Wire cutting magnetic field lines

[Entanglement]

Ok Guys cool...there is one more thing, I don't understand how to calculate the emf when the moving wire connected to a circuit moves by a certain angle with the flux lines " not perpendicularly ". My book says BLV Sin theta, is that sin for the resolution of the velocity vector or for the flux and if it's for the resolution of the flux, does theta represent the angle between the motion of the wire and the flux lines?

 

[Brinx]

So, would it be possible to measure the local charge density at each endpoint of a non-closed length of wire moving through the uniform magnetic field? This should be done in such a way that no closed circuit is formed which the wire would be a part of. Think of a dumbbell with the wire being the central bar and the charge measurements going on in the spheres at either end.

The Lorentz force acting on the charge carriers in the wire would act to separate them out along its length - effectively only pushing the electrons to one end by a bit. The lattice atoms/ions would remain where they are, being part of the lattice. So naively you would expect a potential difference to exist between the ends of the wire - but no further current would flow as the electrons have nowhere to go beyond the wire itself. The motion through the magnetic field would induce a mechanical stress on the wire along its length, proportional to B * v * l.

Does that make sense?

 

[Philip Wood]

ElmorshedyDr. Shouldn't your book tell you what \theta means? Should you consider using a different book?

In fact the question of angles is not entirely straightforward, because three vectors are involved: \vec B, \vec v, \vec L. So two angles are needed to take account of all possible angles between the three vectors.

The thumbnail is my attempt to show this in three dimensions. The portion of the circuit emf due to L moving is equal to the 'volume' of the parallelepiped (wonky cuboid), that is: emf = Lv \mbox{sin} \theta \ B\ \mbox{cos} \phi. In this equation, Lv \mbox{sin} \theta is the area of the face of the parallelepiped containing the \vec v and \vec L vectors. Multiplying by B\ \mbox{cos} \phi gives the volume.

All this talk of areas and volumes is not needed if you are familiar with dot and cross vector products. Forgive me if you are.

 

[Philip Wood]

Brinx. You're absolutely right. There are two cases (as far as I know) in which it's easy to work out the charge density

(1) The wire (length L) runs between two parallel flat plates, perpendicularly to them. The area A of the plates need to be >> L².

In that case, the equilibrium p.d. between the plates will be BLv, assuming all three vectors at right angles to each other, so the charge on either plate will be \frac{\epsilon_{0} A}{L} BLv, so the charge density (on the inner surface of either plate) will be \epsilon_{0} Bv.

(2) The wire (length L) runs between two metal spheres of diameter D. This time D << L. If this condition is met, the spheres behave as isolated spherical capacitors, and I think you'll find that charge density = \frac{L}{D}\epsilon_{0} Bv\.

 

[Entanglement]

ElmorshedyDr. Shouldn't your book tell you what \theta means? Should you consider using a different book?
In fact the question of angles is not entirely straightforward, because three vectors are involved: \vec B, \vec v, \vec L. So two angles are needed to take account of all possible angles between the three vectors.
The thumbnail is my attempt to show this in three dimensions. The portion of the circuit emf due to L moving is equal to the 'volume' of the parallelepiped (wonky cuboid), that is: emf = Lv \mbox{sin} \theta \ B\ \mbox{cos} \phi. In this equation, Lv \mbox{sin} \theta is the area of the face of the parallelepiped containing the \vec v and \vec L vectors. Multiplying by B\ \mbox{cos} \phi gives the volume.
All this talk of areas and volumes is not needed if you are familiar with dot and cross vector products. Forgive me if you are.

What I mean is that there is an angle between the velocity vector and and the flux, simply " not perpendicular "

 

[Entanglement]

The motion through the magnetic field would induce a mechanical stress on the wire along its length, proportional to B * v * l.

Thanks a lot ! But I don't completely understand this part

 

[Philip Wood]

ElmorshedyDr. Did you study the thumbnail as well as the text in my post 53? It shows the angles involved, and answers the question you asked in post 51. I thought you wanted to know what the \theta meant in the formula you quoted. I have shown you.

