[Antimony]

Hi,

I don't really understand this question. There is only an emf induced when there is a rate of change of flux linkage but in this question the wheel is spinning at a constant rate so I thought this meant that there was no change in the amount in flux lines being cut so there is no emf induced but the mark scheme says that there is a rate of change of flux /flux linkage due to spoke's motion.

Could someone please explain why

Thanks

[Stonebridge]

If you have learned this as being due to lines of flux being cut, the rule is that the emf is equal to the flux cut per second. (Not the rate of change of the flux cut)
You calculate the area A swept out by a spoke in one second and multiply it by B to get the flux cut.
As the wheel is spinning uniformly, the spokes cut a constant amount of flux every second. They sweep out a constant area per second.

I find students often get confused by this. You can consider an emf being generated when there is a change of flux linking an area. The emf then depends on the rate of change of flux linkage.
You can also think of the emf as being due to a wire cutting through lines of flux. In this case the emf depends on the rate at which the flux is cut.
You should have been shown at some point that the two ways of thinking of this are equivalent.
In this case, it's easier to think in terms of the amount of flux cut per second.



The direction of the induced emf is given by the Right Hand Rule. The rule gives the direction of the induced emf (centre finger) by considering the field (into the page, pointer finger) with the direction of motion of the spoke (thumb).

[Antimony]

(Original post by Stonebridge)
You can consider an emf being generated when there is a change of flux linking an area. The emf then depends on the rate of change of flux linkage.

This is the reason an emf is induced in transformers right? Linking: no movement?

I'm a bit confused about the difference between flux linkage (= BAN) and flux (=BA) I thought you could think of it in the same way (?).

Also when square coil travels through a uniform magnetic field with constant velocity, there is movement and there is a rate of flux cut per second but how come an emf is induced only when it enters and leaves the field and not when completely inside it? What's the best way to look at it, flux linking or cutting?

Thanks again

[rub em out]

(Original post by Antimony)
This is the reason an emf is induced in transformers right? Linking: no movement?

I'm a bit confused about the difference between flux linkage (= BAN) and flux (=BA) I thought you could think of it in the same way (?).

Also when square coil travels through a uniform magnetic field with constant velocity, there is movement and there is a rate of flux cut per second but how come an emf is induced only when it enters and leaves the field and not when completely inside it? What's the best way to look at it, flux linking or cutting?

Thanks again

emf is induced in transformers as the coil moves relative to the field, cutting the lines of flux. emf can be thought of as the rate at which these lines are being cut. Note that where the sides of the coil move parallel to the field, or the direction of motion of the coil is parallel to the field there will be no induced emf.

Flux and Flux linkage definitely confuse me too and correct me if im wrong but I think of it that flux is the amount of field lines in one coil while flux linkage is just the total amount of flux through 'n' coils.

Not sure what you mean when you talk about the square coil example, as far as i'm aware the position of the coil in the field makes no difference as long as flux is cut at the same rate

[Stonebridge]

(Original post by Antimony)
This is the reason an emf is induced in transformers right? Linking: no movement?

I'm a bit confused about the difference between flux linkage (= BAN) and flux (=BA) I thought you could think of it in the same way (?).

Also when square coil travels through a uniform magnetic field with constant velocity, there is movement and there is a rate of flux cut per second but how come an emf is induced only when it enters and leaves the field and not when completely inside it? What's the best way to look at it, flux linking or cutting?

Thanks again

With transformers there is no movement so it's easiest to think in terms of flux linkage.

Flux linkage is BA times the number of turns in the coil, N. That's why it's BAN if there is more than one turn. The flux linking an area is BA. If there are many turns, then the same flux links all of them and the effect is additive.

The question you are referring to regarding the square coil moving across a field has come up on here many times.
The way to think of emf generation is one of two equivalent ideas. Either flux cutting or change of flux linking an area. The two give the same answer, but one is usually more sensible to use than the other for a given situation.
With a single wire moving through a field it is usually more useful to think in terms of the flux cutting. With a coil that has an area it is usually easier to think in terms of flux linking an area.
In the question with the square coil moving through the field, you can think in terms of the flux linking that coil's area changing as it moves further into the field. This causes an emf. When the coil is fully in the field the flux no longer changes so no emf.
If you think in terms of flux cutting (probably the more tricky way with this particular example) when the coil is fully in the field, the leading and trailing edge both cut through lines of flux and so both have an emf generated in them. If you do the Right Hand Rule you will notice that the emfs are in opposite directions in the coil and cancel out, giving no overall emf in that coil.

[Stonebridge]

(Original post by rub em out)
emf is induced in transformers as the coil moves relative to the field, cutting the lines of flux. emf can be thought of as the rate at which these lines are being cut. Note that where the sides of the coil move parallel to the field, or the direction of motion of the coil is parallel to the field there will be no induced emf.

Flux and Flux linkage definitely confuse me too and correct me if im wrong but I think of it that flux is the amount of field lines in one coil while flux linkage is just the total amount of flux through 'n' coils.

Not sure what you mean when you talk about the square coil example, as far as i'm aware the position of the coil in the field makes no difference as long as flux is cut at the same rate

There is no moving coil in a transformer. You are getting it confused with a dynamo.

