Wire cutting magnetic field lines

[Entanglement]

I have read in my schoolbook that when a wire of length l moves at velocity v in a magnetic field cutting the flux lines an emf is induced which it is calculated from emf = blv, I understand that emf is induced at time variation of magnetic flux across the wire, but I Can't see any change in the magnetic flux here, the magnetic flux should stay the same anywhere the wire moves in the field as long as it not getting nearer to or farther from the magnet. But It seems non logical that the flux changes just because the wire covers a certain area during its movement in the field ! Please, this point is bothering me very much I would be very thankful if anyone could explain it to me, thanks in advance

 

[alva]

There is no magnetic field and there is no electrical field, they are both electromagnetic fields.

If a charge is close to a magnet, the charge will not sense any force, but if the charge is moving next to a magnet it will sense an electrical force, because there is an electrical field.

If a magnet is close to a charge, the magnet will not sense any force, but if the magnet is moving, it will sense a magnetic force, because there is a magnetic field.

And this is all I can help.

 

[tiny-tim]

Hi ElmorshedyDr!

I have read in my schoolbook that when a wire of length l moves at velocity v in a magnetic field … I Can't see any change in the magnetic flux here, the magnetic flux should stay the same anywhere the wire moves in the field as long as it not getting nearer to or farther from the magnet. But It seems non logical that the flux changes just because the wire covers a certain area during its movement in the field !

(We'll assume the magnetic field is uniform.)

It depends what you imagine the wire is connected to, to complete a circuit.

If the whole circuit is moving (in the field), and staying the same shape, then yes there's no change in flux.

But we always imagine a rectangular circuit with one edge fixed, and two edges getting longer and longer, so that the flux through the circuit is increasing!

(Because a potential always has to be measured relative to a fixed point of zero potential!)

 

[Entanglement]

But we always imagine a rectangular circuit with one edge fixed, and two edges getting longer and longer, so that the flux through the circuit is increasing!

!)

Thanks Tim, but I'm unable to visualize what you say please make it clearer

 

[tiny-tim]

You need a fixed point of zero potential, and that fixed point has to be part of the circuit.

Since the wire is moving, that means the whole circuit has to be getting large (or smaller).

To simplify things, we make the circuit a rectangle: one edge is the wire, the opposite edge is fixed, and the two "side" edges are in the direction of motion of the wire.

 

[Entanglement]

You need a fixed point of zero potential, and that fixed point has to be part of the circuit.
Since the wire is moving, that means the whole circuit has to be getting large (or smaller).
To simplify things, we make the circuit a rectangle: one edge is the wire, the opposite edge is fixed, and the two "side" edges are in the direction of motion of the wire.

Yeah I got it, but it contradicts the bases of some problems in my school book, where the problems ask us to calculate the emf induced in the the second clock arrow cutting the magnetic field of the earth, according to what you say there shouldn't be a induced emf in the clock arrow

 

[tiny-tim]

… the second clock arrow cutting the magnetic field of the earth, according to what you say there shouldn't be a induced emf in the clock arrow

If I'm understanding correctly, the circuit here is a piece-of-pie with one edge fixed and the other getting further and further round the circle …

so the area is increasing, and the flux is increasing.

 

[Entanglement]

If I'm understanding correctly, the circuit here is a piece-of-pie with one edge fixed and the other getting further and further round the circle …
so the area is increasing, and the flux is increasing.

The arrow doesn't even make a closed circuit

 

[davenn]

Thanks Tim, but I'm unable to visualize what you say please make it clearer ...

The arrow doesn't even make a closed circuit

I think we are all having difficulty trying to visualise what you are talking about

how about scanning the relevant images and text from your book and showing us
exactly what you are talking about

otherwise perople are just going to waste their time guessing about the setup

cheers
Dave

 

[tiny-tim]

(just got up :zzz:)

The arrow doesn't even make a closed circuit

I'm assuming a circuit around the whole piece-of-pie, ie a sector of a cirucle, comprising one fixed radius, one moving radius (with the arrow), and the arc joining them.

 

[Philip Wood]

Does this help? See thumbnail ...

