On Wed, 25 Jun 2008 04:34:22 -0700 (PDT), Peter <Poakfield@[EMAIL PROTECTED]
>
wrote:
>On Jun 25, 1:34 am, "Don Kelly" <d...@[EMAIL PROTECTED]
> wrote:
>> .
>> ----------------------------"Peter" <Poakfi...@[EMAIL PROTECTED]
> wrote in
message
>>
>>
news:c52fae08-023f-4eaa-bf82-1df097e63ae2@[EMAIL PROTECTED]
>> On 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 d...@[EMAIL PROTECTED]
>> remove the X to answer- Hide quoted text -
>>
>> - Show quoted text -
>
>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
Lorentz force as an explanation of the dynamics of a motor or
generator. Lorentz force is no factor, if only because the force is
normal to the motion.
Secondly, q x v, charge times velocity is not a current; you would
need charge per unit length times velocity to make a current. And you
need to deal in currents.
Study up on Lentz' law and other elementary concepts; the Lorentz
force is in a separate exotic class.
John Polasek
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