diff --git a/_psets/4.md b/_psets/4.md index cfcbc72879a3389f5a75d145378be1be948a0b09..273dc79142e090c612b169cfab497bc446f8ced9 100644 --- a/_psets/4.md +++ b/_psets/4.md @@ -60,9 +60,9 @@ to a box that is bisected by the plane, Gauss' Law tells us that $$2 A E = \sigm so the field strength $$E$$ is $$\sigma / (2 \epsilon_0)$$. Thus between two planes with opposite charge densities $$\sigma$$ and $$-\sigma$$, the field is -$$\sigma / \epsilon_0$$ everywhere. (And outside the planes the field is zero.) This means the -potential difference between the planes is $$d \sigma / \epsilon_0$$. As such the capacitance is $$Q -/ V = A \sigma \epsilon_0 / (d \sigma) = A \epsilon_0 / d$$. +$$\sigma / \epsilon_0$$ everywhere. (Outside the field is zero, since the fields from the two planes +cancel.) This means the potential difference between the planes is $$d \sigma / \epsilon_0$$. As +such the capacitance is $$Q / V = A \sigma \epsilon_0 / (d \sigma) = A \epsilon_0 / d$$. ### (b) @@ -123,11 +123,11 @@ $$ \begin{align*} A &= \frac{C d}{\epsilon_0} \\ &= \frac{\num{7.2e3} \si{F} \cdot 10^{-6} \si{m}}{\num{8.85e-12} \si{F/m}} \\ -&= \num{0.81e9} \si{m^2} +&= \num{8e8} \si{m^2} \end{align*} $$ -This is one quarter of the area of Rhode Island. We'd need $$\num{0.81e11}$$ squares with edge +This is one quarter of the area of Rhode Island. We'd need $$\num{8e10}$$ squares with edge length 10cm to get this area. If the plates have negligible thickness (relative to their spacing), they'd be 81 kilometers high when stacked on top of each other. @@ -140,15 +140,14 @@ they'd be 81 kilometers high when stacked on top of each other. Use Stokes’ Law to find the magnetic field of an infinite solenoid carrying a current I with n turns/meter. -Let the solenoid lie along x axis, with current flowing according to the right hand rule (about the -x axis). Consider an axis aligned square in the xy (or xz) plane, where the edges aligned with the x -axis are one meter long, with one fully inside the solenoid one fully outside. By Ampère's -Law, the integral of the magnetic field along the boundary of this square is nI. This is independent -of the length of the y (or z) aligned edges. So the field must be constant inside and out. Very far -away from the solenoid the field should be zero, since the fields from the opposing sides cancel. -Thus the field outside the solenoid is zero everywhere, and the field inside the solenoid is nI -everywhere (pointing along the x axis). So assuming a vacuum inside the solenoid, we have $$H = nI$$ -and $$B = \mu_0 n I$$. +Let the solenoid lie along the x axis, with right handed current. Consider an axis aligned square in +the xy (or xz) plane, where the edges aligned with the x axis are one meter long, with one fully +inside the solenoid one fully outside. By Ampère's Law, the integral of the magnetic field +along the boundary of this square is nI. This is independent of the length of the y (or z) aligned +edges. So the field must be constant inside and out. Very far away from the solenoid the field +should be zero, since the fields from the opposing sides cancel. Thus the field outside the solenoid +is zero everywhere, and the field inside the solenoid is nI everywhere (pointing along the x axis). +So assuming a vacuum inside the solenoid, we have $$H = nI$$ and $$B = \mu_0 n I$$. ### (b) @@ -175,12 +174,15 @@ $$ \frac{\partial}{\partial r} U &= \frac{\partial}{\partial r} \frac{1}{2} \frac{B^2}{\mu_0} \pi r^2 l \\ &= \frac{B^2}{\mu_0} \pi r l \\ -&= \frac{100 \si{T^2}}{4 \pi \times 10^{-7} \si{H/m}} \cdot \pi \cdot 1 \si{m} \cdot 2 \si{m} \\ -&= \num{50e7} \si{N} +&= \frac{100 \si{T^2}}{4 \pi \times 10^{-7} \si{H/m}} \cdot \pi \cdot 0.5 \si{m} \cdot 2 \si{m} \\ +&= \num{2.5e8} \si{N} \end{align*} $$ -Yes, there is a sign error that I'm ignoring for now. +Yes, there is a sign error that I'm ignoring for now. The force should be the negative of the +potential gradient. I think if we explicitly calculated the potential of the solenoid loop by loop +(i.e. constructing it starting with all the loops infinitely far apart), we'd find that the +potential is -1 times what we found above, thus fixing the problem. ## (6.4) @@ -196,12 +198,12 @@ equal to $$\num{2e-7}$$ newton per metre of length." {:.question} Show that that current at that distance produces that force. -First let's find