 

[Entanglement]

So according to your thumbnail Emf = Blv Cos Phi

 

[Philip Wood]

No, My text tells you that

emf = Lv\ \mbox{sin}\ \theta \ B \mbox{cos}\ \phi.

My thumbnail tells you what \theta\ \mbox{and}\ \phi mean.

\mbox{sin}\ \theta = 1 only if \vec{v} is perpendicular to \vec{L}.

\mbox{cos}\ \phi = 1 only if \vec{B} is perpendicular to the plane containing \vec{v} and \vec{L}.

 

[kira506]

Well , I explain it to myself as follows : Magnetic flux is constant , but the magnetic flux density isn't , say a magnet is placed at position x and the wire is placed perpendicular to the magnetic field , as we get farther from the magnet ,the flux density affecting the wire decreases (B inversely proportional to d ) and not the magnetic flux , the following statement is by all means wrong : say the wire cuts G magnetic flux lines when it is near the magnet ,as it gets farther .the density ( in the form of spacings between flux lines ) decreases , and the wire is of definite length (not an infinitely long one ) so now its definite length is in the range of only 4 magnetic flux lines , here's a rate of change,hope this has helped

 

[kira506]

Your problem lies within the clock hand problem , well , It actually isn't solved by BLVsin theta , but by the rule for average induced EMF , EMF=-N*delta (A*B)/delta time , the problem is just written in an utterly retarded way ,that the change in area is not given by m^2 /s , but only m^2

 

[Philip Wood]

Your problem lies within the clock hand problem , well , It actually isn't solved by BLVsin theta […]

You can use BLVsin theta, but it involves a simple integration, because different parts of the clock-hand move at different speeds. I apply the method in the second part of post 48, so you can see how it works. It's not as easy to use in this case as emf = rate of cutting of flux, but is arguably more fundamental as it shows how the emf arises from the Bqv (Lorentz) forces on the charge carriers.

 

[kira506]

Uhm , I don't know what the Bqv forces are , nor do I know what the charge carriers are XD sorry , I wanted to ask but didn't because this isn't my post ( I'm not sure if creating a subpost within the post is against the rules or no >.> ) although I've read the 4 pages of the post , I didn't understand what the Bqvs or charge carriers are , I just suggested the otherone because I'm not sure if we could consider the clock hand as a straightwire ,because its movement is notin a st. Line , it moves in a circular path which makes me wonder how the clock really works , is it the smae basis of a galvanometer or justt DC from battery converted to kinetic energy through some kind of motor ? Our books don't provide much info about such stuff ...

 

[Philip Wood]

Kira: for explanation of "Bqv forces" try http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html.

The point is that if you move a conductor in a magnetic field, then you are also moving the charge-carriers (usually free electrons) which are in the wire. The charge carriers will experience Bqv forces which tend to drive them through the wire (provided the direction of the wire and the direction of motion of the wire are such that flux is cut). That's how the emf arises.

There we are. I've done my best!

 

[Entanglement]

Kira: for explanation of "Bqv forces" try http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html.
The point is that if you move a conductor in a magnetic field, then you are also moving the charge-carriers (usually free electrons) which are in the wire. The charge carriers will experience Bqv forces which tend to drive them through the wire (provided the direction of the wire and the direction of motion of the wire are such that flux is cut). That's how the emf arises.
There we are. I've done my best!

So if we suppose a car moving in a magnetic field, and its antenna is cutting the magnetic field so the emf will be due to Lorentz force

 

[Philip Wood]

Yes. This would happen, for example if the car were traveling East-West in the Earth's magnetic field.

But that's not how the aerial picks up radio signals. The radio waves have consist of rapidly oscillating electric and magnetic fields. Its the electric field which exerts forces on the free electrons in the aerial, making them oscillate up and down. The car doesn't have to be moving for this to happen.

