11.md

title: Problem Set 11

(14.1)

{:.question} Do a Taylor expansion of equation (14.6) around V = 0.

Equation 14.6 states

E \approx 2 E_F - 2 E_C e^{-2/(N_F V)}

The Taylor expansion of e^x about zero is

e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}

so

E \approx 2 E_F - 2 E_C \sum_{n = 0}^\infty \frac{1}{n!} \left( \frac{-2}{N_F V} \right)^n

But this is an expansion around V = \infty.

If you really want an expansion about V = 0, note that

\frac{d}{dx} e^{x^{-1}} = -x^{-2} e^{x^{-1}}

As we keep taking higher derivatives we'll get more and more negative powers of x, but we'll never get rid of the e^{1/x}. So at x = 0 the latter term dominates, meaning the function and all its derivatives are zero. Thus the Taylor expansion about zero is identically zero.

Plugging this into equation 14.6 gives us the rather uninteresting

E \approx 2 E_F

(14.2)

{:.question} Problem 6.5 showed that for a Kibble balance the current I measured in the dynamic phase and the voltage V measured in the static phase are related to the mass m, gravitational constant g, and velocity v by IV = mgv. Using the inverse AC Josephson effect (equation 14.25) to determine the voltage, and the quantum Hall effect (equation 13.41) along with the inverse AC Josephson effect to determine the current, relate the measurement to fundamental constant(s).

In problem 6.5 in problem set 4 we found that

m = -\frac{I V}{g v}

The AC Josephson effect gives us a relation between voltage and frequency that only depends on fundamental constants (and n, a positive integer).

V = n \frac{h}{2 e} f

The quantum Hall effect can give us a resistance that only depends on fundamental constants (and i, a positive integer).

R_H = \frac{h}{i e^2}

By Ohm's Law IV = V^2/R, so the Kibble balance equation can be written as

mgv = -\frac{V^2}{R}

Plugging in the values above we find

4mgv = -h i n^2 f^2

(14.3)

{:.question} If a SQUID with an area of A = 1 cm^2 can detect 1 flux quantum, how far away can it sense the field from a wire carrying 1 A?

As found in problem 6.4 in problem set 4, the magnitude of the magnetic field a distance r away from an infinitely long and thin conductor carrying a current I is I/(2 \pi r). One flux quantum is \num{2.07e-7} \si{G \cdot cm^2} i.e. \num{2.07e-11} \si{T \cdot cm^2}. So to get one flux quantum over 1 \si{cm^2}, we need a magnetic field of \num{2.07e-11} \si{T}. Thus a one amp current can be detected at a distance of

\begin{align*} r &= \frac{\mu_o I}{2 \pi B} \\ &= \frac{\num{1.26e-6} \si{T m / A} \cdot 1 \si{A}}{2 \pi \cdot \num{2.07e-11} \si{T}} \\ &= \num{9.66e3} \si{m} \end{align*}

(14.4)

{:.question} Typical parameters for a quartz resonator are C_e = 5 \si{pF}, C_m = 20 \si{fF}, L_m = 3 \si{mH}, R_m = 6 \si{\ohm}. Plot, and explain, the dependence of the reactance (imaginary part of the impedance), resistance (real part), and the phase angle of the impedance on the frequency.

The circuit in question is depicted in figure 14.4. There's a capacitance C_e in parallel with a series RLC circuit (R_m, L_m, and C_m). To solve this it's helpful to know the complex impedances of the basic electrical components:

\begin{align*} Z_C &= \frac{1}{i \omega C} \\ Z_L &= i \omega L \\ Z_R &= R \end{align*}

Then we just combine these, using the sum for series connections, and the inverse of the sum of inverses for parallel connections. So the total impedance Z is

\begin{align*} \frac{1}{Z} &= \frac{1}{Z_{C_e}} + \frac{1}{Z_{C_m} + Z_{L_m} + Z_{R_m}} \\ &= i \omega C_e + \frac{1}{1/(i \omega C_m) + i \omega L_m + R_m} \\ &= i \omega C_e + \frac{\omega C_m}{-i + \omega C_m (R_m + i \omega L_m)} \\ Z &= \frac{1}{i \omega C_e + \frac{\omega C_m}{\omega C_m (R_m + i \omega L_m) - i}} \end{align*}

Let's see what this looks like with the given values plugged in. All the x axes are in radians per second.

The resistance (real part) is very small except for a spike near \num{4e6} \si{rad/s}.