Interaction between magnet and wire (Faraday Motor)
Hugo Caballero
I would like to ask you for a further explanation concerning the
interaction of the fields depicted in the following applet:
http://www.magnet.fsu.edu/education/tutorials/java/faradaymotor/index.html
Faraday's experiment, as shown here, consisted of a free hanging wire,
through which a current was forced to flow, and a fixed magnet. The
picture shows the different fields both of the magnet and the wire.
I understand the different patterns of the fields but I find it a bit
difficult to visualise their interaction.
I suppose that the free hanging wire, the field of which decays with
r^2, moves in the direction of the lower concentration of field lines,
i. e., far away from the magnet. But why does this happen? How do these
fields interact? Is it possible to have perpendicular field lines? Does
not the system rearranges to avoid this situation?
Could you please go deeper into the explanation of the resulting field
lines pattern? I suppose that the angle between the wire and the axis of
the magnet depends on the value of the current flowing through the wire,
which determines the intensity of the magnetic field around the wire.
But which is the equation for the force acting on the wire? When is the
steady state reached concerning the angle between the magnet and the
wire? Since the magnetic field is not constant and homogeneous along the
whole wire I find it quite difficult to calculate this problem.
Thank you very much in advance!
glen herrmannsfeldt
> I would like to ask you for a further explanation concerning the
> interaction of the fields depicted in the following applet:
> http://www.magnet.fsu.edu/education/tutorials/java/faradaymotor/index.html
> Faraday's experiment, as shown here, consisted of a free hanging wire,
> through which a current was forced to flow, and a fixed magnet. The
> picture shows the different fields both of the magnet and the wire.
> I understand the different patterns of the fields but I find it a bit
> difficult to visualise their interaction.
> I suppose that the free hanging wire, the field of which decays with
> r^2, moves in the direction of the lower concentration of field lines,
> i. e., far away from the magnet. But why does this happen? How do these
> fields interact? Is it possible to have perpendicular field lines? Does
> not the system rearranges to avoid this situation?
Well, one way is with q v cross B, but if you are following
the fields, then the magnetic field energy density is proportional
to the square of the field.
What do you mean by "perpendicular field lines?" The B field
is a vector field, and adds like any vector.
(snip)
-- glen
Salmon Egg
Hugo Caballero <hugoca...@coit.es> wrote:
You have some of the ideas right but not the details. For example, a
magnet has a dipole field. That means that the field dies away faster
than as r^-2. At great distances, it is more like r^-3. It would also be
best for you to forget about LINES and stick to field intensities.
Current flow through the wire increases the field on one side of the
wire and increases it on the other. Then your argument can be replaced
with a similar one using field intensities rather than line
concentration.
Bill
--
An old man would be better off never having been born.
Szczepan Bialek
"Hugo Caballero" <hugoca...@coit.es> wrote
news:i3chef$7hc$1...@speranza.aioe.org...
In Electrodynamics are the two methods: the field method and the charge
method.
If you want to go deeper you should know the both:
http://www.df.lth.se/~snorkelf/Longitudinal/node4.html
To analise Faraday's experiment it is better substitute the magnet with the
solenoid.
S*
Hugo Caballero
First of all, thank you for the answer. If I try to apply Lorentz force
(q v cross b) I get into trouble. If you look at the field generated by
the magnet, most of the lines have their main component parallel to the
wire (after its decomposition in a suitable reference system, for
example, taking the wire as one dimension), and, therefore, parallel to
the velocity of the charges. Because of this, this component does not
exert any net force. The other component is perpendicular to the wire
and produces a force which is tangential to the trajectory of the wire
shown in the applet. However, the field generated by the magnet cuts the
wire in two different directions: the field lines next to the north pole
of the magnet cut the wire "outwards". However, the lines entering the
south pole cut the wire "inwards". Since the velocity of the charges has
not changed its direction, the force exerted on the wire should be
different in the upper and lower part of the wire (?).
What I meant by perpendicular lines is that the field lines generated by
the wire lie almost on a plane which is perpendicular to any plane
described by the magnet bar (from north-pole to south-pole). I say
almost because there is a certain angle between the wire and the magnet.
But as you have said, this should not bother me, because it is a vector
field and the resulting field will just be the sum of both vector fields.
Thanks again!
Hugo Caballero
You are right with the dipole. I was just referring to the wire.
The only problem I have with the idea of forgetting about lines and
concentrating on intensities is that the direction of the field does
matter.
For example, the intensity around a south pole of a magnet will be equal
to the intesity around the north pole of an equal magnet. Nevertheless,
if I put a thir magnet's nord pole next to each of the poles described
before, the behaviour will be very different. The lines concentration in
the case south-north pole will produce an attraction. However, in the
case north-north pole a repulsion will take place.
Thanks!
Hugo Caballero
> In Electrodynamics are the two methods: the field method and the charge
> method.
> If you want to go deeper you should know the both:
> http://www.df.lth.se/~snorkelf/Longitudinal/node4.html
>
> To analise Faraday's experiment it is better substitute the magnet with the
> solenoid.
> S*
>
>
Thanks for the answer! I will examine it in detail!