 

[Entanglement]

Yeah, we are in no need of talking about radio waves now as it's not our topic. So the emf in the antenna due to Lorentz force is calculated also from Blv?

 

[kira506]

Kira: for explanation of "Bqv forces" try http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html.

The point is that if you move a conductor in a magnetic field, then you are also moving the charge-carriers (usually free electrons) which are in the wire. The charge carriers will experience Bqv forces which tend to drive them through the wire (provided the direction of the wire and the direction of motion of the wire are such that flux is cut). That's how the emf arises.

There we are. I've done my best!

Thank you so much , I think I kinda understand what they are , Bqv , q here stands for charge and charge carriers are electrons , that's why its better to solve it in terms of the Bqv forces , as the L in Blv represents the q of free electrons in the conductor , the actual ones affected by the field , right ? Thank you so much again c:

 

[Philip Wood]

ElmorshedyDr. Yes. If the aerial is vertical but it's moving horizontally, \mbox{sin}\ \theta will be 1 in our previous discussion.

kira506. Bqv is the force on a charge q moving at speed v at right angles to a magnetic field B.

emf is defined as work done per unit charge on a charge moving round a circuit.

If charge q moves through length L of conductor, work done on it will be force x distance = BqvL
So emf = work done per unit charge = BqvL/q = BLv.

I'm assuming for simplicity that the wire is at right angles to its direction of motion.

 

[Entanglement]

emf is defined as work done per unit charge on a charge moving round a circuit.
If charge q moves through length L of conductor, work done on it will be force x distance = BqvL

So emf = work done per unit charge = BqvL/q = BLv.
I'm assuming for simplicity that the wire is at right angles to its direction of motion.

So the emf in the Ariel (Bvql/q) is calculated by the same law of a wire widening and narrowing a loop ( B delta A = Blv ) so that's just a coincidence

 

[Entanglement]

emf is defined as work done per unit charge on a charge moving round a circuit.
If charge q moves through length L of conductor, work done on it will be force x distance = BqvL

So emf = work done per unit charge = BqvL/q = BLv.
I'm assuming for simplicity that the wire is at right angles to its direction of motion.

So the emf in the Ariel (Bvql/q) is calculated by the same law of a wire widening and narrowing a loop ( B delta A = Blv ) so that's just a coincidence?

 

[Philip Wood]

No, it's not a co-incidence! I've taken the view throughout this thread, that emf is only correctly spoken about for a closed loop. If the whole of the loop moves with the aerial (top digram) there is no emf in the loop. If the rest of the loop is stationary (bottom diagram) there is an emf, BLv, which can correctly be written as B delta A/Delta t, in which A is the loop area.

Please ignore this post. It is replaced by a post with thumbnail.

 

[Entanglement]

No, it's not a co-incidence! I've taken the view throughout this thread, that emf is only correctly spoken about for a closed loop. If the whole of the loop moves with the aerial (top digram) there is no emf in the loop. If the rest of the loop is stationary (bottom diagram) there is an emf, BLv, which can correctly be written as B delta A/Delta t, in which A is the loop area.

I understand the moving loop as all or just a part of it, thanks to you in the earlier posts of this thread But in the case of the an actual aerial there's no loop so we calculate the emf from Blv which is derived from Bvql and not B delta A / delta t ?

 

[Philip Wood]

No, it's not a co-incidence! I've taken the view throughout this thread, that emf is only correctly spoken about for a closed loop. If the whole of the loop moves with the aerial (top digram) there is no emf in the loop. If the rest of the loop is stationary (bottom diagram) there is an emf, BLv, which can correctly be written as B delta A/Delta t, in which A is the loop area.

 

[Philip Wood]

But in the case of the aerial there's no loop yet we calculate the emf from Blv which is derived from Bvql and not B delta A / delta t indeed

We've been here before! My view is that strictly it's wrong to talk about the emf in the aerial. It's only correct to talk about the emf in a complete loop. You have to make the aerial part of a loop. Then it does make sense to talk about the loop area changing.

