
magnetic field does no work on a charge and cannot change its energy only the direction of
its velocity.
4
Motion of a charged particle in a magnetic field
We suppose a constant magnetic field in the ẑ direction, and a charged particle of mass, m,
~ moves in this field. The motion is given by the Lorentz force,
charge, q, and velocity, V
which in non-relativistic form, is given by the kinetic equation, F~ = m~a.
~
~ × B]
~
m dV = q[V
dt
Neglect the interaction of the charge with other charges which may be present in a beam of
such particles. In Cartesian coordinates the coupled equation set below can be produced for
~ = B ẑ.
B
3
dVx = V qB
y m
dt
dVy
qB
= −Vx m
dt
dVz = 0
dt
The last equation requires that Vz = constant. The first two equations may be decoupled
giving a second order ode.
d2 Vi = − qB V
m i
dt2
In the above, Vi represents Vx or Vy . The solution is harmonic, and to satisfy both first order
equations;
Vx = V0x cos(ωt + φ)
Vy = −V0x sin(ωt + φ)
qB
Where ω = m and φ is a phase angle to be determined from the initial conditions. Assume that at t = 0 Vx = V0x and Vy = 0. Then φ = 0 and A = V0x . Now integrate these
equations again to get the coordinate trajectories. Let the initial velocity in the ẑ direction
be, Vz = V0z , and the initial position for (x, y, z) is, (0, (V0x /ω), Z0).
z = V0z t + Z0
x = (V0x /ω) sin(ωt)
y = (V0x /ω) cos(ωt)
Substitution verifies these solutions and initial conditions. We observe the motion is a helix in
3-D with a projection of circular motion onto the (x, y) plane. The radius of this circle, R, is ;
x2 + y 2 = R2 = [V0x /ω]2
In the above, R, represents the radius of curvature of the particle in the magnetic field.
Insert the particle momentum projected onto the (x, y) plane, p.
p = mV0x = mωR = qBR
p
R = qB
4
5
Drift Velocity and the Lorentz force
Suppose we apply a constant magnetic field in the ẑ direction and a constant electric field
in the x̂ direction to a charged particle. The particle is initially at rest. The equations of
motion in the (x, y) plane are;
dVx = qB V + (q/m) E
0
m y
dt
dVy
qB
= − m Vx
dt
After application of the initial conditions, the solution has the form, with ω defined in the
previous section;
Vx = (q/mω) sin(ωt)
Vy = (q/mω)[cos(ωt) − E0 /B]
The position is obtained by a second integration;
x = −(q/mω 2 )[cos(ωt)/ω] + X0
y = (q/mω 2)[sin(ωt)/ω − (E0 /B)t] + Y0
In the above (X0 , Y0) is the initial position. This motion is strange as the circle center
moves with constant velocity in the ŷ direction. Now we study this a little further by placing a resistance, proportional to the velocity, to the motion of the charge. This is artificial
because force is proportional to acceleration not velocity, but you have used frictional forces
proportional to velocity in mechanics, and here we want an energy disipating term. Thus
add a term which contains the first odd derivative of the position, velocity. Introduction of
the force term σ V~ into the above equations gives;
dVx + σ/m V = qB V + (q/m) E
0
x
m y
dt
dVy
qB
+ σ/m Vy = − m Vx
dt
Look for the equilibrum solution, ie the solution when the velocity becomes independent of
time so that dVi = 0.
dt
qσ
Vx = 2
E0
σ + (qB)2
q2B
E0
Vy = 2
σ + (qB)2
5
b
a
B
V
dl
I
Figure 3: The geometry describing the Hall effect
The drift angle between the applied field and the motion is called the Lorentz angle and is
given by;
tan(θ) = Vy /Vx = −qB/σ
The particle drifts at a constant velocity at the angle θ with respect to the applied field. The
above examples give a few simple illustrations of the motion of charged particles in magnetic
fields. In general, this topic is treated in magnetohydrodynamics (plasma) for example, and
the motion is highly non-linear and non-intuitive. The understanding of plasma is crucial to
the development of controlled fusion reactors.
6
The Hall effect
We suppose a current through a conducting medium in which there is a magnetic field perpendicular to the current flow and the surfaces of the conductor Figure 3. The current
represents a flow of charge so that there is a deflection of the current due to the Lorentz
force. As previously determined Idl = qV so the force on a small element of current due to
the magnetic field when the magnetic field and the current are perpendicular is;
dFm = IBdl
Charge flows and builds up an electric field on the parallel surfaces of the conductor. When
equilibrum is reached, the electric force cancels the magnetic force.
~ + ~v × B]
~
F~ = 0 = q[E
Use in the above qv = Idl. When the force reaches equlibrium;
6
q(dE) = −IdlB
The charge per unit volume in the conductor as obtained from the figure is ρ = q/ab(dl).
Substitute for q and let dE = V /b, with V the potential between the sides of the conductor.
Then;
V = −IB/(ρa)
The hall effect occurs for current flow in a magnetic field and is used in a number of instruments to measure either currents or magnetic fields.
