Assignment 11, solutions

Problem 1:

Find an expression for the radius of the trajectory of a particle of charge q moving at a speed v at right angles to a uniform magnetic field of magnitude B, if v << c.  Estimate the radius of motion for an electron and also that for a proton assuming that they move at velocity ~0.1 c at right angles to a magnetic field of 5*10−6 G, a typical interstellar magnetic field.  Will such protons and electrons be confined to a galaxy of size 105 light years?
(1G = 10-4 T)

Solution:

bulletConcepts:
The Lorentz force
bulletReasoning:
r = p/(qB) = mv(qB), r = v*1.14*10-2 s for electrons and r = v*20.9 s for protons. 
bulletDetails of the calculation:
With v ~ 0.1 c, the radius of circular motion for electrons is 3.4 *105 m, and the radius for protons is 6.3 *108 m.  These distances are much smaller than the size of the galaxy, so in a uniform magnetic field of this magnitude protons and electrons with v = 0.1 c are confined.

Problem 2:

An infinitely long cylinder with radius a and permeability μ is placed into an initially uniform magnetic field B0 with its axis perpendicular to B0.  Find the resultant field inside and outside of the cylinder.

Solution:

bulletConcepts:
The principle of superposition, boundary conditions, B = μH.
bulletReasoning:
The external magnetic field will magnetize the cylinder.   The total magnetic field will be the superposition of the magnetic field of a cylinder with uniform magnetization M = mk and  the external field B0.
bulletDetails of the calculation:
Assume that the magnetic field will be of the form
Bin = B2(cos(φ)[r/r] - sin(φ)[φ/φ]) = B2k.
Bout = B0 + (B1/r2)(cos(φ)[r/r] + sin(φ)[φ/φ]).
If this form of the field can satisfy the boundary conditions, then the uniqueness theorem guaranties that it is the correct and only solution.
The boundary conditions at r = a are B∙n = continuous, H∙t = continuous.
B∙n = continuous -->  B0cos(φ) + (B1/a2)cos(φ) = B2cos(φ),  B0 + (B1/a2) = B2.
H∙t = continuous -->  -(B00)sin(φ) + (B1/(μ0a2))sin(φ) = -(B2/μ)sin(φ),  -(B00) + (B1/(μ0a2)) = -(B2/μ).
Combining:  -(B00) + (B1/(μ0a2)) = -(B0/μ) - (B1/(μa2)),  (B1/a2)(μ + μ0) = B0(μ - μ0).
B1 = a2B0(μ - μ0)/(μ + μ0),  B2 = B02μ/(μ + μ0).
Bin = [B02μ/(μ + μ0)]k.
Bout = B0 + [B0(μ - μ0)/(μ + μ0)](a2/r2)(cos(φ)[r/r] + sin(φ)[φ/φ]).

Problem 3:

The space between a pair of coaxial cylindrical conductors is evacuated.  The radius of the inner cylinder is a, and the inner radius of the outer cylinder is b as shown in the figure.

The outer cylinder, called the anode, may be given a positive potential V relative to the inner cylinder.  A static homogeneous magnetic field B parallel to the cylinder axis, directed out of the plane of the figure, is also present.  Induced charges in the conductors are neglected.
We study the dynamics of electrons with mass m and charge –qe.  The electrons are released at the surface of the inner cylinder.
(a)   First the potential V is turned on, but B = 0.  An electron is set free with negligible velocity at the surface of the inner cylinder.  Determine its speed v when it hits the anode.  Give the answer both when a non-relativistic treatment is sufficient and when it is not.

For the remaining parts of the problem a non-relativistic treatment suffices.
(b)   Now V = 0 but the homogeneous magnetic field B is present.  An electron starts out with an initial velocity v0 in the radial direction.  For magnetic field magnitudes larger than a critical value Bc, the electron will not reach the anode.  Make a sketch of the trajectory of the electron when B is slightly larger than Bc.  Determine Bc.

From now on both the potential V and the homogeneous magnetic field B arc present.
(c)  The magnetic field will give the electron a non-zero angular momentum L with respect to the cylinder axis.  Write down an equation for the rate of change dL/dt of the angular momentum.  Show chat this equation implies that L - kqeBr2 is constant during the motion, where k is a definite pure number.  Here r is the distance front the cylinder axis.  Determine the value of k.
(d)  Consider an electron. released front the inner cylinder with negligible velocity, that does not reach the anode, but has maximum distance from the cylinder axis equal to rm.  Determine the speed vrm at that point rm where radial distance is maximum, in terms of rm.
(e)  We are interested in using the magnetic field to regulate the electron current to the anode.  For B larger than a critical magnetic field Bc, an electron released with negligible velocity will not reach the anode.  Determine Bc.

