/sci/ - Science & Math

magnetic field in electric motors does not provide any work, the electric current does.

you can't make a 100% efficient motor due to various losses.

bump

Magnetic forces do no work.

>>1602401
>Anyone who once had an industrial magnet in his hands knows they can provide a lot of power.

wat?

>this could be realized with the same logic, the only thing you have to do is put stronger or larger magnets in it

wat?

>>1602442
>>1602427
>>1602420

>>1602427

This is a fact.

>>1602494
this logic makes >>1602483 free energy ?

>>1602510
too bad the
>>1602483
won't work

>>1602483

It works, if anyone thinks it doesn't explain why you think so and I'll tell you why you're wrong.

>and high efficient alternators.
You're a moron. Alternators don't use permanent magnets. GTFO.

>>1602564

Wow, you're pretty stupid.

>>1602547
Guess I'll bite then, magnetic forces cannot alter the speed of an object, only the direction, so even if that thing started out with a speed, friction would make it come to a halt eventually.

>>1602576
Wow... no, that is not why it will not work. Equalizing of forces, try that direction.

>>1602576

>magnetic forces cannot alter the speed of an object

clearly you've never played with magnets before

>>1602597
Clearly you don't know how magnetic forces work.

>more then 100% effeciency
Science does not work like that.

>>1602613

Care to elabroate?

>>1602613

Care to elabroate?

>>1602627
Magnetic forces do not produce any work, so when for example you see a magnetic crane lift a car or something, it's not the magnetic force producing the work, but something else. This may seem illogical, but it's true.

>>1602576

Changing the direction of a moving object does require energy.
So the magnet delivers a force, it works.
/discussion

>>1602624
That picture was on the cover of my statistics textbook.

Ugh, I'll make a new thread later with an animation and everything for the dullards.

>>1602660
>Changing the direction of a moving object does require energy.
...which is provided by electric current, as previously mentioned

>>1602660
>Changing the direction of a moving object does require energy.
>Implying forces perpendicular to displacement do work.
>Implying energy is fed into the Earth-Moon system to keep the moon in orbit.

ITT: this

>>1602718
No no no, it's if they do any work or not, not how they work.

>>1602695
>>1602712

What backward ass country do you people come from?

>>1602728
Let me do the fucking math for you. Let <span class="math">W[/spoiler] denote the work done by the magnetic force. We know from the Lorentz force law that <span class="math">F_m = q\mathbf{v}\cross \mathbf{B}.[/spoiler]. We also define the work done by a force to be <span class="math">\int F\cdot d \mathbf{s}[/spoiler] where the integral is taken along the path <span class="math">\mathbf{s}[/spoiler] taken by the charged particle. Writing it out in full:
<span class="math">W = \int q\mathbf{v}\cross \mathbf{B} \cdot d \mathbf{s}[/spoiler]
But we know from the chain rule that <span class="math">d\mathbf{s} = \frac{d\mathbf{s}{dt}dt = \mathbf{v} dt [/spoiler].
So we rewrite the integral as
<span class="math">W = \int q\mathbf{v}\cross \mathbf{B} \cdot \mathbf{v} dt[/spoiler].
Now what is a property of the cross product of v and B? It is perpendicular to both of them. What is the scalar product of a vector and something that is perpendicular to it? By definition of perpendicular, ZERO. Hence the integrand is 0 for all values of t and thus the integral is identically 0.

tl;dr Magnets do no work.

>>1602401
The energy for a motor comes from the electricity that powers it. And while you can lift large objects with magnets, that energy comes from a reduction in the potential energy of the system. Separating the object from the magnet will require the energy be put back in again. Physics 101.

>>1602427
Magnetic fields can't do work on a plain charge, but can do work on particles with an intrinsic magnetic moment.

>>1602660
I'm going to hope you're trolling.

>>1602728
Let me do the fucking math for you. Let <span class="math">W[/spoiler] denote the work done by the magnetic force. We know from the Lorentz force law that <span class="math">F_m = q\mathbf{v}\times \mathbf{B}.[/spoiler]. We also define the work done by a force to be <span class="math">\int F\cdot d \mathbf{s}[/spoiler] where the integral is taken along the path <span class="math">\mathbf{s}[/spoiler] taken by the charged particle. Writing it out in full:
<span class="math">W = \int q\mathbf{v}\times \mathbf{B} \cdot d \mathbf{s}[/spoiler]
But we know from the chain rule that <span class="math">d\mathbf{s} = \frac{d\mathbf{s}}{dt}dt = \mathbf{v} dt [/spoiler].
So we rewrite the integral as
<span class="math">W = \int q\mathbf{v}\times \mathbf{B} \cdot \mathbf{v} dt[/spoiler].
Now what is a property of the cross product of v and B? It is perpendicular to both of them. What is the scalar product of a vector and something that is perpendicular to it? By definition of perpendicular, ZERO. Hence the integrand is 0 for all values of t and thus the integral is identically 0.

tl;dr Magnets do no work.

Also this forum needs a preview for LaTeX.

>>1602763
I use this:
http://userscripts.org/scripts/show/81801

>>1602813
Oh. That might actually make me install greasemonkey.