7
Example of Ampere’s law and superposition
There is a current flow in a cylindrical conductor of radius, R. The conductor has a hole
of radius, a, displaced a distance, b from the axis of the conductor. The geometry is shown
in Figure 4. It is assumed that the current density is constant in the cylinder. We are to
find the magnetic field in the hole. This is done by superposition of a current in a cylinder
of radius a centered at z = b in opposition to the current in the cylinder, as shown in the
figure. The current in the cylinder without the hole is;
IT = JπR2
The current in the hole flowing in the opposite direction is;
IH = Jπa2
Thus the total current in the cylinder with the hole is IT − Ia = IC
IC = J(π[R2 − a2 ])
J =
IC
π[R − a2 ]
2
Now find the field at point P in Figure 4 by superimposing the contributions to the field
from each of these current densities. This can be done by Ampere’s law.
H
~ · d~l = µ0 I
B
4π
For the conducting cylinder without the hole, the enclosed current is; Ic = Jc (πr 2 ). The
circulation on the left side of the above integral is evaluated to be B(2πr).
7
^
φ
z’
z
b
^
φ
P
θ
r’
b θ r
a
R
Figure 4: The geometry to find the magnetic field inside a cylindrical cavity in a cylindrical
conductor
µ0 I
µ0 Ir
πr 2
2
2 2πr =
π(R − a )
2π(R2 − a2 )
BC =
In the same way the field for the current creating the hole is;
BH =
µ0 Ir ′
2π(R2 − a2 )
The geometry is shown in the figure. Subtract these vector fields. Write the unit vectors
dzẑ = ~r × d~l and use the geometry to obtain z − z ′ = b. The final result is;
B =
µ0 Ib
2π(R2 − a2 )
Thus the magnetic field is constant in the interior and pointed in the ẑ direction.
8
Magnetic pressure and energy
Consider two parallel current sheets separated by a distance, d, with uniform, constant currents flowing in opposite directions, Figure 5. Find the magnetic field due to one of these
sheets on the other. The magnetic field is obtained using Ampere’s law as previously. Because of symmetry the magnetic field must be directed parallel to the sheet, and can only
depend on the perpendicular distance from the sheet to the field point. Evaluation of the
integral form of Ampere’s law gives;
H
~ · d~l = 2BL = µ0 I
B
8
L
I/L
B
B
B
B
B
d
I/L
Figure 5: The magnetic field created by 2 parallel current sheets with currents flowing in
opposi te directions
The factor of 2 comes from the field above and below the sheet, with L the distance parallel
~ I is the current that flows through this Amperian loop.
to the sheet of the path along B.
Thus for one sheet;
B = µ(I/L)/2 = (µ/2) I
In the above we have written I as the current per unit width on the sheet. The direction
of the magnetic field is given by the right-hand-rule. Note that the field is independent of
the distance from the sheet. Thus the fields when superimposed from the two sheets, add
between the sheets and cancel outside the sheets. Finally we also see that the force generated
by the magnetic field on one sheet interacting with the current on the other is repulsive. We
visualize this situation by thinking of the magnetic field as creating a pressure between the
plates tending to expand the distance between them.
Use the Lorentz force to calculate this force. In the equation for the Lorentz force, substitute
IL for qV . The field and current direction are perpendicular, so F2 = I2 LB1 = I2 x2 (µ/2)I1 .
Now the total magnetic field between the plates comes from the superposition of both fields
which add to BT = 2(µ/2)I1,2. Then I1,2 = BL/µ which we substitute for I1 in the force
equation yielding;
F = 1 B2
2µ0
Lx
The above is the force per unit surface area (pressure) of one current sheet on the other. This
pressure attempts to push the plates apart. Now suppose we do work against this pressure
by compressing the plates a distance, d. The movment of the plates removes a volume of the
magnetic field, Lxd, under the Amperian loop. and puts an energy into the system given
by W = F d. We remove the geometry in the equation by dividing by the volume to obtain
the energy per unit volume which we assign to the magnetic field. If we allow the plates to
9
I
B
Figure 6: The geometry to find field enclosed in a torus using Ampere’s law
re-expand, the plates do work removing this energy creating additional field in this volume.
Compare this energy density, (1/2µ0)B 2 , to the energy density of the electric field, (ǫ0 /2)E 2 .
9
9.1
Additional Examples
Field of a torus
The geometry of a torus is illustrated in Figure 6. We assume that a current sheet moving
in a circular direction around the donught is continuous, (ie very tightly wound wires). No
field leaks out of the donught, and the geometry is symmetric in azimuthal angle. Thus we
can use Ampere’s law to get the magnetic field. Use a circular Amperian loop as shown;
H
~ · d~l = µ0 IT
B
µ0
B = 2πr
In this expression IT is the total current flowing. We could write a current per unit width
IT to remove the geometric dependence.
of I = 2πr
10
z
z
dz
I
r
I
y
R1
R2
x
Figure 7: The geometry to find the vector potential created by two long, parallel filaments
of current flowing in opposite directions
9.2
Field of a bipolar filament
We assume two long filaments carring a current I in opposite directions. We are to find the
~ for this geometry, Figure 7. By symmetry, A,
~ must be independent of
vector potential, A,
z. We have;
R
J~
~ = µ0
A
4π dτ r
The current density is in the direction of ẑ and use J~ · d area
~ .
R∞
R∞
dz
dz
~ = µ0 ẑ [
A
−
]
2
4π −∞ (R2 + z 2 )1/2
(R
+
z 2 )1/2
−∞
2
1
q
z + z 2 + R22 ∞
~ = 2 µ0 I ln[
q
A
]0 ẑ
4π
z + z 2 + R12
~ = 2 µ0 I ln[ R1 ]
A
4π
R2
11