Wire cutting magnetic field lines

[Philip Wood]

A potato moves in an enormous magnetic field at very large velocity. Electrical discharges can be observed near the surface of the potato.
Now if we say that no EMF caused those discharges

Not what I'm saying, of course. The surrounding air (more or less stationary?) completes a loop in which there is an emf (so the equivalent circuit would be the one I drew in post 11. Were the air carried along with the potato, I'd say there was no emf in the loop.

 

[jartsa]

Not what I'm saying, of course. The surrounding air (more or less stationary?) completes a loop in which there is an emf (so the equivalent circuit would be the one I drew in post 11. Were the air carried along with the potato, I'd say there was no emf in the loop.

Let's say a lamp L is connected to two metall spheres, and this device moves through different magnetic fields:

                                                        n                                  s                                    n    


O                            
|                          
L         ------->                              
|
O                          

                                      S                                    N                                     S

Lamp glows. No loop.

(Magnetic fields pointing into the screen and away from the screen are difficult to "draw" )

 

[Philip Wood]

I think you may have made your case...

If your sphere-ended conductor started moving through a uniform field electrons would flow on to one sphere and off the other, propelled by Bqv forces. But the flow would soon stop, and the lamp would go out, when electrostatic forces due to charge on the spheres balanced the Bqv forces, even if the wire kept moving and cutting flux.

I agree that the lamp would glow whenever the field reversed, as the Bqv forces would reverse, so the free electrons would shuffle backwards and forwards through the lamp and wires to different spheres. [Perhaps you'd agree that your Ns and Ss should be 'out of the screen' and 'into the screen' rather than as you've drawn them.]

I'm still not sure I'd want to use the term emf here. I note that the most authoritative textbook I have to hand, Panofsky & Phillips, defines emf only for a closed loop, but perhaps I'm being too much of a purist.

 

[utkarshakash]

I'm talking about straight wire not connected to a circuit that isn't a loop moving in a field, in that case will be an induced emf ?

I think I've understood your problem. In your case the induced emf is not due to change in magnetic flux(we don't talk about flux unless we have a loop). The emf is due to Hall Effect. You already know that a neutral metallic wire has an equal no. of electrons and protons. As the wire is moved the magnetic field exerts force on both the charged particles. As a result the electrons get concentrated at one end and protons at the other. This develops an electric field inside the wire and is the cause of induced emf. The emf induced this way is called Motional EMF.
Hope I helped :)

 

[alva]

Does this help? See thumbnail ...

Referring to your drawing. How the top, botton and right wires know that something is happening? They can not sense anything.

 

[alva]

Note that to MEASURE the voltage produced one would have to connect a voltmeter between the inner end (X) of the hand and its outer end (Y). It matters how you do this…

If you could strap a mini-voltmeter to the hand itself, and connect wires from it going along the hand to X and Y, zero voltage would be measured.

What would happen if the wires connected to the voltmeter were magnetic shielded?

 

[Philip Wood]

Referring to your drawing. How the top, botton and right wires know that something is happening? They can not sense anything.

A current will flow through them if the circuit is complete, but I expect this evades the issue prompting your question. A more useful response might be that the other wires, and the fact that they're stationary define the path around which the emf is to be calculated. If the three wires you're referring to were moving at the same velocity as the left hand (moving!) wire, there'd be no emf in the loop.

I'm obstinately insisting - and I'm not alone in this - see post 38 - that emf is only properly defined around a particular path. But, as I admitted in the same post, maybe this is too purist a stance. What I've never challenged is that, in the laboratory frame of reference, the FORCES urging the free electrons through the wires arise in the left hand, moving wire. They are the Bqv forces.

 

[Philip Wood]

What would happen if the wires connected to the voltmeter were magnetic shielded?

I should suppose that there will then be an emf in the loop, and a current will flow. But there may be some unexpected consequence of the magnetic shielding which I haven't spotted! What I'm getting at is that the shielding will affect the field elsewhere, specifically because of the requirement that [itex]div \vec{B} = 0[/itex]. But I'm not clever enough to work out what the consequences will be in this case.

 

[Entanglement]

I think I've understood your problem. In your case the induced emf is not due to change in magnetic flux(we don't talk about flux unless we have a loop). The emf is due to Hall Effect. You already know that a neutral metallic wire has an equal no. of electrons and protons. As the wire is moved the magnetic field exerts force on both the charged particles. As a result the electrons get concentrated at one end and protons at the other. This develops an electric field inside the wire and is the cause of induced emf. The emf induced this way is called Motional EMF.
Hope I helped :)

Is that how electricity is induced in an AC generator

 

[utkarshakash]

Is that how electricity is induced in an AC generator

No. In AC generator the current is induced due to change in flux.

 

[Entanglement]

No. In AC generator the current is induced due to change in flux.

