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184_notes:examples:week14_changing_current_rectangle [2017/11/28 14:32] – [Changing Current Induces Voltage in Rectangular Loop] tallpaul | 184_notes:examples:week14_changing_current_rectangle [2017/11/28 16:16] – tallpaul | ||
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* We represent the situation with the following visual. We arbitrarily choose a direction for the current. | * We represent the situation with the following visual. We arbitrarily choose a direction for the current. | ||
- | {{ 184_notes: | + | {{ 184_notes: |
====Solution==== | ====Solution==== | ||
- | We wish to find the magnetic field in the plane we've shown in the representations. Due to the circular symmetry of the problem, we choose a circular loop in which to situate our integral $\int \vec{B}\bullet\text{d}\vec{l}$. We also choose for the loop to be the perimeter of a flat surface, so that the entire thing lies in the plane of interest, and there is no enclosed current | + | In order to find the induced voltage, |
- | {{ 184_notes: | + | $$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A} = \int B\text{d}A$$ |
- | Below, we also draw the direction | + | Usually, we would pull the $B$ out of the integral, but we cannot do that in this case! That is because $B$ varies for different points in the rectangle' |
- | {{ 184_notes: | + | $$\int_{\text{rectangle}} |
- | We are pretty well set up to simplify our calculation of the integral | + | In the first equality, we just changed |
- | $$\int \vec{B} \bullet \text{d}\vec{l} = \int B(r) \text{d}l = B(r) \int \text{d}l = 2\pi r B(r)$$ | + | {{ 184_notes: |
- | Next, we need to find the changing electric flux in our loop. Since our loop was described with a flat surface, and the electric field is directed parallel to the area-vector of the loop, we can write electric flux as $\Phi_E = \vec{E} \bullet \vec{A} = EA$. This formula will need to be split up for parts of the surface inside | + | At this point, our integral |
- | $$\Phi_\text{E, in} = EA = \frac{Q/A_{\text{plate}}}{\epsilon_0}A_{\text{loop}} = \frac{Q}{\epsilon_0\pi R^2}\pi r^2 = \frac{Qr^2}{\epsilon_0 R^2}$$ | + | \begin{align*} |
- | $$\Phi_\text{E, out} = EA = E_\text{in}A_\text{in} + E_\text{out}A_\text{out} = \frac{Q/A_{\text{plate}}}{\epsilon_0}A_{\text{plate}} + 0 = \frac{Q}{\epsilon_0\pi R^2}\pi R^2 = \frac{Q}{\epsilon_0}$$ | + | \int_{x=d}^{x=d+w} |
+ | &= \frac{\mu_0 I}{2 \pi} \left[\log x \right]_{d}^{d+w} \left[ y \right]_{0}^{y} \\ | ||
+ | &= \frac{\mu_0 I}{2 \pi} \left(\log(d+w) - \log(d)\right) \left( h-0 \right) \\ | ||
+ | &= \frac{\mu_0 I h}{2 \pi} \log\left(\frac{d+w}{d}\right) | ||
+ | \end{align*} | ||
- | Now, if we wish the find the change in flux, we will take a time derivative. Notice that all the terms in the flux expressions above are constant, except for $Q$, which is changing with time as dictated by $I$. | + | At this point, we are equipped to find the induced voltage. Notice that everything |
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0 R^2} = \frac{Ir^2}{\epsilon_0 R^2} \text{, inside, } r<R$$ | + | $$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{\text{d}I}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{I_0}{t_0}$$ |
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0} = \frac{I}{\epsilon_0} \text{, outside, | + | |
- | We can now connect | + | Notice that the induced voltage is negative. This means that the induced current produces a magnetic fields whose corresponding flux is negative flux. Since the area-vector was defined as into the page ($-\hat{z}$), this means that the magnetic field produced by the induced current should be out of the page ($+\hat{z}$). By the right hand rule, this means the induced currents is directed // |
- | $$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 \frac{Ir^2}{R^2} \text{, inside, } r<R$$ | + | {{ 184_notes:14_rectangle_induced_current.png?500 |Induced Current}} |
- | $$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 I \text{, outside, } r>R$$ | + | |
- | + | ||
- | We are ready to write out the magnetic field. | + | |
- | + | ||
- | \[ | + | |
- | B(r) = \begin{cases} | + | |
- | \frac{\mu_0 I r}{2\pi R^2} &&& | + | |
- | \frac{\mu_0 I}{2\pi r} &&& | + | |
- | | + | |
- | \] | + | |
- | + | ||
- | Notice, the distance between the plates has no effect on the magnetic field calculation. Also, the amount of the charge on the plates at a given time does not matter -- we only care about how fast the charge is changing (the current!). Also, it is interesting that outside the plates, the magnetic field is the same as it would be for a long wire. This would be just as if the capacitor were not there, and the wire were connected. Below, we show a graph of the magnetic field strength as a function of the distance from the center of the capacitor. | + | |
- | + | ||
- | {{ 184_notes:14_capacitor_b_field_graph.png?400 |B-Field Strength, Graphed}} | + | |
- | + | ||
- | We have enough information to find the maximum B-field, which is at the edge of the plates: | + | |
- | $$B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$$ | + |