the magnetic field of an infinitely long straight conductor. Considering the -Biot-Savart Law and the symmetry of this problem, the magnetic field must have no component parallel -to the wire, must always be perpendicular to the radial separation between the test point and the -wire, and must have a constant magnitude at each radius. Consider then a circle of radius $$r$$ -centered on the wire. Ampère's Law tells us that the magnitude of the field on this circle is -$$I / (2 \pi r)$$. +First let's find the magnetic field of an infinitely long straight conductor. Let's use a +cylindrical coordinate system along this axis. Considering the Biot-Savart Law and the symmetry of +this problem, the magnetic field must be oriented along $$\hat{\mathrm{d} \theta}$$, with a +magnitude that depends only on $$r$$. Consider then a circle of radius $$r$$ centered on the wire. +Ampère's Law tells us that the magnitude of the field at any point on this circle is $$I / (2 +\pi r)$$. The differential force exerted by this field on a differential piece of current is $$dF = I (dl \times B)$$. In this case the direction of the current and the magnetic field are perpendicular, so @@ -219,8 +221,8 @@ $$ Note that we don't need to multiply this by two to account for the force wire two exerts on wire one. Newton's [third law](https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton's_third_law) guarantees that -this is equal and opposite, so it only makes sense to talk about the force exerted by one on the -other. Maxwell's equations respect this (otherwise it would be easy to build a perpetual motion +this is equal and opposite, so it only makes sense to talk about the force as a pair. Naturally +Maxwell's equations respect this law (otherwise one could build an electromagnetic perpetual motion machine). ### (b) @@ -229,8 +231,10 @@ machine). What is the problem with defining the ampere this way? It's not a very practical experiment. No wires are infinitely long, so you'll necessarily have -fringe fields to factor into your results. No conductors are infinitely thin. You have to hold the -wires and supply current to them somehow, so they can't be totally surrounded by a vacuum. +fringe fields to factor into your results. No conductors are infinitely thin, so their response to +an applied field depends on their specific geometry. And you have to hold the wires and supply +current to them somehow, so they can't be totally surrounded by a vacuum. Not to mention we can't +create perfect vacuums anyway. ## (6.5) @@ -270,8 +274,8 @@ I dl_x \left( B_{0x} + \frac{\partial B_y}{\partial x} x + \frac{\partial B_y}{\ - I dl_y \left( B_{0y} + \frac{\partial B_x}{\partial x} x + \frac{\partial B_x}{\partial y} y \right) $$ -Let's move to radial coordinates to make the integration easier. Let's say our loop has radius -$$r$$. For a point on our loop at angular position $$\theta$$, +Let's move to radial coordinates to make the inevitable integration easier. Let's say our loop has +radius $$r$$. For a point on our loop at angular position $$\theta$$, $$ dl = (-r \mathrm{d} \theta \sin{\theta}, r \mathrm{d} \theta \cos{\theta}, 0) @@ -286,12 +290,12 @@ $$ \frac{\partial B_x}{\partial y} \sin{\theta} \right) $$ -We need to integrate this over $$\theta$$ from 0 to $$2 \pi$$. Though the expression above looks -complicated, the integral ends up being easy. To spare all the algebra, any term that only involves -$$\cos{\theta}$$ or $$\sin{\theta}$$ integrates to zero since we're integrating over one complete -period. The same happens for both terms that have $$\cos{\theta} \sin{\theta}$$. So there are only -two terms left, and they reduce to $$\cos^2{\theta}$$ and $$\sin^2{\theta}$$, both of which -integrate to $$\pi$$. So in the end the $$z$$ component of the force on the loop is +We need to integrate this with respect to $$\theta$$ from 0 to $$2 \pi$$. Though the expression +above looks complicated, the integral ends up being easy. To spare all the algebra, any term that +only involves $$\cos{\theta}$$ or $$\sin{\theta}$$ integrates to zero since we're integrating over +one complete period. The same happens for both terms that have $$\cos{\theta} \sin{\theta}$$. So +there are only two terms left, and they reduce to $$\cos^2{\theta}$$ and $$\sin^2{\theta}$$, both of +which integrate to $$\pi$$. So in the end the $$z$$ component of the force on the loop is $$ -I \pi r^2 \left( \frac{\partial B_y}{\partial y} + \frac{\partial B_x}{\partial x} \right) @@ -306,7 +310,9 @@ $$ mg = I \pi r^2 \frac{\partial B_z}{\partial z} $$ -where $$g = 9.8 \si{m/s^2}$$ is the local acceleration due to gravity. +where $$g = 9.8 \si{m/s^2}$$ is the acceleration due to gravity. (To get a really precise +measurement of $$m$$, you'd want to measure $$g$$ at your lab specifically since it varies a bit +with local geography and geology.) ### (b) @@ -352,16 +358,19 @@ $$ m = -\frac{I V}{g v} $$ -If we use a coil with $$n$$ windings instead of a single loop then we just need to multiply by -$$n$$. +If we use a coil with $$n$$ windings instead of a single loop then we just need to replace $$I$$ +with $$n I$$. ### (a), (b), and (c) without the assumption of linearity -Let's not assume that the magnetic field varies linearly with distance. Instead, let's assume that -the field is radially symmetric. This is a reasonable assumption since if the field isn't at least -periodically radially symmetric then there will be a torque on the loop. Call the radial component -of the field experienced by our loop $$B_r(z)$$. (There must also be some z component, but we won't -end up needing it so we won't give it a name.) +Now let's drop the assumption that the magnetic field varies linearly with distance. After all, +though this assumption is reasonable over small distances, it's hard to justify over the whole +macroscopic length of our current loop. Instead, let's assume that the field is radially symmetric. +This is a reasonable assumption since if the field isn't at least periodically radially symmetric +then there will be a torque on the loop. + +Call the radial component of the field experienced by our loop $$B_r(z)$$. (There must also be some +z component, but we won't end up needing it so we won't give it a name.) Now calculating the force on the loop is easy. The z component of $$B$$ doesn't matter since it just pulls the loop outward. And the radial component produces a force along z. So the force is $$-2 \pi @@ -369,7 +378,9 @@ r I B_r(z)$$. But how can we calculate the flux? Consider a z-aligned cylinder of radius $$r$$ and height $$z$$, with the center of its base at the origin. The magnetic flux through the bottom is some constant -amount $$\Phi_\text{bottom}$$. The flux through the walls is +amount $$\Phi_\text{bottom}$$. (We could arrange for this to be zero, either by placing our origin +very far away, or by using two magnets in a mirrored configuration, as is the case with NIST's +balance. But it doesn't matter for this analysis.) The flux through the walls is $$ \Phi_\text{walls}(z) = 2 \pi r \int_0^z B_r(z') \mathrm{d}z' @@ -412,11 +423,12 @@ Well the voltage is only generated when the coil is moving, but the current is d keeps the coil still. So you can't do both at the same time. You could start the coil moving and then measure the current that keeps it moving at a constant velocity, simultaneously measuring the resulting voltage. But then the voltage is a combination of the applied voltage and the induced -voltage, and ditto for the current. The field is much harder to analyze. +voltage, and ditto for the current. At the end of the day the point is to have the effects of the magnetic field cancel out, so it -doesn't need to be perfect as it did in the previous definition. In other words, we just want to -measure a voltage and a current and not worry about the specifics of the magnetic field. +doesn't need to be perfect as it did in the previous definition of the ampere. In other words, we +just want to measure a voltage and a current and not worry about the specifics of the magnetic +field. ## (6.6) @@ -428,7 +440,8 @@ Assume that sunlight has a power energy density of $$1 \si{kW/m^2}$$ (this is a typical average value in the continental USA is $$\approx 200 \si{W/m^2}$$). Estimate the electric field strength associated with this radiation. -Update: this is all wrong. +Update: this is wrong. I quickly jumped to using $$U$$ since I'd already used it in previous +problems. But I need to take into account the Poynting vector. The energy delivered per square meter in one second by sunlight is the field energy stored in a 1 meter by 1 meter by 1 light second volume. So the average energy density of sunlight must be