Szczepan Bialek
"Hugo Caballero" <hugoca...@coit.es> wrote
news:i3f53h$juv$1...@speranza.aioe.org...
> Hi Glen!
>
> First of all, thank you for the answer. If I try to apply Lorentz force (q
> v cross b) I get into trouble.
Lorentz force is a simplification for the students. It is a special case of
the "angular force law" by Ampere.
In the Faraday's experiment you must use the full equation.
There work the two forces:
The longitudinal force makes the circuit longer (the wire becomes not
vertical = longer).
The angular force try to align the wire with the wires of solenoid (magnet =
solenoid).
The longitudinal force is also omitted in teaching program (to simplicity).
I hope that you know that the electrons (current elements) repel each other
in a wire and that would complicate teaching program
For example in Faraday's paradox you must take into account the mass of
electrons.
S*
Don Kelly
"Hugo Caballero" <hugoca...@coit.es> wrote in message
news:i3f53h$juv$1...@speranza.aioe.org...
magnet. Balance a wire on the top terminal (+) and bend it to contact the
magnet completing the circuit. A bit of trial and error to get this balance
and continuous contact at the bottom but it eliminates the free mercury.
Have fun
--
Don Kelly
cross out to reply
Benj
> First of all, thank you for the answer. If I try to apply Lorentz force
> (q v cross b) I get into trouble. If you look at the field generated by
> the magnet, most of the lines have their main component parallel to the
> wire (after its decomposition in a suitable reference system, for
> example, taking the wire as one dimension), and, therefore, parallel to
> the velocity of the charges. Because of this, this component does not
> exert any net force. The other component is perpendicular to the wire
> and produces a force which is tangential to the trajectory of the wire
> shown in the applet. However, the field generated by the magnet cuts the
> wire in two different directions: the field lines next to the north pole
> of the magnet cut the wire "outwards". However, the lines entering the
> south pole cut the wire "inwards". Since the velocity of the charges has
> not changed its direction, the force exerted on the wire should be
> different in the upper and lower part of the wire (?).
As usual, teachers today are by no means as smart as Faraday was back
then. The "explanation" of this device as given is mostly rubbish and
of course has misled you too. First off, magnetic fields do NOT
"interact". Magnetic fields are completely neutral to each other and
in fact you can have magnetic fields flowing in opposite directions at
the same time without any mutual interaction. It's the basis of the
superposition principle.
The basis of this experiment is the interaction between Fields and
matter. Yes, it is the Lorentz force that makes this work. The central
magnet creates a field about itself. It would be well to draw that
field and what it looks like. There are lots of pictures of such a
field. [oops. I see the applet has it built in if you push a button]
Notice your argument about the fields at the top and bottom of magnet
being "in" and "Out" is valid but doesn't apply here. The magnet is
not placed symmetrically to the wire. Hence mostly the upper part of
the field is creating force, not the reversed bottom.
Note one more fact while we are at it. The magnetic field has NO
component in the theta direction which is to say in a direction of
circles about the magnet. All B vectors are in planes slicing the axis
of the magnet. No vectors have components perpendicular to those
planes. What is the meaning of this? In which direction do we then NOT
have electrical forces on the wire? Figure it out.
As for the wire forces, let us remember that the qV x B forces are
integrated along the entire wire. We can assume the wire is stiff and
hence adds up all forces it experiences. Forces are NOT equal at all
points on the wire. The force at any point is the product of the
current and the COMPONENT of the magnetic field perpendicular to the
wire. A magnetic field is a vector and any vector in a plane can be
resolved into two perpendicular vectors, one parallel to the wire
(giving no force) and one perpendicular to the wire (giving the
motion).
Note that the field lines as shown in the applet are shown more
symmetrical than they should be and even worse, when you turn the
motor "off" the wire suddenly reverses direction and slows down and
stops. Which is a good question: What would have to reverse to reverse
the direction of the "motor"?
And notice that the wire has some kinks in it. Does that have an
effect upon the forces driving the "motor" and hence it's motion and
speed?
It's pretty hard for someone to learn about science when all the
people teaching them haven't a clue either. The blind lead the blind.
Hugo Caballero
delay! I've been away for some days... You'll find my conclusions
underneath your response! I will post a new question soon and I hope to
hear from you again =)
El 14/08/2010 18:40, Benj escribió:
> On Aug 5, 3:57 pm, Hugo Caballero<hugocaball...@coit.es> wrote:
>
>> First of all, thank you for the answer. If I try to apply Lorentz force
>> (q v cross b) I get into trouble. If you look at the field generated by
>> the magnet, most of the lines have their main component parallel to the
>> wire (after its decomposition in a suitable reference system, for
>> example, taking the wire as one dimension), and, therefore, parallel to
>> the velocity of the charges. Because of this, this component does not
>> exert any net force. The other component is perpendicular to the wire
>> and produces a force which is tangential to the trajectory of the wire
>> shown in the applet. However, the field generated by the magnet cuts the
>> wire in two different directions: the field lines next to the north pole
>> of the magnet cut the wire "outwards". However, the lines entering the
>> south pole cut the wire "inwards". Since the velocity of the charges has
>> not changed its direction, the force exerted on the wire should be
>> different in the upper and lower part of the wire (?).
