Differences

This shows you the differences between two versions of the page.

--- 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 15:39] – [Solution] tallpaul
@@ Line 23: / Line 23: @@
 {{ 184_notes:14_wire_rectangle.png?300 |Wire and Rectangle}}
 ====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 (so $I_{enc} = 0$). We show the drawn loop below, split into two cases on the radius of the loop.
+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, we will also define for the area vector to be into the page. Since they both point in the same direction, the dot product simplifies:
-{{ 184_notes:14_capacitor_loops.png?600 |Circular Loops}}
+$$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A} = \int B\text{d}A$$
-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, due to the increasing electric field into the page. This comes from an extension of Lenz's Law, upon which discussion is not needed for this course.
+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's area. In order to see this a little more clearly, we will break the integral down a little and also insert the expression for the magnetic field from a long wire:
-{{ 184_notes:14_capacitor_b_field_loops.png?600 |Circular Loops, with B-field shown}}
+$$\int_{\text{rectangle}} B\text{d}A = \int_{\text{left to right}} \int_{\text{top to bottom}} B\text{d}x\text{d}y = \int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y$$
-We are pretty well set up to simplify our calculation of the integral in the representations, since the B-field is parallel to the loop's perimeter. Below, we show the integral calculation, where the magnetic field at a radius $r$ is displayed as $B(r)$.
+In the first equality, we just changed the integral from over the "rectangle" to be over its two dimensions, "left to right", and "top to bottom". We also changed $\text{d}A$ to be in terms of its two dimensions, $\text{d}A = \text{d}x\text{d}y$. A visual is shown below for clarity.
-$$\int \vec{B} \bullet \text{d}\vec{l} = \int B(r) \text{d}l = B(r) \int \text{d}l = 2\pi r B(r)$$
+{{ 184_notes:14_da_dx_dy.png?600 |Picture of dA Breakdown}}
-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.
+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:
-$$\Phi_\text{E, in} = EA = \frac{Q/A_{\text{plate}}}{\epsilon_0}A_{\text{loop}} = \frac{Q}{\epsilon_0\pi R^2}\pi r^2 = \frac{Qr^2}{\epsilon_0 R^2}$$
+\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} \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*}
-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 in the magnetic flux expression is constant in time, //except// for to current, $I$. Here is the induced voltage:
-$$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}r^2}{\epsilon_0 R^2} = \frac{Ir^2}{\epsilon_0 R^2} \text{, inside, } r<R$$
+$$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{\text{d}I}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{I_0}{t_0}$$
-$$\frac{\text{d}\Phi_E}{\text{d}t} = \frac{\frac{\text{d}Q}{\text{d}t}}{\epsilon_0} = \frac{I}{\epsilon_0} \text{, outside, } r>R$$
-We can now connect the pieces together (remember, $I_{enc}=0$, so we omit it below). We can write:
+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 //counterclockwise//. See below for a visual.
-$$2\pi r B(r) = \int \vec{B}\bullet \text{d}\vec{l} = \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t} = \mu_0 \frac{Ir^2}{R^2} \text{, inside, } r<R$$
+{{ 184_notes:14_rectangle_induced_current.png?600 |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.
-\[
-B(r) = \begin{cases}
-          \frac{\mu_0 I r}{2\pi R^2} &&& r<R \\
-          \frac{\mu_0 I}{2\pi r} &&& r>R
-       \end{cases}
-\]
-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:
-$$B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$$