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184_notes:examples:week14_changing_current_rectangle [2017/11/28 03:15] – [Changing Current Induces Voltage in Rectangular Loop] tallpaul | 184_notes:examples:week14_changing_current_rectangle [2017/11/28 15:34] – [Solution] tallpaul | ||
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===== Changing Current Induces Voltage in Rectangular Loop ===== | ===== Changing Current Induces Voltage in Rectangular Loop ===== | ||
- | 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 height | + | 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 |
- | + | ||
- | {{ 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==== | ||
+ | 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, | ||
+ | |||
+ | $$\Phi_B = \int \vec{B} \bullet \text{d}\vec{A} = \int B\text{d}A$$ | ||
+ | |||
+ | 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' | ||
+ | |||
+ | $$\int_{\text{rectangle}} B\text{d}A = \int_{\text{left to right}} \int_{\text{top to bottom}} B\text{d}x\text{d}y = \int_{x=d}^{x=d+w} \int_{y=0}^{y=h} \frac{\mu_0 I}{2 \pi x}\text{d}x\text{d}y$$ | ||
+ | |||
+ | In the first equality, we just changed the integral from over the " | ||
+ | |||
+ | {{ 184_notes: | ||
+ | |||
+ | 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. | 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. | ||