Solution:

bulletConcepts:
The Lorentz Force
bulletReasoning:
A charged particle moving in the presence of an electric and a magnetic field is acted on by a force F = q(E +v × B). 
bulletDetails of the calculation:
(a)  Potential energy is converted into kinetic energy.
Non-relativistic:  ½mv2 = qeV.  v = (2qeV/m)½.
Relativistic:  mc2/(1 - v2/c2)½ - mc2 = qeV.  v = c[1 – (mc2/( qeV + mc2))2]½.

(b)  F = qev×B.  The electron will move in a circle.  The initial velocity is tangential to the circle.
For the radius of the circle r we have qeBv = mv02/r,  B = mv/(qer).

When r = R, the electron reaches the anode, where (a2 + R2) ½ = b – R.
R = (b2 – a2)/2b.
Bc = 2bmv/(qe(b2 – a2)).

(c)  torque τ = r × F = qer × (E + v × B) = qer × (v × B)
= qe[(r∙B)v – (r∙v)B = qervrB = dL/dt.
(r and v are perpendicular to B, E points radially inward.)
dL/dt = qeBrdr/dt.  d(L – qeBr2/2)/dt = 0,  L – qeBr2/2 = C.  C is a constant and k = ½.

(d)  At the point rm L = mrmvrm, at r = a L = v = 0.
Therefore 0 – qeBa2/2 = mrmvrm - qeBrm2/2.  vrm = qeB(rm2 – a2)(2mrm).

(e)  For the critical magnetic field Bc the maximum radius rm equals b.
vb = qeBc(b2 – a2)(2mb).
But we also have ½mvb2 = qeV, vb = (2qeV/m)½.
Therefore (2qeV/m)½ = qeBc(b2 – a2)(2mb),  Bc = [2b/(b2 – a2)][2mV/qe]½.

Problem 4:

In Bohr's 1913 model of the hydrogen atom, the electron is in a circular orbit of radius 5.29*10-11 m and its speed is 2.19*106 m/s.
(a)  What is the magnitude of the magnetic moment due to the electron's motion?
(b)  If the electron orbits counterclockwise in a horizontal circle, what is the direction of this magnetic moment vector?

Solution:

bulletConcepts:
The magnetic moment m = IAn.
bulletReasoning:
We are given the orbital radius and the speed of the electron.  The electron is a negatively charged particle and therefore its motion about the nucleus constitutes an average current in a current loop.
bulletDetails of the calculation:
(a)  m = IA
The current is defined as the charge that passes a stationary observer per unit time.
I = q/T = qv/(2πR),  T = period.
m = qvR/2 = (1.6*1019)(2.19*106)(5.29*10-11) Am2 = 9.27*10-24 Am2.
We can also write m =  qL/(2me), where L is the angular momentum of the electron about the nucleus.
(b)  Let the horizontal circle lie in the z = 0 plane and let the electron move in the φ direction.  Then the current is flowing in the -φ direction and m points in the -z direction.
m = -9.27*10-24 Am2 k.

Problem 5:

Find the magnetic field B at the center of a flat spiral with current I.  The spiral is contained between two radii r and R and has N turns.  Do not consider the effect of the connecting wires.


Solution:

bulletConcepts:
The Biot-Savart law
bulletReasoning:
For filamentary currents we have.  B(r) = (μ0/(4π))∫vI dl' × (r-r')/|r-r'|3.   (SI units).
bulletDetails of the calculation:
Place the center of the spiral at the origin.  Assume the current flows counterclockwise.
Let D = (R - r)/N be the pacing between turns and let he spiral starts at φ = 0.  (0 ≤ φ ≤ N2π)
Consider a section of the spiral a distance r’ = R – (D/2π)φ from the center of length dl = r’dφ. 
The magnetic field at the origin due to this section is
dB = (μ0/(4π))I dφ /r’ = (μ0I/(4π))dφ/(R – (D/2π)φ). (Biot-Savart law)
The magnetic field produced by the spiral at the origin is
B = (μ0I/(4π))∫0N2πdφ/(R – (D/2π)φ) = (μ0I/(2D))ln(R)/ln(R - ND)
= (μ0IN/(2(R - r)))ln(R/r).
B points perpendicular to the plane of the spiral.  If the current flows counterclockwise, B points out of the page.
or
For a single loop the field at the center of the loop of radius r' is B = μ0I/(2r'), normal to the plane of the loop.
Let I = kdr', k = NI/(R - r).  Here k is the magnitude of the surface current density.
Then for the spiral we have
B = μ0rRkdr'/(2r') = (μ0k/2) ln(R/r) = (μ0IN/(2(R - r)))ln(R/r).