On Jun 25, 1:34 am, "Don Kelly" <d... at (no spam) shaw.ca> wrote:
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----------------------------"Peter" <Poakfi... at (no spam) msn.com> wrote in message
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On Jun 24, 1:33 am, "Don Kelly" <d... at (no spam) shaw.ca> wrote:
----------------------------"Peter" <Poakfi... at (no spam) msn.com> wrote in message
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Hi! Can someone please help me? I understand the force in an electric
generator is F = B (q x v). Is the work done on the generator equal to
W = B(q x v) (vt), where vt is the distance the electrons in the wire
move because of the rotation of the rotor? If the angular velocity of
the rotor is doubled, does the work done on the generator increase
four times, because it would then be W = B (q x 2v) (2vt); thus the
energy produced by the generator also increases four times? Thank.
Please excuse me if I am saying dumb things. I may be missing
something. I am trying to learn. Thanks.
No. Note that force is q(v x B) not B (q x v) There is a difference. The x
is not a multiplication but a vector cross product. B and v are vectors
while q is a scalar (-ve for electrons).
It is easier to consider the case of a conductor of length l which is at
right angles to the field and to the motion- then this is simplified.
Then there is a force acting on the conductor of F=Bli where i is the
current (qv) and acting in a direction to oppose the motion - at right
angles to the current.
You can start from the same basis or use Faraday to determine the voltage
V=Blv
The power of the generator will be Vi which can be expressed as Fv and the
work in a given time will be the integral of this.
If you double the velocity, you will double the voltage but the current
will depend on what is connected to the terminals of the generator. If
open or short circuited, the work will be 0.
In between with some load there will be some power.
In the case of a fixed resistance load, doubling the speed, doubles the
voltage and so quadruples the power (and energy in a given time). You are
right with respect to that although your approach is not correct.
Don Kelly d... at (no spam) shawcross.ca
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Thank you. That helped a lot. Then, is it correct if I say the
instantaneous work done is W = q(v x B)(vt), where vt is the distance
and t is the time? (This is in analogy to W = force x distance.)
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No. Note that the force is perpendicular to the velocity -not in the
direction of the velocity-not co-linear as in the mechanical analogy.
Don Kelly d... at (no spam) shawcross.ca
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Thanks. Then, what would be the work done in terms of B, q, v, and t?
Peter, you would do yourself a favor if you didn't try to coerce the