>
> As usual, teachers today are by no means as smart as Faraday was back
> then. The "explanation" of this device as given is mostly rubbish and
> of course has misled you too. First off, magnetic fields do NOT
> "interact". Magnetic fields are completely neutral to each other and
> in fact you can have magnetic fields flowing in opposite directions at
> the same time without any mutual interaction. It's the basis of the
> superposition principle.
Alright! The resulting magnetic field is the superposition (linear
addition) of all the fields.
> The basis of this experiment is the interaction between Fields and
> matter. Yes, it is the Lorentz force that makes this work. The central
> magnet creates a field about itself. It would be well to draw that
> field and what it looks like. There are lots of pictures of such a
> field. [oops. I see the applet has it built in if you push a button]
> Notice your argument about the fields at the top and bottom of magnet
> being "in" and "Out" is valid but doesn't apply here. The magnet is
> not placed symmetrically to the wire. Hence mostly the upper part of
> the field is creating force, not the reversed bottom.
Fine!
> Note one more fact while we are at it. The magnetic field has NO
> component in the theta direction which is to say in a direction of
> circles about the magnet. All B vectors are in planes slicing the axis
> of the magnet. No vectors have components perpendicular to those
> planes. What is the meaning of this? In which direction do we then NOT
> have electrical forces on the wire? Figure it out.
Let's use cylindrical coordinates, where theta describes the angle swept
arount the magnet, rho, the radius from the axis of the magnet to a
certain coordinate and z, the direction of the axis of the magnet. I
hope this is sufficiently clear.
The rotation of the field around the magnet on any perpendicular plane
to the axis of the magnet is zero. All the lines "travel" from the north
pole to the south pole. The force generated by a B field is always
perpendicular to that B field, so that there won't be any force exerted
in the direction of the B field. The components of the B field lie on
the plane z-rho, that is, there is no theta component. Then any vector b
has the following general form b_i=(bz_i, brho_i, 0).
In order to find out feasible directions for the force, we calculate the
scalar product and force it to be zero, so that both vectors are
perpendicular.
b_i·f_i=(bz_i*fz_i+brho_i*frho_i)
Therefore:
frho_i = - fz_i * bz_i / brho_i
Then:
f_i = (fz_i, -fz_i * bz_i / brho_i, ftheta)
for any given vector b_i. The theta direction of the force is
independent (since there are infinite perpendicular vectors to any
single vector).
However, when we include the wire, which also lies in the z-rho
direction, we just have one possible solution for the direction of the
force.
With the velocity of the charges being given by v_i = (vz_i, vrho_i, 0),
the direction of the force can be extracted from the following equation,
using the vector product:
v_i x b_i = (0, 0, vz_i*brho_i-vrho_i*bz_rho)
This implies that the force will just have a component different from
zero in the theta coordinate, that is, it will only exist on a circle
around the magnet.
Is this correct?
> As for the wire forces, let us remember that the qV x B forces are
> integrated along the entire wire. We can assume the wire is stiff and
> hence adds up all forces it experiences. Forces are NOT equal at all
> points on the wire. The force at any point is the product of the
> current and the COMPONENT of the magnetic field perpendicular to the
> wire. A magnetic field is a vector and any vector in a plane can be
> resolved into two perpendicular vectors, one parallel to the wire
> (giving no force) and one perpendicular to the wire (giving the
> motion).
That's clear!
> Note that the field lines as shown in the applet are shown more
> symmetrical than they should be and even worse, when you turn the
> motor "off" the wire suddenly reverses direction and slows down and
> stops. Which is a good question: What would have to reverse to reverse
> the direction of the "motor"?
To reverse the direction of the wire, the force needs to be reversed.
There are two possibilities in order to change the direction of the
wire: either changing the polarity of the source, so that the current
flows in the opposite direction or rotating 180 degrees the magnet, so
that the magnetic field changes its direction.
> And notice that the wire has some kinks in it. Does that have an
> effect upon the forces driving the "motor" and hence it's motion and
> speed?
I think so. If you introduce kinks, the angle between the magnetic field
and the velocity of the charges is changed and therefore the direction
and/or magnitude of the resulting force changes as well. In fact, if we
bend the wire in such a way that we introduce, for example, a trajectory
parallel to theta, we will create forces that will move a section of the
wire inwards or outwards. This can yield very funny movements. The more
straight the wire, the smoother the movement.
> It's pretty hard for someone to learn about science when all the
> people teaching them haven't a clue either. The blind lead the blind.
Please, let me know if I should go through something again! And thank
you very much for your support!
Szczepan Bialek
"Hugo Caballero" <hugoca...@coit.es> wrote
news:4C6AD22A...@coit.es...
>
> El 14/08/2010 18:40, Benj escribió:
>>
>>Which is a good question: What would have to reverse to reverse
>> the direction of the "motor"?
>
> To reverse the direction of the wire, the force needs to be reversed.
Not in this case.
In the demo, after each stop, the wire travels to the centre position.
Before or behaind the magnet. So the wire reverse or continue the movement
to reach the nearest starting position.
S*
>