What about the clock hand question that I have referred to in a previous post, how Can I calculate the induced EMF in a clock hand cutting a magnetic field ?

 

[utkarshakash]

What about the clock hand question that I have referred to in a previous post, how Can I calculate the induced EMF in a clock hand cutting a magnetic field ?

Use the formula (vxB).l

 

[Philip Wood]

See my post 15 (especially last paragraph) for how the emf must be measured.

Suppose that the number of revolutions of the hand per unit time is f. [So [itex]f=\frac{1}{60}\ \mbox{s}^{-1}[/itex] for a seconds-hand, and so on.]

So emf = work done per unit charge = rate of cutting of flux by hand (length L) = B x area swept out per unit time = [itex]B\pi\ L^{2}\ f[/itex].

Or, as frequently mentioned in earlier posts, we can get the same result by direct consideration of the Bqv forces on the charge carriers in the wire…

emf = [itex]\frac{1}{q} \int_{0}^{L}Bqv\ \mbox{d}r = B \omega \int_{0}^{L}r\ \mbox{d}r =B\ 2 \pi f \int_{0}^{L}r\ \mbox{d}r = B\pi\ L^{2}\ f[/itex]

because [itex]v = r \omega = r\ 2 \pi f. [/itex]

 

[Entanglement]

See my post 15 (especially last paragraph) for how the emf must be measured.
Suppose that the number of revolutions of the hand per unit time is f. [So [itex]f=\frac{1}{60}\ \mbox{s}^{-1}[/itex] for a seconds-hand, and so on.]
So emf = work done per unit charge = rate of cutting of flux by hand (length L) = B x area swept out per unit time = [itex]B\pi\ L^{2}\ f[/itex].
Or, as frequently mentioned in earlier posts, we can get the same result by direct consideration of the Bqv forces on the charge carriers in the wire…



emf = [itex]\frac{1}{q} \int_{0}^{L}Bqv\ \mbox{d}r = B \omega \int_{0}^{L}r\ \mbox{d}r =B\ 2 \pi f \int_{0}^{L}r\ \mbox{d}r = B\pi\ L^{2}\ f[/itex]
because [itex]v = r \omega = r\ 2 \pi f. [/itex]

I know the emf induction equations pretty well, but I don't understand where the circuit or loop is in the clock and how its area increases and decreases ?

 

[Philip Wood]

Again, may I refer you to my post (15). As I explained there, the clock hand is not of itself in ANY circuit; you have to construct your own circuit. I discussed two examples of circuits, one in which there would be no emf, and one in which there would be an emf.

Even with the 'right' circuit, it is difficult or impossible to apply the idea of a continuously increasing area. I pointed this out in the first para of my post 15. The difficulty is acknowledged in the excellent Wiki article on electromagnetic induction

That's why, in post 48, I used emf = rate of CUTTING of flux, not rate of change of flux linkage. Alternatively, go back to basics, that is the Bqv forces on the charge carriers, as in my second method in post 48.

 

[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 [itex]\theta[/itex] means? Should you consider using a different book?

In fact the question of angles is not entirely straightforward, because three vectors are involved: [itex]\vec B[/itex], [itex]\vec v[/itex], [itex]\vec L[/itex]. 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 = [itex]Lv \mbox{sin} \theta \ B\ \mbox{cos} \phi[/itex]. In this equation, [itex]Lv \mbox{sin} \theta[/itex] is the area of the face of the parallelepiped containing the [itex]\vec v[/itex] and [itex]\vec L[/itex] vectors. Multiplying by [itex] B\ \mbox{cos} \phi[/itex] 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 [itex]\frac{\epsilon_{0} A}{L} BLv[/itex], so the charge density (on the inner surface of either plate) will be [itex]\epsilon_{0} Bv[/itex].

(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 = [itex]\frac{L}{D}\epsilon_{0} Bv\ [/itex].

 

[Entanglement]

ElmorshedyDr. Shouldn't your book tell you what [itex]\theta[/itex] means? Should you consider using a different book?
In fact the question of angles is not entirely straightforward, because three vectors are involved: [itex]\vec B[/itex], [itex]\vec v[/itex], [itex]\vec L[/itex]. 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 = [itex]Lv \mbox{sin} \theta \ B\ \mbox{cos} \phi[/itex]. In this equation, [itex]Lv \mbox{sin} \theta[/itex] is the area of the face of the parallelepiped containing the [itex]\vec v[/itex] and [itex]\vec L[/itex] vectors. Multiplying by [itex] B\ \mbox{cos} \phi[/itex] 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 [itex]\theta[/itex] 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 = [itex]Lv\ \mbox{sin}\ \theta \ B \mbox{cos}\ \phi[/itex].

My thumbnail tells you what [itex]\theta\ \mbox{and}\ \phi[/itex] mean.

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

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

 

[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, [itex]\mbox{sin}\ \theta[/itex] 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