Modern family: Addition

State level Olympiad (5)

Q 4b. Diamagnetism.
We model a diamagnetic substance to have all atoms oriented so that the electron orbits are in the x-y plane, exactly half of which are clockwise and half counterclockwise when viewed from the positive z axis looking toward the origin. Some substances are predominantly
diamagnetic.

i. Calculate the total magnetic moment of a diamagnetic substance with N atoms.
Write your answer in terms of any or all of the following parameters:  e, m_e, R, N, μ_0 .

ii. An external magnetic field
~ B_0 = B_0 zˆ  is applied to the substance. Assume that the
introduction of the external field doesn’t change the fact that the electron moves in a circular orbit of radius R. Determine Δω , the change in angular velocity of
the electron, for both the clockwise and counterclockwise orbits. Throughout this
entire problem you can assume that Δω << ω_0 . Write your answer in terms of
e, m_e, and B_0 only.

iii. Assume that the external field is turned on at a constant rate in a time interval Δt . That is to say, when t = 0 the external field is zero and when t = Δt the external field is ~ B_0. Determine the induced emf E experienced by the electron. Write your
answer in terms of any or all of the following parameters: e, m_e,  R, N, B_0, ω_0, and μ_0 .

iv. Verify that the change in the kinetic energy of the electron satisfies ΔK = e E.
This justifies our assumption in (ii) that R does not change.

v. Determine the change in the total magnetic moment Δm for the N atoms when the
external field is applied, writing your answer in terms of e, m_e,  R, N, μ_0, and B_0 .

vi. Suppose that the uniform magnetic field used in the previous parts of this problem
is replaced with a bar magnet. Would the diamagnetic substance be attracted or repelled by the bar magnet? How does your answer show this?

Solution 4b : Diamagnetism

i. If half go one way and half go the other, M = 0.

ii. Additional force from magnetism,
F_B = qvB_0 = eRωB_0  (Q 4b-1)
modifies previous central force problem to give
m_eRω^2 = {e^2/(4π €_0 R^2)}
± eRω_0B_0,     (Q 4b-2)
where the positive sign corresponds to anticlockwise motion, the negative to clockwise motion.

A little math,
m_e R (ω^2 − ω_0^2) = ±eRωB_0, (Q 4b-3)

m_e (ω − ω0)(ω + ω0) = ±eωB_0,
(Q 4b-4)

m_e(∆ω)(2ω0) = ±eω_0B_0,
(Q 4b-5)

where in the last line we have used the approximation ω ≈ ω_0. Then
∆ω = ± (eB_0)/(2m_e) . (Q 4b-6)

iii. The emf is given by
E = n∆Φ/∆t = (∆n/∆t)Φ, (Q 4b-7)
but ∆n/∆t is a measure of the number of turns made by the electron in a time interval
∆t, so ∆n/∆t = ω_0R/2πR
= ω_0/2π . (Q 4b-8)

Then
E = (ω_0/2π) B_0πR^2
= (1/2) (ω_0 B_0 R^2) . (Q 4b-9)

iv. The change in kinetic energy is given by
∆K = ∆ {(1/2) (m_eω^2 R^2)} ,
(Q 4b -10)

= (m_e R^2 ω ∆ω),.  (Q 4b-11)
≈ (m_e R^2 ω_0 ∆ω), (Q 4b-12)
= (m_e ω_0 R^2) (± eB_0/2m_e)
, (Q 4b-13)
= eE.  (Q 4b-14)

v. ∆M = N δm, where N is the number of atoms, and ∆m the change in magnetic moment
in each. The change is
∆m = ∆ (eω_0R/2) . (Q 4b-15)

= (eR/2) ∆ω, (Q 4b-16)
= (e^2 R^2 B_0)/(4m_e) ,
(Q 4b-17)

so,
∆M = N (e^2R^2B_0)/(4m_e) .
(Q 4b-18)

vi. Repelled, by Lenz’s law.

Q 5. A group of 12 resistors is arranged along the edges of a cube as shown in the diagram below.
The vertices of the cube are labeled a- h.

a.  The resistance between each pair of vertices is as follows:
R_ab = R_ac = R_ae = 3.0 Ω
R_cg = R_ef = R_bd = 8.0 Ω
R_cd = R_bf = R_eg = 12.0 Ω
R_dh = R_fh = R_gh = 1.0 Ω
What is the equivalent resistance between points a and h?

b. The three 12.0 Ω resistors are replaced by identical capacitors. C_cd = C_bf = C_eg = 15.0 μF.
A 12.0 V battery is attached across points a and h and the circuit is allowed to
operate for a long period of time. What is the charge (Q_cd, Q_bf, Q_eg) on each capacitor
after this long period of time?

Solution 5 :
a. There is a high degree of symmetry present. Points b, c, and e are at the same potential;
similarly, points d, f, and g are at the same potential. The circuit then reduces to a series
connection of three parallel resistor clusters.
The three parallel clusters have effective resistances of 1 Ω, 8/5 Ω, and 1/3 Ω. The effective
resistance of the circuit is then 44/15 Ω.

b. After a long time no current will flow through the branches of the circuits containing capaci-
tors. The circuit then reduces to a parallel connection of three series resistor clusters.
The effective resistance of the circuit is 4 Ω, the current through the circuit is then
(12 V)/(4 Ω) = 3A.    (Q 5-1)

The three branches are identical; each then carries 1A.
The potential drop across each capacitor is the same as the potential drop across the 8 Ω
resistors, so
V_C = (1A)(8 Ω) = 8 V.  (Q 5-2)
Finally, the charge on each capacitor is
Q = (15 µF)(8 V) = 120 µC.

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These were the questions asked in the exam. As always you can see, they are pretty simple compared to the problems I have practiced so there wasn't much to use brain. As always, I completed my paper earlier than time allotted still I checked my answers twice for any error or omissions.

After that I exited the exam hall and went back to my hotel room along with my father. Had light meal and rested. Now there is only one exam ie. Chemistry is remained. So far there has no obstacles/difficulties. But I have to set my other important plans in motion.

Time awaits no one and I will not allow myself to take it for granted in this life. There is so much to do, so much to build, so much to achieve etc. I have potential, opportunity and resources are also being accumulated; if I can't achieve the highest ever possible, if I can't fullfill my ambitions with so many favourable things on my side; due to, be it anything say my laziness or someone else machinations or interference; it will be worse than pouring best wine in sewage.