If you choose a loop ALL of which moves with the car (see my top diagram in post 74) there is no emf, and there is no rate of change of loop area, so we have consistency.

[But as discussed earlier, rate of change of loop area is hard or impossible to apply in the case of the clock-hand, even when the hand is included in a suitable circuit.]

 

[Entanglement]

We've been here before! My view is that strictly it's wrong to talk about the emf in the aerial. It's only correct to talk about the emf in a complete loop. You have to make the aerial part of a loop. Then it does make sense to talk about the loop area changing.

So practically speaking No loop -------> no emf ??

But I'm not talking about a changing area I'm talking about the emf induced due to "Lorentz force" even if the loop doesn't exist as I understand from the earlier posts

 

[Philip Wood]

I do understand where you're coming from here. You can indeed calculate the work done per unit charge on the moving charges in the aerial. It is BLv. And this would be measured in volts, like emf. I'd call it a CONTRIBUTION to the emf. That's because I think emf should only be applied to the complete amount of work done per unit charge on a charge going round a complete circuit. If you stick to this idea of emf, then BLv DOES equal B times dA/dt, i.e. rate of change of flux linking the circuit.

If, loosely, you talk about BLv as the emf in just the aerial, then, as you say, rate of change of flux doesn't make sense. My recommendation is to be rigorous and only talk about emfs in complete circuits. If there is no obvious circuit - imagine one! That will also help you to understand how you could MEASURE the emf.

 

[Entanglement]

I do understand where you're coming from here. You can indeed calculate the work done per unit charge on the moving charges in the aerial. It is BLv. And this would be measured in volts, like emf. I'd call it a CONTRIBUTION to the emf. That's because I think emf should only be applied to the complete amount of work done per unit charge on a charge going round a complete circuit. If you stick to this idea of emf, then BLv DOES equal B times dA/dt, i.e. rate of change of flux linking the circuit.
If, loosely, you talk about BLv as the emf in just the aerial, then, as you say, rate of change of flux doesn't make sense. My recommendation is to be rigorous and only talk about emfs in complete circuits. If there is no obvious circuit - imagine one! That will also help you to understand how you could MEASURE the emf.

So in the case of just an aerial there is an emf due to Lorentz force calculated from blv but we shouldn't talk about an emf here because there is no practical circuit and the emf should be the total work done on a coulomb through a full cycle in the loop. Isn't that correct ?

 

[Philip Wood]

Yes, that's my view. I'd be happy about calling BLv a contribution to the circuit emf. And I have to admit that most people probably refer to BLv as the emf in the aerial. This is fine as long as it's understood to be shorthand for the contribution that the aerial would make to the emf in a circuit of which it forms part!

 

[tiny-tim]

Hi Philip! Hi ElmorshedyDr!

So practically speaking No loop -------> no emf ??

You can indeed calculate the work done per unit charge on the moving charges in the aerial. It is BLv. And this would be measured in volts, like emf. I'd call it a CONTRIBUTION to the emf.

That's because I think emf should only be applied to the complete amount of work done per unit charge on a charge going round a complete circuit. If you stick to this idea of emf, then BLv DOES equal B times dA/dt, i.e. rate of change of flux linking the circuit.

If, loosely, you talk about BLv as the emf in just the aerial, then, as you say, rate of change of flux doesn't make sense. My recommendation is to be rigorous and only talk about emfs in complete circuits …

So in the case of just an aerial … we shouldn't talk about an emf here because there is no practical circuit and the emf should be the total work done on a coulomb through a full cycle in the loop. Isn't that correct ?

