..
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"Peter" <Poakfield@[EMAIL PROTECTED]
> wrote in message
news:c52fae08-023f-4eaa-bf82-1df097e63ae2@[EMAIL PROTECTED]
Jun 24, 1:33 am, "Don Kelly" <d...@[EMAIL PROTECTED]
> wrote:
> ----------------------------"Peter" <Poakfi...@[EMAIL PROTECTED]
> wrote in message
>
>
news:346cd631-7159-497e-a7aa-834fda352c9b@[EMAIL PROTECTED]
>
> > 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...@[EMAIL PROTECTED]
> remove the X to answer
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.)
----------------------------------------
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 dhky@[EMAIL PROTECTED]
the X to answer
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