Problems in Magnetostatics


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Problems in Magnetostatics

8th February 2007

Some of the later problems are quite challenging. This is characteristic of problems in magnetism. There are trivial problems and there are tough problems. Very few problems lie in between.

1. Which of these Magnetic fields can exist? Determine the current density that created the valid fields. What Vector Potential corresponds to these fields?
(a) B(r) = e−y2 xˆ: Divergence of the vector gives
∇ · B = ∂xe−y2 = 0

The field does not blow up anywhere. Hence this is a valid magnetic field. (It does not go to zero for large x but small |y| though).
The curl gives the current:

1 ∇ × B = − zˆ ∂ e−y2 = 2ye−y2 zˆ

µ0

µ0 y

µ0

The vector potential can be found directly from B = ∇ × A. Let A = Ayyˆ. Then
Bx = −∂zAy
Hence, Ay = −ze−y2 , i.e., A = −ze−y2 yˆ

(b) B(r) = e−x2 xˆ: Divergence of this field is not zero. Hence it cannot be a valid magnetic field.
(c) B(r) = sin (kr) rˆ (r, θ, z coordinate system): Again, divergence of this field is
∇ · B = ∂r sin (kr) = k cos (kr)
is not zero. Hence not a valid field.

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(d) B(r) = rθˆ (r, θ, z coordinate system):
The divergence of this field is zero since Bθ does not depend on θ. The current density is given by

j = zˆ 1 ∂r r2 = 2zˆ

µ0 r

µ0

From Stokes theorem H B·dl = 2πr2 = πr2 j0µ0 which confirms this answer. The vector potential is again easiest to get from B:

A = rzrˆ

2. The following current densities are given. Determine the Magnetic Fields.

(a) j(r) = re−rθˆ:

From symmetry, we expect ∂z, ∂θ = 0. So, B = Bz(r)zˆ. From Ampere’s
Law, we have µ0re−r = −∂rBz

Hence,

Bz(r)

=

Bz(0)



µ0

Rr
0

r

e−r

dr

.

Since

the

total

current

is

bounded

and goes quickly to zero along rˆ, limr→∞ Bz = 0. Hence,

Z∞ Bz(r) = µ0 r e−r dr
r

The vector potential is similarly got from B = ∇ × A. Clearly, A = Aθ(r)θˆ.

Z Bz(r) = µ0


r e−r dr

=

1 ∂r (rAθ)

r

r

This can again be integrated to obtain A. Note that the integrals are obtainable via integration by parts, but that is not the point of this course.

(b) j(r) is a square of current of 1 Ampere with its sides along the z axis, the y axis, y = 1 and z = 1. The current is flowing along +zˆ up the z−axis. (It is sufficient to give the answers in terms of integrals, if you cannot simplify them).:
For any line current I extending from r0 to r0 + Lzˆ, the Vector Potential is given by

A = µ0 IzˆZ L 4π 0

dζ (x − x0)2 + (y − y0)2 + (z − z0 − ζ)2

Similarly for a line current along yˆ,

A = µ0 IyˆZ L 4π 0

dζ (x − x0)2 + (y − y0 − ζ)2 + (z − z0)2

Hence the total Vector Potential is given by

A = µ0I zˆZ 1 4π 0

1− x2 + y2 + (z − ζ)2

1 dζ x2 + (y − 1)2 + (z − ζ)2

+ µ0I yˆZ 1 4π 0

1− x2 + (y − ζ)2 + (z − 1)2

1 dζ x2 + (y − ζ)2 + z2

2

which can easily be expressed in closed form. The curl of this can now be taken.
The far field due to this (or any) loop, can be obtained by Taylor expansion (see Far field of an arbitrary current loop in the site)

A(r) = µ0 Z √ j(r )

dV

4π r2 + r 2 − 2r · r

µ0 1 Z

r·r

4π r j(r ) 1 + r2 dV

µ0 m × r = 4π r3

The derivation is lengthy and is given separately. The important thing to note is that this formula now applies to any loop, even one not in the plane. The magnetic dipole moment is given by

I m = r × j(r )dV

3. A current loop is elliptical in shape and lies in the y − z plane with its major axis along yˆ and its centre at the origin. What components of A and B are present along the x-axis? What components are present along the y-axis? Hint: If a function is symmetric about a point, its derivitive at that point goes to zero.:
For each element of current in z < 0, there is a symmetrical element with z > 0. Combining them yields a net j along zˆ. Hence A is along zˆ everywhere on the x − y plane, including along the x- and y-axes. Thus, B has only x and y components.
B = xˆ∂yAz − yˆ∂xAz

By symmetry we expect B to be along xˆ on the y-axis and along xˆ on the x-axis (this is obviously true for the special case of a circular loop).

