Solve the problem

(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 $(1+x)^{-1} \approx 1 - x + x^2$. $R(t) = R_0(1 - \beta \cos(\omega t) + \gamma \cos^2(\omega t) ...)$ is not quite right; we must expand $I = V/R$. $I(t) = \frac{V}{R_0} [1 + \frac{2\epsilon}{\pi} \cos(\omega t) + (\text{higher order terms})]$. Wait, the resistance itself was an integral. Let's re-evaluate $I(t) = V / [\int dR]$. $R(t) = \frac{R_0}{L} \int_0^L \frac{dx}{1 + \epsilon \sin(kx) \cos(\omega t)}$. Using $(1+u)^{-1} \approx 1 - u + u^2$: $R(t) \approx \frac{R_0}{L} \int_0^L [1 - \epsilon \sin(kx) \cos(\omega t) + \epsilon^2 \sin^2(kx) \cos^2(\omega t)] dx$. $R(t) = R_0 [1 - \frac{2\epsilon}{\pi} \cos(\omega t) + \frac{\epsilon^2}{2} \cos^2(\omega t)]$. Now, $I(t) = \frac{V}{R(t)} \approx \frac{V}{R_0} [1 - (-\frac{2\epsilon}{\pi} \cos(\omega t) + \frac{\epsilon^2}{2} \cos^2(\omega t)) + (-\frac{2\epsilon}{\pi} \cos(\omega t))^2]$. $I(t) = \frac{V}{R_0} [1 + \frac{2\epsilon}{\pi} \cos(\omega t) - \frac{\epsilon^2}{2} \cos^2(\omega t) + \frac{4\epsilon^2}{\pi^2} \cos^2(\omega t)]$. Time average $\langle \cos(\omega t) \rangle = 0$ and $\langle \cos^2(\omega t) \rangle = 1/2$. $\langle I \rangle = \frac{V}{R_0} [1 - \frac{\epsilon^2}{4} + \frac{2\epsilon^2}{\pi^2}]$. $R_0 = \frac{0.850}{(4.50 \times 10^{16})(1.6 \times 10^{-19})(0.450)(1.20 \times 10^{-4})} = \frac{0.850}{3.888 \times 10^{-7}} = 2.186 \times 10^6\ \Omega$. $I_{base} = V/R_0 = 12.0 / 2.186 \times 10^6 = 5.489 \times 10^{-6}\text{ A}$. Correction factor: $1 + \epsilon^2(\frac{2}{\pi^2} - \frac{1}{4}) = 1 + (0.02)^2(0.2026 - 0.25) = 1 - 0.0000189$. $\langle I \rangle \approx 5.49 \times 10^{-6}\text{ A}$ (The shift is very small, but the derivation is the key). (d) $I_{max} \approx I_{base}(1 + \frac{2\epsilon}{\pi}) = 5.489 \mu\text{A} \times (1 + \frac{0.04}{\pi}) = 5.489 \times 1.0127 = 5.559 \mu\text{A}$. Using Ampere's Law: $B = \frac{\mu_0 I}{2\pi r}$. $B_{max} = \frac{4\pi \times 10^{-7} \times 5.559 \times 10^{-6}}{2\pi \times 0.00500} = \frac{2 \times 10^{-7} \times 5.559 \times 10^{-6}}{0.00500} = 2.22 \times 10^{-10}\text{ T}$. (e) 1. Frequency: Since $c_s = f \lambda$ and $\lambda$ is fixed by $L$, if $c_s$ increases, $f$ must increase ($f \propto \sqrt{T}$). 2. Current: Mobility $\mu$ usually decreases with temperature in ionized gases (due to increased collision frequency). If $\mu$ decreases, $R_0$ increases, so $\langle I \rangle$ decreases.