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    When reading my textbook I came across this example of a dynamo
    Name:  Dynamo image.png
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    Now according to the conservation of energy when the dynamo spins and transfers energy to the lamp work must be done to keep the magnet spinning (as it is losing energy).

    Now when thinking of it in terms of forces I seem to be picturing it wrong so I was wondering if anyone could point out the flaw in my reasoning.

    Here it goes
    The spinning dynamo on a wire with no lamp there will be a larger current induced and therefore a greater magnetic field created in the coil to oppose this change (lenz's law). But when the lamp is connected the current should be lower and therefore a weaker magnetic field and hence a weaker reaction force on the dynamo. This doesnt make sense in terms of the conservation of energy as it implies that more energy is needed for a wire with no lamp compared to a wire with a lamp connected.
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    (Original post by JakeRStudent)
    When reading my textbook I came across this example of a dynamo
    Name:  Dynamo image.png
Views: 14
Size:  12.8 KB

    Now according to the conservation of energy when the dynamo spins and transfers energy to the lamp work must be done to keep the magnet spinning (as it is losing energy).

    Now when thinking of it in terms of forces I seem to be picturing it wrong so I was wondering if anyone could point out the flaw in my reasoning.

    Here it goes
    The spinning dynamo on a wire with no lamp there will be a larger current induced and therefore a greater magnetic field created in the coil to oppose this change (lenz's law). But when the lamp is connected the current should be lower and therefore a weaker magnetic field and hence a weaker reaction force on the dynamo. This doesnt make sense in terms of the conservation of energy as it implies that more energy is needed for a wire with no lamp compared to a wire with a lamp connected.
    with no lamp connected there is no circuit and therefore no current. You would get a pd developed across the dynamo terminals but because P=IV there is no power output.
    practically you get some losses caused by eddy currents in the dynamo but the effort required to turn the dynamo is definitely less when the bulb is disconnected (as was commonly noticed by cyclists when dynamo lights were more common)
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    (Original post by Joinedup)
    with no lamp connected there is no circuit and therefore no current. You would get a pd developed across the dynamo terminals but because P=IV there is no power output.
    practically you get some losses caused by eddy currents in the dynamo but the effort required to turn the dynamo is definitely less when the bulb is disconnected (as was commonly noticed by cyclists when dynamo lights were more common)
    When I said a wire without a lamp I meant a complete circuit. So there is a current.
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    (Original post by JakeRStudent)
    When I said a wire without a lamp I meant a complete circuit. So there is a current.
    OK ten there would be a high current which in reality would cause heating, especially in the coils of the dynamo
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    (Original post by Joinedup)
    OK ten there would be a high current which in reality would cause heating, especially in the coils of the dynamo

    That doesn't really answer my question
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    (Original post by JakeRStudent)
    That doesn't really answer my question
    It is more difficult to turn the rotor with the dynamo terminals wired directly together than when driving a small load.
 
 
 
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