4. Consider the solution for A and B for a finite solenoid. Obtain the radial component of the B field near the axis. Hint: Use the divergence theorem and collect terms order by order in smallness and find the lowest order radial term.:

From the derivation of the field for the finite solenoid, we have an expression for B = Bz(z)zˆ on axis. Let

∑ Br =


Brk(z)rk

k=0

∑ Bz =


Bzk(z)rk

k=0

We have expressions for Bz0 and Br0 = 0. Now apply the divergence theorem to a cylinder of height dz and radius r

∑ ∑ ∞

rk+2 (Bzk(z + dz) − Bzk(z)) 2π

+



Brk(z)rk2πrdz

=

0

k=0

k + 2 k=0

3

∑ ∑ ∞




(∂zBzk(z)) rk+2 + 2πBrk(z)rk+1

=

0

k=0 k + 2

k=0

∑∞
2π rk+1
k=0

1 ∂zBz,k−1 + Brk k+1

=0

Working order by order in r, this yields

Br0 = 0 Br1 = − 1 ∂zBz0
2
etc. Using (from the solenoid problem)

 Bz0 = µ0nI0 

L 2−z −
L 2 − z 2 + a2

 −L 2 − z
, −L 2 − z 2 + a2

ra

Br1 can now be calculated for any z.
5. A solid wire carries a D.C. current I. How is this current distributed? What happens to the qv × B force acting on conduction electrons?: Due to symmetry, j depends on r but not on θ or z. Applying divergence theorem to a cylinder of radius r and arbitrary height, we obtain that
2πr jr = − dQencl = 0 dt
Thus, j = jθ(r)θˆ + jz(r)zˆ, which automatically implies that A also lies in the θ-z plane (why? remember the circular loop derivation). Now inside a conductor we must have

F = qen E + ve × B = qenE + j × B

or

j ∝ qenE + j × B

(1)

Now, A lies in the θ-z plane and depends only on r. Hence B = ∇ × A also lies in the θ-z plane and depends only on r. So j × B is the cross product of two vectors, both of which are in the θˆ-zˆ plane. The direction of this term is rˆ. Thus, Eq. 1 can be written as

jr = 0
jθ = qen ve × B θ = 0 jz = σEzzˆ

This, in turn implies A = Az(r)zˆ and B = Bθ(r)θˆ. Since the magnetic field is static, Faraday’s law (which we do not officially know about yet) does not apply,
and the Electric field remains curl free. What this means is that

ZL

1Z L

0 Ezdz = σ 0 jzdz = Applied Voltage

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is independent of r. Note that this last step would fail if E = −∇φ. Thus, j = jzzˆ is uniform over the cross-section. How is this possible? After all we have an Idl × B force that would tend to push the current towards the center of the wire (this is called a pinch effect, and is real). The answer is that static Electric field builds up so that at each radius r
Er = vzBθ
Then, the total force on conduction electrons becomes
F = qe Ezzˆ + Errˆ + ve × B = qeEzzˆ
It is as if the magnetic field never was. Note: Earlier we used j = σE and here we are violating that. The actual equation is j = σ E + v × B . Since B is along θ and v is along z only the r component sees this extra term. This equation is known as generalized Ohm’s law.
6. A pipe of irregular cross-section carries a D.C. current I. Determine how the current is distributed.: This is a more complicated problem. The cross-section is not given to be axially symmetric in θ. But it is given to be a cylinder, i.e., ∂z = 0. Inside the metal, the steady state momentum equation for electrons yields
0 = qe E + ve × B − meνeve
Thus, the current j must satisfy
j = neqeve = neq2e E + νe × B = α neqeE + j × B meνe
Since j is orthogonal to j × B, the j × B term cannot cause j. This is actually the immediate proof that no steadystate current can exist in the absence of an Electric Field. Let E = E jˆ+ E⊥. Then,
0 = qeneE⊥ + j × B j = σE jˆ ∇ × B = µ0 j
Now, it is quite possible that part of the applied Electric Field goes towards building up E⊥. However, even when that is so, it must be the case that E is the uncancelled portion of the applied (external) Electric Field.
• But this means that E is along zˆ since the applied field is along zˆ. • Then j must also be along zˆ since j = σE jˆ. • It immediately follows that A = Az(x, y)zˆ. • Taking the curl, then, B = Bxxˆ + Byyˆ.
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To summarize:

