deadlyboringmath.us/site/posts/physics-exam-3-review.html

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<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>
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[=title "Physics Exam 3 Review"]
[=author "Tyler Clarke"]
[=date "2025-4-11"]
[=subject "Physics"]
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