 

[Entanglement]

I think we are all having difficulty trying to visualise what you are talking about
how about scanning the relevant images and text from your book and showing us

exactly what you are talking about
otherwise perople are just going to waste their time guessing about the setup
cheers

Dave

Dave its, what I'm talking about is a clock arrow cutting the magnetic field it ask me to calculate the Emf in the arrow I can't understand how can an emf get induced in the arrow?

 

[Entanglement]

Does this help? See thumbnail ...

Actually I don't understand why is a emf induced when a wire cuts a magnetic field although the shape of the wire didn't change besides the wire doesn't go nearer to or farther from the wire so delta BA should be zero

 

[Philip Wood]

BA is the flux LINKED WITH (going through) the circuit. Because, in my diagram, the area of the circuit gets less as the wire moves to the right, BA (that is the flux linked) gets less.

It's convenient to calculate the emf from the rate of change of flux linked, because the equation can be applied to both types of e-m induction (roughly speaking those due to A changing and to B changing). But it doesn't, imo, shed much light on WHY the emf is induced. One way to understand why, in the case of moving conductors, is to realize that there are motor effect (Bev) forces on the charge carriers in the wires, when these charge carriers move because the wire itself is moved.

 

[Philip Wood]

I agree that the case of the clock-hand (I take it that's what you mean by 'clock arrow') is a difficult one. It's not easy to apply the concept of change of flux linkage to it. [Indeed, the closely related rotating disc is one of the awkward cases cited in the Wiki article on electromagnetic induction that vanHees2 justifiably recommends.]

Nonetheless one can still apply the idea of flux CUTTING by the hand, or go back to basics and use the BEv method I mentioned in my previous post.

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. One way of looking at this is that the voltages induced in the hand and in the wires will be equal and in the same direction in space as each other, so in terms of the complete circuit, will cancel each other, being in opposite senses.

To measure a voltage you'd have to let Y brush against a stationary conductor and connect the voltmeter between this and X, so the voltmeter and the wires are stationary (in the laboratory frame of reference), while just the hand itself moves. [Principle of Faraday disc generator]

 

[Entanglement]

BA is the flux LINKED WITH (going through) the circuit. Because, in my diagram, the area of the circuit gets less as the wire moves to the right, BA (that is the flux linked) gets less.

It's convenient to calculate the emf from the rate of change of flux linked, because the equation can be applied to both types of e-m induction (roughly speaking those due to A changing and to B changing). But it doesn't, imo, shed much light on WHY the emf is induced. One way to understand why, in the case of moving conductors, is to realize that there are motor effect (Bev) forces on the charge carriers in the wires, when these charge carriers move because the wire itself is moved.

Thanks very much for your GREAT explanation, So you mean that the moving wire is connected to a circuit, and what happens when I move the wire, is that the area of the circuit increases or decreases, which certainly causes a Delta A x B

 

[Entanglement]

I agree that the case of the clock-hand (I take it that's what you mean by 'clock arrow') is a difficult one. It's not easy to apply the concept of change of flux linkage to it. [Indeed, the closely related rotating disc is one of the awkward cases cited in the Wiki article on electromagnetic induction that vanHees2 justifiably recommends.]
Nonetheless one can still apply the idea of flux CUTTING by the hand, or go back to basics and use the BEv method I mentioned in my previous post.
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. One way of looking at this is that the voltages induced in the hand and in the wires will be equal and in the same direction in space as each other, so in terms of the complete circuit, will cancel each other, being in opposite senses.
To measure a voltage you'd have to let Y brush against a stationary conductor and connect the voltmeter between this and X, so the voltmeter and the wires are stationary (in the laboratory frame of reference), while just the hand itself moves. [Principle of Faraday disc generator]

I'm unable to visualize that when a hand of a clock is rotating the area of a circuit increases? Where is the circuit in the clock?

 

[Philip Wood]

Re-read my last paragraph! You have to incorporate the hand in a circuit of your own making. And as I acknowledged in my first paragraph, even if you do this, applying to the circuit the idea of flux linkage changing is not straightforward and may lead to contradictions. But you can apply the concept of the hand CUTTING flux.