j = jzzˆ = σEzzˆ A = Azzˆ B = Bxxˆ + Byyˆ

This solution requires that we be able to find a curl-free Electric Field that satis-
fies j
E⊥ + qene × B = 0

A solution will exist if ∇ × E⊥ = 0:

∇×E = 1 ∇× j×B qene

=1 qene

∇·B j− ∇· j B

=0

since both j and B are divergence free.
Thus we now have a complete answer to the problem, since a uniform j fully specifies all the currents present, which allows us to calculate E⊥ and B as well.

7. A conducting fluid is forced across a uniform magnetic field. Show that a voltage develops across the terminals (with area A, kept at x = 0 and x = L). Assume that the terminal at x = 0 is grounded. If a resistor R is connected across the terminals, determine how this potential relates to the velocity profile of the fluid, u = uz(x)zˆ
and the strength of the B = B0yˆ field. You may assume j = σ E + u × B .:

The fluid has a conductivity σ and flows with a velocity uz(x)zˆ across the terminals. The terminals are asssumed to be at x = 0 (grounded) and x = L (at some voltage V ), and have an area A. A uniform magnetic field B0yˆ is present. The terminals are connected across a resistor R, and carry a current V R. This current must be the same at all x positions by charge continuity. Hence,

jx = σ (Ex − uzB0) = − I A

We can therefore calculate V :

ZL V = − Exdx
0

ZL I

=−
0

Aσ + uzB0

IL

ZL

= Aσ − B0 0 uz

= −IR

where the − sign in the last term was due to the fact that the direction of current was into the positive terminal of the MHD battery. Hence,

VL

ZL

V = − R Aσ − B0 0 uz

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i.e., V = −B0 R0L uzdx = −B0 R0L uzdx 1 + L RAσ 1 + Ri R

8. Show that the magnetic field cannot do any work on a particle, i.e., cannot change its energy.:
The force equation is dv
m =q E+v×B dt
The work done by the applied force is given by
dW = F·v=q E+v×B ·v
dt = qE · v

Hence, the magnetic field cannot directly do any work on a charged particle. It can however do work via an induced electric field.

9. Given

µ0 m × r A = 4π r3

determine B(r). Then obtain ∇ × B and show that it is zero except at the origin.

10. An axially symmetric (∂θ = 0) magnetic field is given by z2
Bz = B0 1 + 2 1 + z2

A proton (charge e, mass mp) h√as an initial position at the origin and an initial velocity of v = (3xˆ + zˆ) × 104 10 metres per second.

(a) Determine Br and Bθ. Sketch them. This is what is called a “magnetic mirror” where the axial magnetic field is stronger at two ends of a mag-

netic field. Examples include the earth’s magnetic field which is strongest

near the ends, i.e., the poles and weaker in the middle which is high in the

magnetosphere.

(b) Let the intial velocity of the proton be 104 metres per second along xˆ. Solve

for the trajectory of the proton in time. Remember that the equation to be

solved is

mpv˙ = ev × B

The proton moves in circles in a uniform magnetic field.

(c) Consider the actual trajectory. Assume (as is quite valid) that the distance travelled along zˆ is negligible during a single gyroperiod (T = 2π Ωp, ΩP = eBz(z) mp). At any given position in z, determine the force on the
proton due to v × B. Write down the equation for the evolution of vz(t).

(d) Invoke conservation of energy (see previous problem) to obtain the modified vr(t). Compute µ = mv2⊥ 2Bz. How does it evolve in time? Note: µ is what is called an adiabatic invariant, and it is an extraordinarily important
quantity that is instrumental in helping us understand waves in magnetised
gases (eg., solar wind, ionosphere, solar corona etc.)

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Problems in Magnetostatics