Flux through rotating ring

  • Thread starter dikmikkel
  • Start date
  • #1
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Homework Statement


A ring can rotate about a horizontal axis(x), and a diameter placed on the x-axis. A uniform field is perpendicular to the ring -B0*y. The diameter of the ring is D. it spins with constant angular velocity ω around the x-axis. At at time t = 0 the ring is entirely in the xy plane

How do i find the flux change per time? And what does it mean that it can rotate around a diameter.
I've attached a figure.


Homework Equations


[itex] Flux = \int \vec{B}\cdot d\vec{a}[/itex]


The Attempt at a Solution


At a time t = 0 the flux through the loop is 0 and i tried to write a solution using that B is constant:
[itex] Flux = B\int sin(\theta (t))d\vec{a}[/itex]
[itex] Flux/dt = B\int cos(\theta(t) ) \omega d\vec{a}[/itex]
 

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Answers and Replies

  • #2
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Nwm.
I found out.
 

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