Wire cutting magnetic field lines

[Entanglement]

I don't understand why a straight moving wire doesn't induced emf while a changing area loop does induce emf ?

 

[Philip Wood]

There IS an emf due to a moving wire cutting magnetic flux, provided that, by moving, the wire changes the area of the circuit it is in. This may seem puzzling, so I'll try and explain…

Suppose we have a wire loop of fixed shape, at right angle to a uniform magnetic field. If we move the loop through the field, always keeping it at right angles to the field, there will be no emf induced. This is because the Bqv forces on the charges in the wire are in different senses as we go round the loop, and the line integral of the force per unit charge is zero for the complete loop.

On the other hand, there will be an emf due to Bqv forces in the wire in the set-up I sketched in post 11, because the forces are not canceled out, when we form the line integral, by others in the rest of the loop, as the rest of the loop is stationary.

 

[Entanglement]

There IS an emf due to a moving wire cutting magnetic flux, provided that, by moving, the wire changes the area of the circuit it is in. This may seem puzzling, so I'll try and explain…
Suppose we have a wire loop of fixed shape, at right angle to a uniform magnetic field. If we move the loop through the field, always keeping it at right angles to the field, there will be no emf induced. This is because the Bqv forces on the charges in the wire are in different senses as we go round the loop, and the line integral of the force per unit charge is zero for the complete loop.
On the other hand, there will be an emf due to Bqv forces in the wire in the set-up I sketched in post 11, because the forces are not canceled out, when we form the line integral, by others in the rest of the loop, as the rest of the loop is stationary.

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 ?

 

[Philip Wood]

I don't like talking about emf unless for a loop (real or imaginary). You need a second (and third ….) opinion on this.

 

[jartsa]

I don't like talking about emf unless for a loop (real or imaginary). You need a second (and third ….) opinion on this.

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, then we are saying something which is wrong, just intuitively it seems wrong.

 

[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 div \vec{B} = 0. 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 f=\frac{1}{60}\ \mbox{s}^{-1} 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 = B\pi\ L^{2}\ f.

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 = \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

because v = r \omega = r\ 2 \pi f.

 

[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 f=\frac{1}{60}\ \mbox{s}^{-1} 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 = B\pi\ L^{2}\ f.
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 = \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
because v = r \omega = r\ 2 \pi f.

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 \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