On Jun 25, 1:34=A0am, "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
messag=
e
>
>
>news:346cd631-7159-497e-a7aa-834fda352c9b@[EMAIL PROTECTED]
>
> > > Hi! Can someone please help me? I understand the force in an
electric
> > > generator is F =3D B (q x v). Is the work done on the generator
equal=
to
> > > W =3D B(q x v) (vt), where vt is the distance the electrons in the
wi=
re
> > > 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 =3D 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.
Th=
e 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
a=
t
> > right angles to the field and to the motion- then this is simplified.
> > Then there is a force acting on the conductor of F=3DBli 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
volta=
ge
> > V=3DBlv
> > 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
a=
re
> > 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 =3D q(v x B)(vt), where vt is the distance
> and t is the time? (This is in analogy to W =3D 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?
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