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184_notes:examples:week14_changing_current_rectangle [2017/11/28 01:11] – created tallpaul | 184_notes:examples:week14_changing_current_rectangle [2017/11/28 15:39] – [Solution] tallpaul | ||
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===== Changing Current Induces Voltage in Rectangular Loop ===== | ===== Changing Current Induces Voltage in Rectangular Loop ===== | ||
- | Suppose you have a parallel plate capacitor that is charging with a current | + | Suppose you have an increasing current through |
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- | {{ 184_notes: | + | |
===Facts=== | ===Facts=== | ||
- | * The capacitor | + | * The current in the long wire increases with time and is $I(t) = I_0 \frac{t}{t_0}$. |
- | * $R=10 \text{ m}$ | + | * The rectangle has dimensions |
- | * $d=1 \text{ cm}$ | + | * The rectangle and the wire lie in the same plane, and are separated by a distance |
- | * The capacitor is charging with a current | + | |
===Lacking=== | ===Lacking=== | ||
- | * A description | + | * $V_{ind}$. |
+ | * Direction | ||
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * We are only concerned about a snapshot in time, so the current | + | * The long wire is infinitely long and thin and straight. |
- | * The electric field between | + | * There are no external contributions to the B-field. |
- | * The electric field outside the plates is zero. | + | |
===Representations=== | ===Representations=== | ||
- | * We represent the electric | + | * We represent the magnetic |
- | * We can represent | + | * We represent magnetic |
- | $$\int \vec{B}\bullet \text{d}\vec{l} = \mu_0 I_{enc} + \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t}$$ | + | * We can represent induced voltage as $$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t}$$ |
- | * We represent the situation with the following visual: | + | * We represent the situation with the following visual. We arbitrarily choose a direction for the current. |
- | {{ 184_notes:14_capacitor_side_view.png?300 |Plane in which we wish to find B-field}} | + | {{ 184_notes:14_wire_rectangle.png?300 |Wire and Rectangle}} |
====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, |
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- | {{ 184_notes: | + | |
- | + | ||
- | 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 | + | |
- | + | ||
- | {{ 184_notes: | + | |
- | + | ||
- | We are pretty well set up to simplify our calculation of the integral | + | |
- | + | ||
- | $$\int \vec{B} \bullet \text{d}\vec{l} = \int B(r) \text{d}l = B(r) \int \text{d}l = 2\pi r B(r)$$ | + | |
- | 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 the plates versus outside, since the electric field is different. | + | $$\Phi_B = \int \vec{B} \bullet |
- | $$\Phi_\text{E, in} = EA = \frac{Q/ | + | Usually, we would pull the $B$ out of the integral, but we cannot do that in this case! That is because |
- | $$\Phi_\text{E, out} = EA = E_\text{in}A_\text{in} + E_\text{out}A_\text{out} = \frac{Q/ | + | |
- | Now, if we wish the find the change in flux, we will take a time derivative. Notice that all the terms in the flux expressions above are constant, except for $Q$, which is changing with time as dictated by $I$. | + | $$\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 |
- | $$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0 R^2} = \frac{Ir^2}{\epsilon_0 R^2} \text{, inside, } r<R$$ | + | In the first equality, we just changed the integral from over the " |
- | $$\frac{\text{d}\Phi_E}{\text{d}t} | + | |
- | We can now connect the pieces together (remember, $I_{enc}=0$, so we omit it below). We can write: | + | {{ 184_notes:14_da_dx_dy.png? |
- | $$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$$ | + | At this point, our integral is set up enough that we can crunch through the analysis. We can pull out constants to the front and calculate to the end: |
- | $$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. | + | \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) - \log(d)\right) \left( h-0 \right) \\ | ||
+ | &= \frac{\mu_0 I h}{2 \pi} \log\left(\frac{d+w}{d}\right) | ||
+ | \end{align*} | ||
- | \[ | + | At this point, we are equipped to find the induced voltage. Notice that everything in the magnetic flux expression is constant in time, //except// for to current, $I$. Here is the induced voltage: |
- | B(r) = \begin{cases} | + | |
- | \frac{\mu_0 | + | |
- | \frac{\mu_0 I}{2\pi r} &&& | + | |
- | | + | |
- | \] | + | |
- | 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 | + | $$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{\text{d}I}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{I_0}{t_0}$$ |
- | {{ 184_notes: | + | 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 // |
- | We have enough information to find the maximum B-field, which is at the edge of the plates: | + | {{ 184_notes: |
- | $$B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$$ | + |