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Direction of Induced Current

  1. Nov 5, 2005 #1
    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.
    I'm in urgent so I have to post here.

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  3. Nov 5, 2005 #2

    Doc Al

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    Staff: Mentor

    OK. So the induced current in the circuit will be due to the motional EMF generated by moving a conductor through a magnetic field. The magnetic field exerts a force on the moving charges (in the wire); the direction of that force ([itex]\vec{F} = q\vec{v} \times \vec{B}[/itex]) is given by a right hand rule or by Fleming's rule, if you prefer.
    AD is moving through a magnetic field. (But the magnetic flux through the entire loop is changing. Another, more general, way to find the direction of the current is using Lenz's law.)
    Note: AD and BC are not loops, they are just sections of the loop. The loop is the complete circuit: A-B-C-D.

    Fleming's rule is for finding the direction of the induced EMF in a wire moving through a magnetic field. There's no magnetic field at BC.
    Right. The induced EMFs on AD and BC would cancel. Note that the change in magnetic flux through the loop would be zero; thus no induced current.
    Yes. The shape of the loop doesn't matter. If the flux changes, a current will be induced. (The direction of the current can be found using Lenz's law.)
    Last edited: Nov 5, 2005
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