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# Flemming's Left Hand Rule Confusion watch

1. I really dont understand flemmings left hand rule, usually they give you one plane it acts down so surely it would rotate around that plane?

see question

So my first finger is parallel to the lines my thumb is pointing at 90 degrees but it rotates around the first finger so is not always in one direction, and if I rotate around the first finger my second can end up where my thumb is, this makes no sense.
2. I'm thinking it's the first finger, as you said, parallel to the lines shown, then if the electron is moving up the page, the current is moving downwards (I think?). Do this, and you should have your thumb (the force) should be coming out of the screen towards you.
3. (Original post by vandub)
I'm thinking it's the first finger, as you said, parallel to the lines shown, then if the electron is moving up the page, the current is moving downwards (I think?). Do this, and you should have your thumb (the force) should be coming out of the screen towards you.
Oh like the electrons moving from negative to positive.
4. Q5 doesnt even work too, you need to find r, they dont give v. So I made a subsitution which doesnt work and got me 10^-25 out.
5. The current is conventional current flow (positive to negative) so the flow of current is opposite to the way the electron is moving as the electron is negatively charged. Therfore by flemmings left hand the force is coming out of the paper.

Not enough information to help question 5.
6. (Original post by Oh my Ms. Coffey)
Oh like the electrons moving from negative to positive.
Yep.
And that bit about Q5 would be easier to help with if you showed the question it's related to.
7. (Original post by vandub)
Yep.
And that bit about Q5 would be easier to help with if you showed the question it's related to.
Question

My working

8. (Original post by Oh my Ms. Coffey)
Question
Okay, so I did Q2 without problem (I think) and got F=0.655N.

For Q5, I thought it'd be to do with circular motion, so I started with the equation F=ma=mv^2/r, then r=mv^2/F. Knowing m and F, I went the same way you did, to get v=F/QB, buuut, this is where I wasn't sure. The force I found earlier, is dependant on the current, yet now we're using the charge on a single electron? Anyway, I went ahead and tried it, putting r=mF^2/(B^2Q^2F)=mF/(B^2Q^2).

Soo, I checked the units at this point to see if they're consistent, and they aren't. So I'm completely stumped on this.

Sorry I couldn't help. Out of interest, what exam board are you on?!

Oh, and sorry for the late reply to this.
This is clearly a misprint, it should be in Qu 3.
Then it's just a standard case of centripetal force = Bqv

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