Differences

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

--- 184_notes:examples:week14_changing_current_rectangle [2017/11/28 01:11] – created tallpaul
+++ 184_notes:examples:week14_changing_current_rectangle [2017/11/28 15:34] – [Solution] tallpaul
@@ Line 1: / Line 1: @@
 ===== Changing Current Induces Voltage in Rectangular Loop =====
-Suppose you have a parallel plate capacitor that is charging with a current $I=3 \text{ A}$. The plates are circular, with radius $R=10 \text{ m}$ and a distance $d=1 \text{ cm}$. What is the magnetic field in the plane parallel to but in between the plates?
+Suppose you have an increasing current through a long wire, $I(t) = I_0 \frac{t}{t_0}$. Next to this wire, there is a rectangular loop of width $w$ and height $h$. The side of the rectangle is aligned parallel to the wire so that the rectangle is a distance $d$ from the wire, and they are both in the same plane. What is the induced voltage in the rectangle? In what direction is the induced current in the rectangle?
-{{ 184_notes:14_capacitor_picture.png?400 |Charging Capacitors}}
 ===Facts===
-  * The capacitor is a parallel plate capacitor with circular plates.
+  * 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 $w$ by $h$, and a side with length $h$ is parallel to the wire.
-  * $d=1 \text{ cm}$
+  * The rectangle and the wire lie in the same plane, and are separated by a distance $d$.
-  * The capacitor is charging with a current $I=3 \text{ A}$.
 ===Lacking===
-  * A description of the magnetic field.
+  * $V_{ind}$.
+  * Direction of $I_{ind}$.
 ===Approximations & Assumptions===
-  * We are only concerned about a snapshot in time, so the current is $I$, even though this may change at a later time as the capacitor charges.
+  * The long wire is infinitely long and thin and straight.
-  * The electric field between the plates is the same as the electric field between infinite plates.
+  * There are no external contributions to the B-field.
-  * The electric field outside the plates is zero.
 ===Representations===
-  * We represent the electric field in a parallel plate capacitor as $$\vec{E} = \frac{Q/A}{\epsilon_0} \hat{x}$$ where $Q$ is the charge on a plate, $A$ is the area of the plate, and $\hat{x}$ is directed from one plate to the other.
+  * We represent the magnetic field from a very long straight wire as $$B = \frac{\mu_0 I}{2 \pi r}$$ where direction is determined based on the right hand rule
-  * We can represent the magnetic field from a changing electric field as
+  * We represent magnetic flux as $$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A}$$
-$$\int \vec{B}\bullet \text{d}\vec{l} = \mu_0 I_{enc} + \mu_0\epsilon_0\frac{\text{d}\Phi_E}{\text{d}t}$$
+  * 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====
+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:
+$$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A} = \int B\text{d}A$$
+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:
+$$\int_{\text{rectangle}} B\text{d}A = \int_{\text{left to right}} \int_{\text{top to bottom}} B\text{d}x\text{d}y = \int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y$$
+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.
+{{ 184_notes:14_da_dx_dy.png?600 |Picture of dA Breakdown}}
+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:
+\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:
+$$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{\text{d}I}{\text{d}t} = -\frac{\mu_0 I h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{I_0}{t_0}$$
 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.