Primarygun Posted November 5, 2005 Posted November 5, 2005 In my figure, the whole circuit is being pulled to the right. AD experiences no change in magnetic field. Why do we apply the Fleming's right hand rule on the loop AD instead of BC? And, if the whole circuit is moving in the uniform magnetic field, if the shape of the circuit is like the above one, there's no induced current. Is this also true for a circular one? Thanks for kind attention to my thread.
Tom Mattson Posted November 5, 2005 Posted November 5, 2005 Why do we apply the Fleming's right hand rule on the loop AD instead of BC? I don't get it. AD and BC are both in the loop. If you apply the right hand rule to the loop' date=' then you are applying it to every segment, BC included. And, if the whole circuit is moving in the uniform magnetic field, if the shape of the circuit is like the above one, there's no induced current. Is this also true for a circular one? It's not true for either shape. You do get an induced current in both cases because the magnetic flux through the loop varies with time.
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