The Problem of the Dangling Magnet
Here’s a variation on a physics problem we discussed in my first-ever Skills in Physics Tutorial at the University of Sussex. I hadn’t realized that solutions were provided for Tutors so had to exercise my enfeebled brain in finding a solution. You’ll probably find it a lot easier…
A rectangular bar magnet hangs vertically from a pivot at one of its ends. When gently displaced the magnet undergoes small oscillations either side of the vertical with a period of one second. A horizontal magnetic field is then applied so that the equilibrium orientation of the magnet is 45° to the vertical. If the magnet is gently displaced from this new position, what is the new period of oscillation?
Comment: you do not need any further information about the size, shape or mass of the magnet in order to solve this problem.
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This entry was posted on February 20, 2013 at 6:21 pm and is filed under Cute Problems with tags Bar Magnet, cute problems, Physics, problems, tutorial. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.
In the Dark 
February 20, 2013 at 7:42 pm
It helps to remember that the energy of a dipole in a magnetic field: https://en.wikipedia.org/wiki/Magnetic_moment#Effects_of_an_external_magnetic_field_on_a_magnetic_moment
February 20, 2013 at 11:00 pm
0.84089641525371454303 seconds?
February 20, 2013 at 11:20 pm
I don’t think all your figures are significant…
February 21, 2013 at 11:06 am
Yes, it’s just the default precision of my ‘SpeedCrunch’ calculator :o)
But I would disagree with your statement. At first, we don’t know the precision of the measurement (it says mysteriously: ‘one second’) and second (another second this time, oh Lord), as far as I know there is still a bit of a discussion about the numbers of figures one should show for a single particular measurement, since for the error we have only an estimate…
February 21, 2013 at 11:58 am
“One” has only one significant figure
February 21, 2013 at 12:09 pm
With this poor resolution they would get a new period of 1 second again. But I don’t want to bother you anymore with this staff. Sorry, the answer should be .
February 21, 2013 at 12:12 pm
I meant.
February 21, 2013 at 5:27 pm
““One” has only one significant figure”
I would think that “one” can have any number of significant figures. It could mean 1 or 1.00 or 1.00000000000. If it only has one significant figure then my answer to the question is also 1(1sf), otherwise I agree with 2^{-1/4} 😉
February 21, 2013 at 5:42 pm
Indeed you’re probably right I think “1” and “one” both imply an exact integer whereas 1.0 would imply some rounding….
February 21, 2013 at 12:06 am
1.68 s
February 21, 2013 at 10:21 am
2^{-1/4}s ?
February 22, 2013 at 1:23 pm
Inverse fourth root of two?
February 23, 2013 at 3:09 pm
As the person who set this problem, I would accept the answer 2^{-1/4}s ~= 1s, but not 1s! I hope that you enjoyed it. Please if you have any similar problems, send them my way. Thinking up interesting but accessible problems is really hard.
Prof. P.
February 23, 2013 at 3:10 pm
You could have a shufty at my “Cute Problems” category….