Electromagnetism (Generator Effect)

[Jefferson]

Hi all,
Could I receive some clarity for the following question (in attachment).

"For each wire ... label the +ve and -ve ends".

Taking (i) as an example,

• The conductor is moving upwards.
• Magnetic field is into the page.
• Therefore, the direction of the induced current is from right to left (using whichever hand rule).

From this point, I'm not sure which side will be positive/negative.
Given that electrons flow from the left to right (opposite to conventional current), is it:

(a) The right side will be negative, because that's where all the electrons will end up (i.e. they move to the right to make it negative).
∴ Left is positve, right is negative.

or

(b) The induced emf makes the right side positive, which is why the electrons will move there (i.e. the electrons are attracted to the right because it's positive)
∴ Left is negative, right is positive.

Also, does the "assume the circuit is completed externally" play any part in this answer?

[fun_jirachi]

Hey there!

Not too sure, but for i) the velocity is up, the field is into the page and then by the right hand palm rule the conventional current moves to the right. Since the flow of conventional current is right, then I'd assume the right is negative and the left is positive.

[Jefferson]

Hey there!

Not too sure, but for i) the velocity is up, the field is into the page and then by the right hand palm rule the conventional current moves to the right. Since the flow of conventional current is right, then I'd assume the right is negative and the left is positive.

Hi, I think the right hand palm rule is used to determine the force produced from the motor effect.
To get an induced current, you would use another rule (generator effect).

[fun_jirachi]

I still use the right hand rule for that, but I orientated my hand wrong.

So basically, the conventional current should go left. (my bad)
In that case, the opposite is true; ignore everything I just said . What this means is that because conventional current goes left, the flow of negative charge is to the right, and in that case b) is true.

Sorry for the mistake

[Jefferson]

I still use the right hand rule for that, but I orientated my hand wrong.

So basically, the conventional current should go left. (my bad)
In that case, the opposite is true; ignore everything I just said . What this means is that because conventional current goes left, the flow of negative charge is to the right, and in that case b) is true.

Sorry for the mistake

Hi again,
That's what I thought as well, but I've been getting conflicting answers from different people.
One diagram that supports (a) is this one (in attachment), which shows the electron moving towards the positive pole of the conductor.

Edit: Attachment 2 says otherwise.
Edit 2: Attachment 3 comes with an explanation :O.

[jamonwindeyer]

Hey Jefferson! Using (i) as the example, I think your understanding is correct either way. You know the electrons are moving to the right - You can determine that by right hand slap rule. Field into the page, motion upwards - Resultant force on a positive change would be to the left, so the electrons experience a force to the right. That bit is without confusion, and that's the key bit tbh! Like, all of your diagrams say the exact same thing in terms of where the electrons are moving, I don't think they conflict each other at all.

The issue isn't really about your knowledge of induction, it's about the meaning of a '+'ve end and a '-'ve end. The electrons accumulate on the right hand side of the rod, meaning it will carry a negative charge on that side. Thus, that side is negative, the '-' end. In the context of electrical circuits though, you'd consider where the electrons are flowing to be the '+' end, because electrons always flow towards the positive terminal.

The "circuit completed externally" bit actually matters here, because if we have a circuit charge won't accumulate anywhere. It will just flow around the circuit in a certain direction (in this case, from right to left through the rod). In that case I guess the argument could be made that the '+' end is the right side of the rod, but I don't feel like that is what the question is driving at? I think it wants charge accumulation, not polarity.

Again, semantics. I think the question is a little poorly worded (it's tricky to word these things sometimes)

[Jefferson]

Hey Jefferson! Using (i) as the example, I think your understanding is correct either way. You know the electrons are moving to the right - You can determine that by right hand slap rule. Field into the page, motion upwards - Resultant force on a positive change would be to the left, so the electrons experience a force to the right. That bit is without confusion, and that's the key bit tbh! Like, all of your diagrams say the exact same thing in terms of where the electrons are moving, I don't think they conflict each other at all.

The issue isn't really about your knowledge of induction, it's about the meaning of a '+'ve end and a '-'ve end. The electrons accumulate on the right hand side of the rod, meaning it will carry a negative charge on that side. Thus, that side is negative, the '-' end. In the context of electrical circuits though, you'd consider where the electrons are flowing to be the '+' end, because electrons always flow towards the positive terminal.

The "circuit completed externally" bit actually matters here, because if we have a circuit charge won't accumulate anywhere. It will just flow around the circuit in a certain direction (in this case, from right to left through the rod). In that case I guess the argument could be made that the '+' end is the right side of the rod, but I don't feel like that is what the question is driving at? I think it wants charge accumulation, not polarity.

Again, semantics. I think the question is a little poorly worded (it's tricky to word these things sometimes)

Thank you so much for the detailed explanation, That makes a lot of sense.