From 0bcddda73448c4cb03983f3e5d67c4afed2d968d Mon Sep 17 00:00:00 2001 From: Tyler Clarke Date: Fri, 7 Mar 2025 14:45:31 -0500 Subject: [PATCH] another dbmus post --- compass-angle-hell.html | 345 ++++++++++++++++++++++++++++++++++++++++ index.html | 4 + 2 files changed, 349 insertions(+) create mode 100644 compass-angle-hell.html diff --git a/compass-angle-hell.html b/compass-angle-hell.html new file mode 100644 index 0000000..2456908 --- /dev/null +++ b/compass-angle-hell.html @@ -0,0 +1,345 @@ + + + + + + + +- no title specified + + + + + + + + + + + + + + + + + + + + + +

Compass Angle Hell

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by Tyler Clarke

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Well, je suis retourné! The exam went well, and although I’m not sure what I got exactly, I think it was probably good. One thing I noticed was an uncomfortable amount of trigonometry: I had to derive polar coordinate forms for a translated circle, among other things. Fortunately, trig was always a strong suit.

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My other classes seem to be picking up on the trig craze too, which is in fact why I’m here: the very first question on my homework set for this week in Physics (yeah, I know I’m doing it pretty late, shoot me) requires finding the magnetic dipole of a bar magnet… knowing nothing but the local strength of the earth’s magnetic field, and the angle of a compass affected by both.

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The problem is not too hard from a trigonometry perspective, but there are a few stumbling blocks. Rather than trust my geometric intuition, I drew this handy lil’ diagram:

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We know that the magnetic field of the Earth is 3e-5 Tesla at the observation point (the origin), the bar magnet is at + + + + [ + + + + − + 0.189 + + , + 0 + , + 0 + + + ] + + , and the angle between the total field and the magnet is 13.4deg. To find the magnetic dipole, we’ll work backwards from magnetic field and distance. The equation relating the values is + + + + tan + + + ( + + θ + + ) + + = + + + b + earth + + + b + mag + + + + +  (one of the basic trig relationships). Because we know + + + + b + earth + + , and we know + + + θ + , this can be algebra’d to get + + + + + b + mag + + = + + + b + earth + + + tan + + ( + + θ + + ) + + + + = + + 3e-5 + + tan + + ( + + 13.4 + + ) + + + + + . This evaluates “nicely” to 1.2592681e-4. Great.
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Unfortunately, this isn’t the final answer: we need the magnitude of the magnetic dipole moment, not the magnetic field. There’s a handy fact about magnets that becomes useful here: for a distance + + + + r + ≫ + s + +  (where s is the length of the magnet), the magnetic field on the parallel axis is just + + + + + k + m + + + + 2 + u + + + r + 3 + + + + , where u is the magnitude of the magnetic dipole moment and r is the distance (which, handily, we know). This works out to an equation + + + + + 1.2592681e-4 + = + + k + m + + + + + 2 + u + + + r + 3 + + + → + + u + = + + 1.2592681e-4 + + 2 + + k + m + + + + + + r + 3 + + + . + + + + k + m + +  is just the magnetic constant + + + + + + u + 0 + + + 4 + π + + + = + 1e-7 + + , and r is -0.189 (see above), so the magnitude of the dipole moment must be + + + + + u + = + + + ( + + + − + 0.189 + + + ) + + 3 + + + + + 1.2592681e-4 + 2e-7 + + = + + − + 4.25082884311 + + + + . Yay for horrible trigonometry!
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