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Changing Potential Energy of a Magnetic Coil

  • Thread starter clope023
  • Start date
  • #1
987
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Homework Statement



A coil with magnetic moment 1.40 Am^2 is oriented initially with its magnetic moment antiparallel to a uniform magnetic field of magnitude 0.830 T.

What is the change in potential energy of the coil when it is rotated 180 degrees, so that its magnetic moment is parallel to the field?

Homework Equations



u = IA = magnetic moment

U = -uBcos(phi)

The Attempt at a Solution



U1 = -uBcos(180) = -1.162J
U2 = -uBcos(360) = 1.162J

deltaU = U2-U1 = 1.162J - (-1.162J) = 2.324J

just wondering if I did the problem correctly, I'm not sure I have the correct angles; any help is greatly appreciated.
 

Answers and Replies

  • #2
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It's just [tex]\Delta U = \mu B[/tex] since [tex]U=-\mu B cos(\phi)=-\mu B cos(180) =-\mu B*(-1)=\mu B[/tex]
 
  • #3
987
123
It's just [tex]\Delta U = \mu B[/tex] since [tex]U=-\mu B cos(\phi)=-\mu B cos(180) =-\mu B*(-1)=\mu B[/tex]
I actually tried that before and it came out incorrect.

edit: which is strange since reviewing my concepts again showed that the antiparallel (perpendicular) potential energy should have been 0 so more than likely it should have turned out like you said.

would it be possible that the trig function as changed?
 
Last edited:
  • #4
987
123
it turns out I was correct (according to masteringphysics in this case the first antiparallel angle was 180 degrees) but the sign was negative, turns out it was like this:

U = Uf-Ui

= (-1.4*.830*cos(360))-(-1.4*.830*cos(180)) = -2.324J

thanks for the help anyway dude.
 

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