[Antimony]

Sorry for bringing this up again but I went over the question again and I don't understand the part a) (i) From FRHR, the second finger points upwards so the negative should be upwards right as it conventional? but apparently the centre is negative

[Stonebridge]

(Original post by Antimony)
Sorry for bringing this up again but I went over the question again and I don't understand the part a) (i) From FRHR, the second finger points upwards so the negative should be upwards right as it conventional? but apparently the centre is negative

The pointer finger (field) points into the page. The question says the field "acts into and perpendicular to the paper".
If you consider a spoke at the bottom of the wheel, it is moving to the left. Thumb points to the left.
The centre finger gives the direction of the (+ive) emf, which is downwards towards the rim.

If you do the same for any spoke, the emf is always towards the rim.

[Antimony]

(Original post by Stonebridge)
The pointer finger (field) points into the page. The question says the field "acts into and perpendicular to the paper".
If you consider a spoke at the bottom of the wheel, it is moving to the left. Thumb points to the left.
The centre finger gives the direction of the (+ive) emf, which is downwards towards the rim.

If you do the same for any spoke, the emf is always towards the rim.

Thanks, doesn't the centre finger indicate the direction of conventional current? Or is that just for FLHR and for FRHR, the centre finger indicate the +ve end of emf and so flow of electrons?

[Stonebridge]

(Original post by Antimony)
Thanks, doesn't the centre finger indicate the direction of conventional current? Or is that just for FLHR and for FRHR, the centre finger indicate the +ve end of emf and so flow of electrons?

In the LHR the centre finger points in the direction of flow of conventional (+ive) current. This rule is nearly always then used to find the direction of the force (thumb) on that current when in a field (pointer finger). That's why it is often referred to as the LH (Motor) Rule.
In this rule the current is already flowing and we want the force acting on it.

In the RH Rule the centre finger gives the direction of the conventional (+ive) induced emf, and the direction of flow of any positive current if there is a circuit for it to move around. The thumb is the direction of motion of the wire that is being moved across the field (pointer finger).
This rule is used to find the direction of induced (+ive) emfs when you move a wire through a magnetic field. That's why it's often called the RH (Dynamo) Rule.

[Antimony]

(Original post by Stonebridge)
In the LHR the centre finger points in the direction of flow of conventional (+ive) current. This rule is nearly always then used to find the direction of the force (thumb) on that current when in a field (pointer finger). That's why it is often referred to as the LH (Motor) Rule.
In this rule the current is already flowing and we want the force acting on it.

In the RH Rule the centre finger gives the direction of the conventional (+ive) induced emf, and the direction of flow of any positive current if there is a circuit for it to move around. The thumb is the direction of motion of the wire that is being moved across the field (pointer finger).
This rule is used to find the direction of induced (+ive) emfs when you move a wire through a magnetic field. That's why it's often called the RH (Dynamo) Rule.

Thank you

[Parthenon93]

(Original post by Stonebridge)
The pointer finger (field) points into the page. The question says the field "acts into and perpendicular to the paper".
If you consider a spoke at the bottom of the wheel, it is moving to the left. Thumb points to the left.
The centre finger gives the direction of the (+ive) emf, which is downwards towards the rim.

If you do the same for any spoke, the emf is always towards the rim.

I get the emf towards the rim. But if the induced/conventional current is towards the rim, shouldn't the rim be negative? Isn't conventional current from positive end of battery towards negative end of battery?

[Stonebridge]

(Original post by Parthenon93)
I get the emf towards the rim. But if the induced/conventional current is towards the rim, shouldn't the rim be negative? Isn't conventional current from positive end of battery towards negative end of battery?

This does seem to cause a problem.
Rather than try to explain this here now, it will probably be more useful if you take a look at this current (just finished!) thread where the same question has come up and there has been plenty of discussion, with, I hope, a satisfactory explanation.
http://www.thestudentroom.co.uk/show....php?t=2025296
If it still doesn't make sense come back here and ask again.

[Parthenon93]

(Original post by Stonebridge)
This does seem to cause a problem.
Rather than try to explain this here now, it will probably be more useful if you take a look at this current (just finished!) thread where the same question has come up and there has been plenty of discussion, with, I hope, a satisfactory explanation.
http://www.thestudentroom.co.uk/show....php?t=2025296
If it still doesn't make sense come back here and ask again.

The line - if there is no battery, the rod IS the battery - helped everything.

Using right hand rule, induced/conventional current is from P to Q. So electrons must move in opposite direction from Q to P.

So if electrons are moving from Q to P, P would become negative and hence have lower potential.

Oooooh now I get this one!!!

Since induced current is from hub to rim, electrons must move in the opposite direction from rim to hub. Considering the spoke as the battery itself - the hub becomes negative due to the collection of electrons.

Many many many many many thanks!!! God know how many marks I might have lost due to this!!! I think I might love you :') Thanks again!

PS: I think I faced the problems because I considered the conventional current direction as flowing outside the spoke/rod. In a real battery, the electrons are prevented from flowing directly between the terminals and so electrons have to flow externally . Whereas, the rod/spoke is conducting and so electrons can flow internally - and hence all the confusion