Yes, that's my view. I'd be happy about calling BLv a contribution to the circuit emf.
And I have to admit that most people probably refer to BLv as the emf in the aerial.
This is fine as long as it's understood to be shorthand for the contribution that the aerial would make to the emf in a circuit of which it forms part!

personally, i think of Blv as representing an imaginary battery in the aerial, and i see nothing wrong with calling the work done by moving 1 C along the aerial "the emf induced in the aerial"

an actual battery in the aerial would produce no current (since there's no circuit)

if you complete the loop with a moving circuit (Philip's first diagram), then there's an identical imaginary battery on the other vertical side of the loop: the two batteries cancel, so the emf (of the whole circuit) is 0

if you complete the loop with a circuit with one end fixed (Philip's second diagram), then there's only one imaginary battery, and the emf (of the whole circuit) is equal to the voltage of that imaginary battery, Blv

does the imaginary battery create a voltage difference in the same way as a real battery does?

no

voltage difference is a difference in potential energy, and potential energy is defined as (minus) the work done by a conservative force … but (although the electric field of a battery is conservative) the electric field "induced" by a changing magnetic field is not conservative

this is not mere semantics … the voltage difference from point A to point B (like gravitational potential difference) should not depend on the path taken

in a real battery circuit, it doesn't: going all the way round the circuit takes you through the battery, which cancels out the gain you made from the battery: the work done going either clockwise or anti-clockwise is 0, and the voltage difference from point A to point B (as in gravity) is unambiguous

but in a imaginary battery circuit (ie in an https://www.physicsforums.com/library.php?do=view_item&itemid=294flux-induced circuit), nothing cancels as you go all the way round: the work done going clockwise is minus the work done going anti-clockwise, and both are non-zero … and so the work done from point A to point B clockwise is ambiguous: in other words, it (and the voltage) cannot be defined

this inability to define a voltage in a flux-induced circuit requires a new concept, called emf, defined of course as the work done on 1 C around the circuit in a particular direction

(see also http://en.wikipedia.org/wiki/Electromotive_force#Electromotive_force_and_voltage_difference)

example: suppose we have three vertical aerials, joined by two horizontal wires at top and bottom: so we have one outer loop and two inner loops

suppose the magnetic field is not uniform, so that dflux/dt = emf through the outer loop is 6V, and through the inner loops are 2V and 4V (all the same way): then my imaginary batteries would be 7V 5V and 1V (all the same way)

if they were real batteries, and if the three parts of the circuit each have resistance 1Ω, then the currents upward through the three batteries would be 8/3 A, 2/3 A, and -10/3 A

the result for the flux-induced set-up is the same

it seems to me far easier to work this out by finding the imaginary batteries first, ie what we're not supposed to call "the emf induced in the aerials"!

 

[Philip Wood]

tiny-tim: I see exactly where you're coming from. My wish to cling on to emf as a whole-loop line integral is partly motivated by my wish to use Faraday's law in the form emf = rate of change of flux linkage, which then neatly works for either moving circuit boundaries or changing flux densities: one equation applying to two phenomena. Without a circuit we can't do this (one of ElmorshedyDr's concerns). We can, of course, still talk about rate of flux CUTTING, but that doesn't have the same uniting quality.

Another thing I like about insisting on a complete circuit is that it forces one to think about how you'd MEASURE the induced voltage; for example a voltmeter moving with the car and connected across the aerial would give zero. [I'm not advancing this as a major argument for showing that emf can only be applied to a complete loop.]

 

[kira506]

ElmorshedyDr. Yes. If the aerial is vertical but it's moving horizontally, \mbox{sin}\ \theta will be 1 in our previous discussion.

kira506. Bqv is the force on a charge q moving at speed v at right angles to a magnetic field B.

emf is defined as work done per unit charge on a charge moving round a circuit.

If charge q moves through length L of conductor, work done on it will be force x distance = BqvL
So emf = work done per unit charge = BqvL/q = BLv.

I'm assuming for simplicity that the wire is at right angles to its direction of motion.

sometimes is inclined by angle theta but in most cases it is perpendicular to field and direction of motion
thanks , that was really helpful , thanks to your awesome explanation .I now know how this rule was derived,you should consider being a physics teacher (that's if you're not already one c: )