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184_notes:examples:week14_changing_current_rectangle [2017/11/28 14:39] – [Solution] 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==== | ||
In order to find the induced voltage, we will need the magnetic flux. This requires defining an area-vector and determining the magnetic field. We can use the right hand rule to determine the the magnetic field from the wire is into the page ($-\hat{z}$) near the rectangle. For convenience, | In order to find the induced voltage, we will need the magnetic flux. This requires defining an area-vector and determining the magnetic field. We can use the right hand rule to determine the the magnetic field from the wire is into the page ($-\hat{z}$) near the rectangle. For convenience, | ||
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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' | 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' | ||
- | $$\int_{\text{rectangle}} B\text{d}A = \int_{\text{left to right}} \int_{\text{top to bottom}} B\text{d}x\text{d}y$$ | + | $$\int_{\text{rectangle}} B\text{d}A = \int_{\text{left to right}} \int_{\text{top to bottom}} B\text{d}x\text{d}y = \int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y$$ |
- | 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 | + | In the first equality, |
- | {{ 184_notes:14_capacitor_loops.png?600 |Circular Loops}} | + | {{ 184_notes:14_da_dx_dy.png?500 |Picture of dA Breakdown}} |
- | Below, we also draw the direction of the magnetic field along the loops. We know the magnetic field is directed along our circular loop -- if it pointed in or out a little bit, we may be able to conceive of the closed surface with magnetic flux through it, which would imply the existence of a magnetic monopole. This cannot be the case! We also know that the field is directed counterclockwise, | + | At this point, our integral is set up enough that we can crunch through |
- | {{ 184_notes: | + | \begin{align*} |
+ | \int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y &= \frac{\mu_0 I}{2 \pi} \int_{d}^{d+w} \frac{\text{d}x}{x} \int_{0}^{y} \text{d}y \\ | ||
+ | &= \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) | ||
+ | &= \frac{\mu_0 I h}{2 \pi} \log\left(\frac{d+w}{d}\right) | ||
+ | \end{align*} | ||
- | We are pretty well set up to simplify our calculation of the integral | + | At this point, we are equipped |
- | $$\int \vec{B} \bullet | + | $$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} |
- | Next, we need to find the changing electric | + | Notice that the induced voltage is negative. This means that the induced current produces a magnetic fields whose corresponding flux is negative |
- | $$\Phi_\text{E, | + | {{ 184_notes:14_rectangle_induced_current.png?500 |Induced Current}} |
- | $$\Phi_\text{E, | + | |
- | + | ||
- | 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 $I$. | + | |
- | + | ||
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0 R^2} = \frac{Ir^2}{\epsilon_0 R^2} \text{, inside, } r<R$$ | + | |
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0} = \frac{I}{\epsilon_0} \text{, outside, } r>R$$ | + | |
- | + | ||
- | We can now connect the pieces together (remember, $I_{enc}=0$, | + | |
- | + | ||
- | $$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 \frac{Ir^2}{R^2} \text{, inside, } r<R$$ | + | |
- | $$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 I \text{, outside, } r>R$$ | + | |
- | + | ||
- | 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: | + | |
- | $$B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$$ | + |