physics exam 3 review

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+[!]
+[=post-]
+<p>
+    Hello, everyone! Welcome back to Physics DBMUS, the second least updated part of the site. Really, even the AI section gets more content than this.
+    Today we're covering a <i>whole lot</i> of material: specifically, everything that'll be on my physics 2 exam 3. Hooray.
+    The previous post in the Physics section, <a href="[^baseurl]/posts/inductance-hell.html">Inductance Hell</a>, is a long and pretty-good Faraday problem that I recommend reading through.
+    The Faraday problems listed here are much more simple.
+</p>
+<h2>Magnetism 1</h2>
+<p>
+    As I wrote the above paragraph, someone on my Physics discussion board posted an actually-quite-interesting example problem from one of the old exams.
+    There's a system with three objects:
+</p>
+<ul>
+    <li>A positive charge `q_1 = Q` at the position `r_1 = \[-d, d, 0]` moving with velocity `v_1 = \[-v, 0, 0]`</li>
+    <li>A negative charge `q_2 = -Q` at the position `r_2 = \[0, 3d, 0]`, moving with veloicty `v_2 = \[0, -v, 0]`</li>
+    <li>A long straight wire parallel to the x-axis, defined by `y = -d`, with a current `I` in the positive x direction</li>
+</ul>
+<p>
+    We need to find the magnetic field at the origin produced by all of these objects. We know the Biot-Savart law for a moving point charge is `B = frac {u_0} {4pi} q frac {v times r} {r^3}`,
+    where `r` is the vector pointing <i>from</i> the charge <i>to</i> the observation point. We also know that the magnitude of the field from a wire is `B = frac {u_0} {2pi} frac I r`,
+    in the direction of `I times r` (right hand rule tells us this is just positive z, out of the page). Because superposition applies to magnetic fields, we just need to calculate
+    the field due to all of those different elements at the origin, and add them together. Oh joy of joys.
+</p>
+<ul>
+    <li>
+        For `q_1`, we have `B = frac {u_0} {4pi} Q frac {\[-v, 0, 0] times \[d, -d, 0]} {sqrt(2d^2)^3}`. That cross product is pretty easy, and works out to `\[0, 0, dv]`.
+        Multiplying in the constants (note: `frac {u_0} {4pi} = 1e-7`) gives us `B_1 = \[0, 0, frac {dvQ} {sqrt(8)d^3} * 10^{-7}]`. Why did I multiply `r_1` by `-1`? Because
+        we need `q_1` to the origin, not the origin to `q_1`.
+    </li>
+    <li>
+        For `q_2`, this problem is deceptively simple: because it's moving towards the origin, it doesn't affect the magnetic field at origin at all!
+    </li>
+    <li>
+        For the wire, the equation is really extremely easy: we have `\[0, 0, frac {u_0} {2pi} frac I d]` immediately.
+    </li>
+</ul>
+<p>
+    Adding these together gives us a net magnetic field `\[0, 0, frac {u_0} {2pi} frac I d + frac {dvQ} {sqrt(8)d^3} * 10^{-7}]`. Not too hard!
+</p>
+<h2>Magnetism 2</h2>
+<p>
+    This is a short one, but I thought it was conceptually interesting. We're given two current loops a short distance away from each other along the x-axis (the first one more left, the second more right).
+    The currents are flowing in opposite directions, and the direction of current in the segments of each wire closest to the other is down. Do they attract, or repel?
+</p>
+<p>
+    For this one we apply a quick Right Hand Rule: the direction of the B field at the first wire is going to be the direction of the curl of your fingers if your thumb is pointing
+    down along the second wire: `-hat z`. Then, the force on the first wire is `I times B`, to which you can also apply right-hand rule to determine that this wire is being pulled
+    towards the other wire. They attract.
+</p>
+<h2>Magnetism 3</h2>
+<p>
+    We're given a system with a long straight wire described by `y = R`, with a current `I` flowing in the positive x-direction. There is also a loop with radius `R` centered on the origin,
+    with a current `I` flowing through it clockwise. We need to find the magnetic field at the origin. 
+</p>
+<p>
+    This is really just a case of simple superposition. The magnetic field produced by the wire is `frac {u_0} {2pi} frac I R`, and the magnetic field produced by the
+    loop is `frac {u_0} {2} frac I R`. Right-hand rule tells us these fields point in the same direction (into the page), so we can simply add them together to get
+    `B_{"origin"} = \[0, 0, -frac {u_0} {2pi} frac I R - frac {u_0} {2} frac I R]`.
