# Chapter 30 - Electromagnetic Induction - Exercises and Problems - Page 875: 81

$0.25$ m

#### Work Step by Step

From mechanics, we know the period ($T$) is $\displaystyle T = \frac{2\pi}{\omega}$ and that period is the inverse of frequency $\displaystyle T = \frac{1}{f}$ where $f$ is the frequency. $\displaystyle f = \frac{1}{T} = \frac{\omega}{2\pi}$ We also know that for $LC$ circuits, that is, circuits with only a single capacitor and inductor in them, that the angular frequency is $\omega = \sqrt{\displaystyle \frac{1}{LC}} = \displaystyle \frac{1}{\sqrt{LC}}$. $\displaystyle f =\frac{1/(\sqrt{LC})}{2\pi} = \frac{1}{2\pi\sqrt{LC}}$ However, it is more helpful to us if this equation is given to us in terms of inductance $L$ $\displaystyle 2\pi\sqrt{LC} = \frac{1}{f}$ $\displaystyle \sqrt{LC} = \frac{1}{2\pi f}$ $\displaystyle LC = \frac{1}{(2\pi f)^2} = \frac{1}{4\pi^2f^2}$ $\displaystyle L = \frac{1}{4\pi^2f^2C}$ Chapter 30.8 gives us the equation for the inductance of a solenoid as $L_{solenoid} = \displaystyle \frac{\mu_0N^2A}{l}$ where $l$ is the length of the solenoid, $N$ is the number of turns of wire around the solenoid, and $A$ is the cross-sectional area of the solenoid (note that this cross-sectional area does not include the diameter of the wire; it is merely the cross-sectional area of the cylindrical solenoid). This means $A_{solenoid} = \displaystyle \pi(\frac{d_{solenoid}}{2})^2$ $\displaystyle L = \frac{\mu_0N^2A}{l} = \frac{1}{4\pi^2f^2C}$ Isolating for $l$ we get $l = 4\pi^2f^2C\mu_0N^2A$ We cannot get the correct answer from this equation though: we do not know the correct number of turns of the wire around the cylinder. You may think "sure we do, the question says $2$ layers....." but then immediately after that you will think "oh wait, it says $2$ $layers$ of wire, not $2$ turns of wire. So, how do we find $N$? A bit of conceptualization is required. In an example that is not related to this question, imagine a cylinder that is 1 meter long and has a wire that has a 0.1 meter (10 cm) long diameter. We can conceptualize to see that we can wrap the wire around the cylinder 10 times. Thus, the relationship is $\displaystyle N = \frac{l}{d_{wire}}$ where $N$ is the number of turns, $d_{wire}$ is the diameter of the wire, and $l$ is the length of the solenoid. Now we can plug this equation in to find the length of the solenoid. Don't forget that there are 2 layers of wire, so for this question $N = 2\displaystyle \frac{l}{d_{wire}}$ $\displaystyle l = 4\pi^2f^2C\mu_0N^2A = 4\pi^2f^2C\mu_0(\frac{2l}{d_{wire}})^2A$ $\displaystyle \frac{d_{wire}^2}{16\pi^2f^2C\mu_0A} = \displaystyle \frac{d_{wire}^2}{16\pi^2f^2C\mu_0(\pi(\frac{d_{solenoid}}{2})^2)} = l$ $l = \displaystyle \frac{d_{wire}^2}{16\pi^2f^2C\mu_0(\pi(\frac{d_{solenoid}}{2})^2)} = \frac{(0.25\cdot10^{-3})^2}{16\pi^2(1.0)^2(1.0)\mu_0(\pi(\frac{4\cdot10^{-2}}{2})^2)} = 0.250634$ m With two significant figures in the question, our final answer is $l_{solenoid} = 0.25$ m

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