Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision Next revisionBoth sides next revision | ||
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 [2021/07/13 13:18] – schram45 | ||
---|---|---|---|
Line 1: | Line 1: | ||
+ | [[184_notes: | ||
+ | |||
===== 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 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? | 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? | ||
Line 12: | Line 14: | ||
===Approximations & Assumptions=== | ===Approximations & Assumptions=== | ||
- | * The long wire is infinitely long and thin and straight. | + | * The long wire is infinitely long and thin and straight: With these assumptions the magnetic field produced by the wire only depends on the radial distance away from the wire and the current in it. This also allows us to use a simplified magnetic field equation from the notes. |
- | * There are no external contributions to the B-field. | + | * There are no external contributions to the B-field: We are not told about any other external moving charges or currents that could also produce a magnetic field that would effect the flux through our loop, so we will assume the only contribution is from the long wire. |
===Representations=== | ===Representations=== | ||
Line 21: | Line 23: | ||
* We represent the situation with the following visual. We arbitrarily choose a direction for the current. | * We represent the situation with the following visual. We arbitrarily choose a direction for the current. | ||
- | {{ 184_notes: | + | [{{ 184_notes: |
====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, |
- | + | ||
- | {{ 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 be the case! We also know that the field is directed counterclockwise, | + | |
- | + | ||
- | {{ 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}^{h} \text{d}y \\ | ||
+ | &= \frac{\mu_0 I}{2 \pi} \left[\log x \right]_{d}^{d+w} \left[ y \right]_{0}^{h} \\ | ||
+ | &= \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*} | ||
- | \[ | + | At this point, we are equipped to find the induced voltage. Notice that everything in the magnetic flux expression is constant with respect to time, //except for the current $I$//. The induced voltage can then be found by taking the time derivative |
- | B(r) = \begin{cases} | + | |
- | \frac{\mu_0 I r}{2\pi R^2} &&& | + | |
- | \frac{\mu_0 I}{2\pi r} &&& | + | |
- | | + | |
- | \] | + | |
- | 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: |
- | $$B_{\text{max}} = \frac{\mu_0 I}{2\pi R} = \frac{4\pi \cdot 10^{-7} \text{Tm/A} \cdot 3\text{ A}}{2\pi \cdot 10 \text{ m}} = 60 \text{ nT}$$ | + |