 

[Entanglement]

Re-read my last paragraph! You have to incorporate the hand in a circuit of your own making. And as I acknowledged in my first paragraph, even if you do this, applying to the circuit the idea of flux linkage changing is not straightforward and may lead to contradictions. But you can apply the concept of the hand CUTTING flux.

Can I An emf be induced without a flowing current if the circuit is open ?

 

[Philip Wood]

Yes. There will be no current if the circuit is open, but there will be an emf.[ Strictly you don't even have to have a physical circuit!, just a defined path - but I wouldn't start worrying about this!]

 

[vanhees71]

Sure, the magnetic field leads to the Lorentz force on the conduction electrons in the wire, leading to a charge separation due to the moving electrons,
\vec{F}=\frac{q}{c} \vec{v} \times \vec{B}.

 

[Entanglement]

[ Strictly you don't even have to have a physical circuit!, just a defined path - but I wouldn't start worrying about this!]

What do you mean ?

 

[alva]

From http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/genwir2.html

When a conductor moves through a magnetic field, there will be a generated motional emf. This is one example of Faraday's Law and it arises from the magnetic force (??). The voltage generated in a length of wire, presuming that the entire length moves through a uniform field, is given below.

If the velocity is perpendicular to the magnetic field then the generated voltage is given by the simple product:
Generated voltage = emf = Velocity x B-field x Length

You do not need any closed loop, nor any changing area enclosing the B field.

 

[Entanglement]

From http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/genwir2.html
You do not need any closed loop, nor any changing area enclosing the B field.

Do you mean just a moving wire ?

 

[alva]

Sure, the magnetic field leads to the Lorentz force on the conduction electrons in the wire, leading to a charge separation due to the moving electrons,
F⃗ =qcv⃗ ×B⃗ .

Assuming the force F is produced by an electric field, this field should be:

E = v/c \times B (1)

because F = E * q.

I have been looking for (1) but I could not find any reference. Could anyone help?

 

[alva]

Do you mean just a moving wire ?

Yes.

 

[Entanglement]

Yes.

how can I calculate the emf then ?

 

[alva]

how can I calculate the emf then ?

Applying the formula "Generated voltage = emf = Velocity x B-field x Length" to each dl of your clock-hand and computing the integral of this dV along the length of the clock-hand.

And do not ask me about maths.

 

[alva]

From http://web.mit.edu/sahughes/www/8.022/lec10.pdf

10.6
What is the magnetic field?
This “magnetic field” is, so far, just a construct that may seem like I’ve pulled out of the air.
I haven’t pulled it out of the air just for kicks — observations and measurements demonstrate
that there is an additional field that only acts on moving charges. But what exactly is this
field? Why should there exist some field that only acts on moving charges?
The answer is to be found in special relativity. The defining postulate of special relativity
essentially tells us that physics must be consistent in every “frame of reference”. Frames of
reference are defined by observers moving, with respect to each other, at different velocities.
Consider, for example, a long wire in some laboratory that carries a current I. In this
“lab frame”, the wire generates a magnetic field. Suppose that a charge moves with velocity
v parallel to this wire. The magnetic field of the wire leads to an attractive force between
the charge and the wire.

Suppose we now examine this situation from the point of view of the charge (the “charge
frame”). From the charge’s point of view, it is sitting perfectly still. If it is sitting still, there
can be no magnetic force! We appear to have a problem: in the “lab frame”, there is
an attractive magnetic force. In the “charge frame”, there can’t possibly be an attractive
magnetic force. But for physics to be consistent in both frames of reference, there must be
some attractive force in the charge frame. What is it?
There’s only thing it can be: in the charge’s frame of reference, there must be an
attractive ELECTRIC field. In other words, what looks like a pure magnetic field in one
frame of reference looks (at least in part) like an electric field in another frame of reference.
To understand how this happens, we must begin to understand special relativity. This is our
next topic.

And this electric field is what we call "emf".

 

[Philip Wood]

Beg to differ: (1) emf is work per unit charge, not force per unit charge (2) emf induced in a loop is line integral of force per unit charge around the loop, and an attractive force between wires has no component parallel to the wires!

 

[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.