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--- 184_notes:examples:week14_ac_graph [2017/11/28 16:34] – created tallpaul
+++ 184_notes:examples:week14_ac_graph [2018/08/09 19:19] (current) – curdemma
@@ Line 1: / Line 1: @@
-===== Changing Current Induces Voltage in Rectangular Loop =====
+[[184_notes:ac|Return to Changing Flux from an Alternating Current notes]]
-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?
+===== Analyzing an Alternating Current Graph =====
+Suppose you are given the following graph of current over time. You can see that the first peak is at the point where $t=0.01\text{ s}$, and $I=0.3\text{ A}$. The graph is shown below. Find the amplitude, period, and frequency of the current, and give an equation that describes the alternating current.
+[{{ 184_notes:14_ac_graph_given.png?400 |Graph of Alternating Current}}]
 ===Facts===
-  * The current in the long wire increases with time and is $I(t) = I_0 \frac{t}{t_0}$.
+  * The first peak in the graph is at $(t=0.01\text{ s}, I=0.3\text{ A})$.
-  * The rectangle has dimensions $w$ by $h$, and a side with length $h$ is parallel to the wire.
-  * The rectangle and the wire lie in the same plane, and are separated by a distance $d$.
 ===Lacking===
-  * $V_{ind}$.
+  * Amplitude, period, frequency.
-  * Direction of $I_{ind}$.
+  * Equation for alternating current.
 ===Approximations & Assumptions===
-  * The long wire is infinitely long and thin and straight.
+  * The graph of the current is a sine wave.
-  * There are no external contributions to the B-field.
+  * As indicated on the graph, at $t=0$, we have $I=0$.
+  * The current is centered about $I=0$, that is, the sine wave has not been shifted vertically at all.
 ===Representations===
-  * 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 represent the graph as given in the example statement.
-  * We represent magnetic flux as $$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A}$$
-  * We can represent induced voltage as $$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t}$$
-  * We represent the situation with the following visual. We arbitrarily choose a direction for the current.
-{{ 184_notes:14_wire_rectangle.png?500 |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:
+Since the graph has not been shifted vertically, we can say that the peak we see is also the amplitude of the graph. So the amplitude of the alternating current is simply $0.3\text{ A}$.
-$$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A} = \int B\text{d}A$$
-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?500 |Picture of dA Breakdown}}
+The period of the graph is given by the time between peaks. One can see that the graph can be split into vertical slices as shown below.
-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:
+[{{ 184_notes:14_ac_graph_slices.png?400 |Vertical Time Slices}}]
-\begin{align*}
+It should be easy to see based on the visual above that the time between peaks is just four times the time value at the first peak. So then the period is $0.04\text{ s}$.
-\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:
+The frequency is just the reciprocal of the period:
-$$V_{ind} = -\frac{\text{d}\Phi}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{\text{d}I}{\text{d}t} = -\frac{\mu_0 h}{2 \pi} \log\left(\frac{d+w}{d}\right)\frac{I_0}{t_0}$$
+$$f = \frac{1}{\tau} = \frac{1}{0.04\text{ s}} = 25\text{ Hz}$$
-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.
+We are also now equipped to write a equation for the alternating current. We will use a sine function rather than a cosine function, since the current begins at $I=0$, and increases. The amplitude will provide the factor out front:
-{{ 184_notes:14_rectangle_induced_current.png?500 |Induced Current}}
+$$I=I_0\sin(2\pi \cdot f \cdot t) = (0.3\text{ A}) \cdot \sin(2\pi \cdot 25\text{ Hz} \cdot t)$$