+</p>
+<h2>Magnetism 4</h2>
+<p>
+    I'm gonna wrap up the magnetism practice with a hard one. A magnetic dipole with moment `mu_d hat x` is at rest in the magnetic field produced by a bar magnet `d hat x` away with
+    magnetic dipole moment `mu_m hat x`. Both are aligned with the x-axis. The bar magnet is rotated until it faces in the opposite direction. How much work has been done on the dipole `mu_d`?
+</p>
+<p>
+    We know that the magnetic field at the location of `mu_d` is `frac {u_0} {2pi} frac {mu_m} {d^3} hat x`. The potential energy of a magnetic dipole is always `-mu cdot B`.
+    Initially, then, the potential energy of `mu_d` is `-frac {u_0} {2 pi d^3} mu_m mu_d`. When we've finished turning, the magnetic field is reversed, so the potential energy
+    becomes `frac {u_0} {2 pi d^3} mu_m mu_d`. Subtracting the initial state from the final gives us work done: `frac {u_0} {pi d^3} mu_m mu_d`. Not actually all that hard!
+</p>
+<h2>Faraday 1</h2>
+<p>
+    Let's do a problem on Faraday's law (induction). We're given a coil of wire with length `L`, `N` turns, and radius `r_s`. There's a current `I_0` running through it that drops linearly
+    in a given time `tau`. We need to find the induced voltage in a radius `r_w` loop around the solenoid.
+</p>
+<p>
+    Faraday's law simply states that `emf = - frac {d phi} {dt}`: the negative of the change in magnetic flux. To find magnetic flux, we need to find magnetic field.
+    There's a handy formula for the magnetic field anywhere inside a solenoid: `u_0 * I * frac N L`. The area that this field passes through is not `pi r_w^2`, because there's
+    no field outside the solenoid; the area is just the area of the solenoid cross-section, which is `pi r_s^2`. So we have the formula for flux: `phi = u_0 pi r_s^2 I frac N L`.
+    We also know that `I(t) = I_0 - I_0 frac t {tau}`. `frac {d phi} {dt} = u_0 pi r_s^2 frac N L frac {dI} {dt}`. `frac {dI} {dt}` is easy to read off: `-frac {I_0} {tau}`.
+    Plugging into Faraday's law gives us `emf = u_0 pi r_s^2 frac N L frac {I_0} {tau}`. Not too hard!
+</p>
+<h2>Faraday 2</h2>
+<p>
+    A classmate sent a most excellent application of Faraday's law in the discussion board: an electric fence. The setup is simple: there's a long transmission line stretched around the entire
+    fence, with an AC current given by the formula `I(t) = I_0 cos(wt)`, suspended a height `z` over the center of a rectangular loop of height `h` and length `L`. We need to find the induced voltage
+    around the loop. We'll do this twice: once assuming that `z >> h`, so the magnetic field at any point inside the rectangular fence is approximately the same, and once by correctly
+    integrating the varying magnetic field.
+</p>
+<h3>The Easy Way</h3>
+<p>
+    Step one is to find a formula for magnetic field with respect to time. Because this is a "long, straight wire", the magnetic field is `frac {u_0} {2pi} frac I z`.
+    We can then find flux: it's just the magnetic field (which we can pretend is uniform) times the area `Lh`. Sub'ing in, we get a formula for flux with respect
+    to time: `phi = frac {u_0} {2pi} frac {I_0 cos(wt)} z Lh`. We already know that `emf = -frac {d phi} {dt}`, so we take the derivative to get
+    `emf = -frac {d phi} {dt} = frac {u_0} {2pi} frac {I_0 wsin(wt)} z Lh` (where'd the negative go? `frac {d} {dx} cos(x) = -sin(x)`). Pretty easy.
+</p>
+<h3>The Correct Way</h3>
+<p>
+    To do this right, we need to find `phi(t)` as a surface integral. Fortunately, the field <i>is</i> horizontally uniform, so the magnetic flux in any single horizontal line comprising
+    the surface is `phi_h(y) = frac {u_0} {2pi} frac {I_0 cos(wt)} {z + y} L`. We need to integrate this in terms of `y`. Because `z` starts at the center of the loop,
+    we're integrating from `-frac h 2` to `frac h 2`: `int_{-frac h 2}^{frac h 2} frac {u_0} {2pi} frac {I_0 cos(wt)} {z + y} L dy`. Because most of this is constant, we can u-sub
+    pretty easily to get `|_{-frac h 2}^{frac h 2} frac {u_0} {2pi} I_0 cos(wt) ln abs {z + y} L`. That got messy fast. Our final result:
+    `frac {u_0} {2pi} I_0 cos(wt) L ln(frac {abs {z + frac h 2}} {abs {z - frac h 2}})`
+</p>
+[/]
+[=title "Physics Exam 3 Review"]
+[=author "Tyler Clarke"]
+[=date "2025-4-11"]
+[=subject "Physics"]
+[#post.html]