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Rotating square coil in constant magnetic field

  • Thread starter snufkin
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  • #1
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Homework Statement



There is one square coil of area A that rotates with speed [tex]\omega[/tex] around axis y. The magnetic field is uniform and points to the z direction. Show that the torque is [tex]\frac{A^2B^2N^2\omega sin\omega t}{R}[/tex].

Homework Equations


[tex]T = |\underline{m} \times \underline{B}|[/tex]
[tex]\underline{m} = N I A[/tex]
Thus [tex]T = NIABcos(\omega t)[/tex]

Unsure about these:
[tex]EMF = NIR[/tex]
[tex]EMF = - N\frac{d \Phi}{d t}[/tex]

The Attempt at a Solution



Now I have read many similar questions on this forums, and I think I am close to fully understand the problem itself, but there is a piece missing, and I don't see from my initial equations where I am going wrong.

I calculate I via the EMF, substitute it into T. But with this I end up with one N, whereas the supposed solution that I need to prove in the exercise is [tex]T = \frac{A^2B^2\omega sin^2\omega t}{R}[/tex].

I am assuming one of the initial equations wrong, I suspect the EMF section, because I not understand it. Any help would be appreciated.
 
Last edited:

Answers and Replies

  • #2
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Thinking of V that is probably wrong, the number of loops has nothing to do with the potential, so [tex]V = IR[/tex].

So EMF is [tex]-N \cdot \frac{d \Phi}{d t}[/tex] because we take [tex]\Phi[/tex] at each loop, and just sum those up. I am wondering now why can we actually do that, since in a different alignment the flux in each loop must be different.
 
  • #3
Redbelly98
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The flux is the same for each loop, so yes EMF = -N dΦ/dt
 

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