Question

Question dealing with magnetic field strength, posted below in a moment. (Sometimes people fall asleep after midnight, it happens.)

Answer 1

Where is it?

Answer 2

3hrs = 1 moment

Answer 3

@Kainui , feel free to have some chuckles at me when finals are over, which is pretty close. I'll post a different problem, though.

Answer 4

The prompt is here: http://i.imgur.com/0Kag1Ww.png

Answer 5

\[\frac{d \Phi_{B}}{dt}=-\mathcal{E}=Blv=L\frac{dI}{dt}\]

Answer 6

This seems like it very obviously includes the term Blv, just not sure what other term on the other side of the equality would be of much benefit.

Answer 7

I'm also just confused about the direction of the current, I understand how to get to the magnitude now using Blv and (stylized) E, but I'm not sure how to determine current direction in a case like this.

Answer 8

My best guess at the direction would be to just use the classic right hand rule.

If you put your fingers in the direction of 'v' and then curl your fingers into 'B' which is into the plane of the screen, you end up with a the force 'F' acting upwards.

My intuition seems to believe that since the bar is a conductor, it obviously contains free charges. These charges will be moving with the bar, so the charges are also moving in the direction of 'v'.

So if there is a magnetic force acting upwards, then it makes me think that the charges will be pushed upwards through the sliding bar. This would cause a counter-clockwise current. So the current would move from top-to-bottom through the resistor R.

I'm unsure if this explanation is physically accurate. I'll keep digging.

Answer 9

One second, just saw your reply. Reading it now.

Answer 10

It depends on how we see things, but I thought the cross hatches meant the field is going "downwards", into the page; I was taught using an arrow metaphor where the x's appear like the back of fins from a shot arrow traveling directly away from you.

Answer 11

Yes, that is what I did.

Answer 12

(Checking my answer to see if the right hand rule thing checks out)

Yeah, I gotchu, heh, I did the right-hand rule as you described and (assuming I'm not an alien/don't have polio) got the same result just confused about how you talked about it.

I tried to do the right-hand rule thing by myself earlier, but I know there are multiple versions of it and was a little worried about which was right.

Answer 13

(Oh, well.....I can't check for the answer since it's an even problem, heh, but this sounds right! Thank you.)

Answer 14

Yeah the best way is to actually curl your hand. Rather than pointing three fingers out, that always used to confuse me as well.

Just look at the cross-product and do it exactly in that order.

you need v x B

So finger go along v

Cross (curl) your fingers into B

You get F on your thumb

Answer 15

Alright. Also, could the work done by the agent pulling the bar (part b) be fairly describe as \[W = (qV \times B)(d)\]
For d being the distance traveled?

Answer 16

Nevermind, my book's answers says that the rate at which work is done by the force moving to the right is zero...I don't get why that is.

Answer 17

Zero... I'm not certain.

Answer 18

Oh, nevermind! It's not in the back of the book, but my instructor included the homework answers with workings online, we can definitely figure out what this is, one second.

Answer 19

http://i.imgur.com/ZThYXDT.png

Answer 20

I was gonna say, there's no way the work is zero haha

Answer 21

Alright, this confuses me, both problem 44 and problem 45 (exact same diagram, the only introduction in the second thing is instead of having that resistor, you have an ideal voltmeter in its place)-

They both include that part b question identically, asking the rate at which work is done by the agent pulling the bar;

45.) in the back of the book clearly states zero as the answer for part B, 44, pretty much the exact same problem with a resistor instead of a voltmeter, states that there is a rate at which work is being done, let me screencap all of this to make sure I'm not crazy.

Answer 22

Problems 44 & 45: http://i.imgur.com/MOjutAM.png

Answer Key, Ch. 27, Problem 45: http://i.imgur.com/Aw7DJK6.png

Solutions Manual, Ch. 27, Problem 44: http://i.imgur.com/ZThYXDT.png

Answer 23

I'm thinking, if you're simply replacing the resistor with a voltmeter attached exactly the same way (in series like the resistor), then there would be no work done moving the bar because there would be no current flowing through the voltmeter. An ideal voltmeter has infinite resistance, so no current will flow.

If no current flows, then there is no force acting on the metal bar against its motion (opposite the direction of v). So there is no force (in direction of v) needed to keep the bar moving at constant velocity.

Work is F dot dl. If F is 0, then work is 0.

Answer 24

Oh, derp. Thanks, that makes sense, I forgot about that entirely.

Answer 25

Thanks so much with this problem in general, by the way, you've been a huge help.

Answer 26

Always welcome

Answer 27

(Also, are you good with ODE stuff? I have a conceptual question more than a "do the math" one.)

Answer 28

I'm not sure what level "good" is, but I can take a look at it.

Answer 29

Alright, cool, I just tagged you in it.