Waves, Electricity problem-solving question

A specialized "plasma-acoustic" sensor consists of a thin cylindrical tube of length $L = 0.850\text{ m}$ and cross-sectional area $A = 1.20 \times 10^{-4}\text{ m}^2$. The tube is filled with a partially ionized gas where the number density of free electrons is $n_e = 4.50 \times 10^{16}\text{ m}^{-3}$. A constant potential difference $V = 12.0\text{ V}$ is maintained across the ends of the tube. A longitudinal sound wave of frequency $f = 440\text{ Hz}$ is launched into the gas, creating a standing wave with a node at each end of the tube. The local density of the gas $\rho(x, t)$ fluctuates, causing the local number density of electrons $n(x, t)$ to vary proportionally: $n(x, t) = n_e [1 + \epsilon \sin(kx) \cos(\omega t)]$, where $\epsilon = 0.020$ is the amplitude of the density perturbation, $k$ is the wavenumber, and $\omega$ is the angular frequency. (a) Determine the harmonic number $m$ of the standing wave and calculate the speed of sound $c_s$ in this gas. [3] (b) Show that the total resistance $R(t)$ of the gas column can be expressed in the form $R(t) = \frac{R_0}{1 + \beta \cos(\omega t)}$ and find an expression for the constant $\beta$ in terms of $\epsilon$. You may assume the electron mobility $\mu$ (where drift velocity $v = \mu E$) is constant throughout the gas. [5] (c) Using the approximation $(1+x)^{-1} \approx 1 - x + x^2$ for small $x$, determine the time-averaged current $\langle I \rangle$ flowing through the tube. The electron mobility is $\mu = 0.450\text{ m}^2\text{V}^{-1}\text{s}^{-1}$. [5] (d) The fluctuating current $I(t)$ induces a magnetic field. Calculate the maximum magnetic flux density $B_{max}$ at a radial distance $r = 5.00\text{ mm}$ from the center of the tube at the moment the current is at its peak. [4] (e) If the temperature of the gas increases, the speed of sound $c_s$ increases. Qualitatively describe how the frequency of the $m$-th harmonic and the time-averaged current would change, assuming the number density $n_e$ remains constant. [3]

(a) For a tube of length $L$ with nodes at both ends, the wavelengths are $\lambda_m = \frac{2L}{m}$. The speed of sound is $c_s = f \lambda_m$. Given $L = 0.850\text{ m}$ and $f = 440\text{ Hz}$. For the fundamental ($m=1$), $\lambda = 1.70\text{ m}$ and $c_s = 440 \times 1.70 = 748\text{ m/s}$. Assuming the gas is a typical light gas (like heated Helium or a plasma mix), $m=1$ is the most physically reasonable assumption for a sensor unless specified. $c_s = 748\text{ m s}^{-1}$. (b) The resistance of a differential element $dx$ is $dR = \frac{\rho_{res} dx}{A}$, where resistivity $\rho_{res} = \frac{1}{\sigma} = \frac{1}{n(x,t) e \mu}$. $dR = \frac{dx}{n_e [1 + \epsilon \sin(kx) \cos(\omega t)] e \mu A}$. Total resistance $R(t) = \int_0^L \frac{dx}{n_e e \mu A [1 + \epsilon \sin(kx) \cos(\omega t)]}$. Let $R_0 = \frac{L}{n_e e \mu A}$. Since $\epsilon \ll 1$, use $(1+u)^{-1} \approx 1 - u$: $R(t) \approx \frac{1}{n_e e \mu A} \int_0^L [1 - \epsilon \sin(kx) \cos(\omega t)] dx$. $R(t) = \frac{1}{n_e e \mu A} [L - \epsilon \cos(\omega t) \int_0^L \sin(\frac{m\pi x}{L}) dx]$. The integral $\int_0^L \sin(\frac{m\pi x}{L}) dx = \frac{L}{m\pi} [-\cos(m\pi) + \cos(0)]$. For $m=1$, this is $\frac{2L}{\pi}$. $R(t) = \frac{L}{n_e e \mu A} [1 - \frac{2\epsilon}{\pi} \cos(\omega t)]$. Comparing to $R(t) = \frac{R_0}{1 + \beta \cos(\omega t)} \approx R_0(1 - \beta \cos(\omega t))$, we find: $\beta = \frac{2\epsilon}{\pi}$. (c) $I(t) = \frac{V}{R(t)} = \frac{V}{R_0} (1 + \beta \cos(\omega t))$. However, the question asks for the second-order effect using $(1+x)^{-1} \approx 1 - x +...

(a) [3 marks] - 1 mark for identifying $\lambda = 2L/m$. - 1 mark for $m=1$ (fundamental). - 1 mark for $c_s \approx 748\text{ m/s}$. (b) [5 marks] - 1 mark for $dR = \rho dx / A$. - 1 mark for substituting $n(x,t)$ into the resistivity expression. - 1 mark for setting up the integral for $R(t)$. - 1 mark for performing the integral of $\sin(kx)$. - 1 mark for correctly identifying $\beta = 2\epsilon/\pi$. (c) [5 marks] - 1 mark for using the Taylor expansion for the integrand. - 1 mark for correctly averaging $\cos(\omega t)$ and $\cos^2(\omega t)$. - 1 mark for calculating $R_0$ correctly ($2.19\text{ M}\Omega$). - 1 mark for calculating $I_{base}$ ($5.49\ \mu\text{A}$). - 1 mark for the final value of $\langle I \rangle$ (must be close to $5.49\ \mu\text{A}$ due to small $\epsilon$). (d) [4 marks] - 1 mark for $I_{max}$ formula (base current + fluctuation). - 1 mark for numerical $I_{max} \approx 5.56\ \mu\text{A}$. - 1 mark for $B = \mu_0 I / 2\pi r$. - 1 mark for final $B...

Hint 1 (mild nudge): For part (a), remember that a tube with nodes at both ends behaves like a string fixed at both ends. For part (b), the total resistance is the sum (integral) of the resistances of infinitesimal slices of the gas. Hint 2 (partial direction): In part (b), the number density $n$ is inside the denominator of the resistivity formula. Use the binomial expansion $(1+x)^{-1} \approx 1-x$ before integrating to simplify the expression. Hint 3 (key insight): For part (c), the time-average of a $\cos(\omega t)$ term is zero, but the time-average of $\cos^2(\omega t)$ is $1/2$. This means the "vibrating" gas has a slightly different average resistance than the "still" gas.