Sample Test
NOT FINISHED Answer: (D)
9.25 Thermal & Electrical Conductivity
A metal is a lattice of atoms, each with a shell of electrons. This forms a positive ionic lattice where the outer electrons are free to dissociate from the parent atoms and move freely through the lattic as a ‘sea’ of electrons. When a potential difference is applied across the metal, the electrons drift from one end of the conductor to the other under the influence of the electric field. It is this free moving electron ‘sea’ that makes a metal an electrical conductor.
These free moving electrons are also efficient at transferring thermal energy for the same reason. Thermal and electrical conductivity in metals are closely related to each other as outlined in the Wiedemann-Franz Law.
κ
σ = LT (9.25.1)
where the Lorenz number, L = 2.44 × 10−8WΩK−1 and κ and σ are the thermal and electrical conductivities respectively. This corelation does not apply to non-metals due to the increased role of phonon carriers.
Answer: (E)
DRAFT
Nonconservation of Parity in Weak Interactions 81
9.26 Nonconservation of Parity in Weak Interactions
Of the four interations, electromagnetism, strong, weak and gravity, parity is conserved in all except for the weak interaction. To examine violations of these interactions we must look at the helicity of our particles and see whether they are “left-handed” or
“right-handed”.
A particle is said to be “right-handed” if the direction of its spin is the same as the direction as its motion. It is “left-handed” if the directions of spin and motion are opposite to each other. Thus the helicity of a particle is the projection of the spin vector onto the momentum vector where left is negative and right is positive.6
h ≡ S · p
Particles are not typically characterized as being “left-handed” or “right-handed”. For example, an electron could have both its spin and momentum pointing in the same direction to the right and hence be classified as “right-handed”. But from the reference frame of someone travelling faster than the speed of the electron, would see the electron travelling to the left and hence conclude the electron is “left-handed”.
Neutrinos, on the other hand, travel very close to the speed of light and it would be very difficult to accelerate to a point where one would be able to change the “handedness”
of the neutrino. Thus, we say that the neutrino has an intrinsic parity, all of them being left-handed. Anti-neutrinos on the other hand are all right-handed. This causes weak interactions, neutrino emitting ones in particular, to violate the conservation of parity law.7
For the pion decay,
π+→ µ++ υµ+ (9.26.2)
It is very difficult to detect and measure the helicity of the neutrino directly but we can measure it indirectly through the above decay and hence demonstrate nonconservation of parity.
If the pion is at rest and has spin-0, the anti-muon and neutrino will come out in opposite directions.8 In the figure below, the anti-muons are observed with their z-component of angular momentum given by mµ = −12. Angular momentum conversation then implies mυ= +12 for the neutrino.
It is very difficult if not impossible to detect neutrinos in a typical laboratory setting but we can detect muons and measure their helicity.
Choice (A) The Q-value is the kinetic Energy released in the decay of the particle at rest. Parity deals with mirror symmetry violations and not energy.
6Draw Helicity Diagrams
7Add section explaining parity
8Draw Diagram Here
©2009 David S. Latchman
DRAFT
82 Sample Test
Choice (B) We measure for violations of parity conservation by measuring the longi-tudinal polarization of the anti-muon. We choose this answer.
Choice (C) The pion has spin-0 and is stationary. So it won’t be polarized. Measuring this gives us no information on our decay products.
Choice (D) The angular correlation would be difficult as neutrinos are difficult to detect.
Choice (E) Parity deals with spatial assymetry and has nothing to do with time. We can eliminate this choice.
Answer: (B)
9.27 Moment of Inertia
The moment of inertia is
I = Z
r2dm (9.27.1)
In the case of a hoop about its center axis,
I = MR2 (9.27.2)
From eq. (9.27.1), we see that the moment of inertia deals with how the mass is distributed along its axis.9 We see that
Ais equivalent to A
Figure 9.27.1: Hoop and S-shaped wire
Thus, see fig. 9.27.1, the moment of inertia of our S-shaped wire can be found from a hoop with its axis or rotation at its radius. This can be calculated by using the Parallel Axis Theorem
I= ICM+ Md2 (9.27.3)
where d2is the distance from the center of mass. This becomes
I = MR2+ MR2 = 2MR2 (9.27.4)
Answer: (E)
9The moment of inertia of a 1 kg mass at a distance 1 m from the axis of rotation is the same as a hoop with the same mass rotating about its central axis.
DRAFT
Lorentz Force Law I 83
9.28 Lorentz Force Law I
I
+ + + + + + + + + + + + + + + + +
u B FE
FB
Figure 9.28.1: Charged particle moving parallel to a positively charged current carrying wire
The force on the charged particle is determined by the Lorentz Force Equation
F= e [E + u × B] (9.28.1)
where FE = eE and FB = e(u × B). For our charged particle to travel parallel to our wire, FE = FB.10
E= λ`
2π0r (9.28.2)
and the magnetic field can be determined from Amp`ere’s Law I
B · ds = µ0Ienclosed (9.28.3) In the case of our wire
B= µ0I
2πr (9.28.4)
Plugging eqs. (9.28.2) and (9.28.4) into eq. (9.28.1) gives F= e
"
λ`
2π0r+ uµ0I 2πr
#
= 0 (9.28.5)
For the particle to be undeflected, FE+ FB = 0 FE+ FB = 0 λ`
2π0r −uµ0I
2πr = 0 (9.28.6)
Now we can go about eliminating choices.
10Add derivation in a section
©2009 David S. Latchman
DRAFT
84 Sample Test
Doubling charge per unit length We see from eq. (9.28.6), halving the current, I, and doubling the linear charge density, λ, will not allow the particle to continue undeflected.
2λ`
2π0r−uµ0I/2
2πr = 2FE− FB
2 , 0 (9.28.7)
Doubling the charge on the particle We see from, eq. (9.28.6) that the charge on the particle, e, has no effect on the particle’s trajectory. We would be left with
λ`
2π0r −uµ0I/2
2πr = FE−FB
2 , 0 (9.28.8)
Doubling both the charge per unit length on the wire and the charge on the particle As shown above, the particle’s charge has no effect on the trajectory. This leaves us with the charge per unit length,λ and as we have seen before, this will change the particle’s trajectory, see eq. (9.28.7).
Doubling the speed of the particle If we double the particle’s speed we will get λ`
This is our answer
Introducing an additional magnetic field parallel to the wire Recalling eq. (9.28.1), the force due to the magnetic field is a cross product between the velocity and the field. A charged particle moving in the same direction as the field will experience no magnetic force.
FB = e [u × B]
= uB sin 0
= 0 (9.28.9)
Answer: (D)
9.29 Lorentz Force Law II
As we can see from eq. (9.28.5), the forces due to the electric and magnetic fields are equal.
DRAFT
Nuclear Angular Moment 85
If we move our charged particle a distance 2r from the wire with a speed nu, eq. (9.28.5) becomes
The speed of the particle is u.
Answer: (C)
9.30 Nuclear Angular Moment
NOT FINISHED