- #1
RoyalCat
- 671
- 2
Hello everyone,
This question is the third question from the 2001 IPhO:
Problem text: http://www.jyu.fi/tdk/kastdk/olympiads/2001/IPhO2001-THEO.pdf
Solution text: http://www.jyu.fi/tdk/kastdk/olympiads/2001/IPhO2001-THEO-SLN.pdf
I'm having trouble understanding how they came to the conclusion that the current is in the [tex]-\hat y[/tex] direction. (Avoidance of a runaway process aside)
I came to the conclusion that it's in the [tex]+\hat y[/tex] direction.
My thought process is as follows:
[tex]\vec F_{Lorentz}=0[/tex]
[tex]\vec E = qvB \hat y[/tex]
[tex]\vec J = \frac{\vec E}{\rho}[/tex]
[tex]I=\vec J \cdot \vec A = \frac{E}{\rho}Lh[/tex]
Now all this is still in the [tex]+\hat y[/tex] direction.
I feel like I should somehow acknowledge the fact that the circuit is shorted out outside the material, and that that's the reason the current in the material goes in the opposite direction, but that doesn't seem right.
Any advice on what I've missed would be greatly appreciated. :)
This question is the third question from the 2001 IPhO:
Problem text: http://www.jyu.fi/tdk/kastdk/olympiads/2001/IPhO2001-THEO.pdf
Solution text: http://www.jyu.fi/tdk/kastdk/olympiads/2001/IPhO2001-THEO-SLN.pdf
I'm having trouble understanding how they came to the conclusion that the current is in the [tex]-\hat y[/tex] direction. (Avoidance of a runaway process aside)
I came to the conclusion that it's in the [tex]+\hat y[/tex] direction.
My thought process is as follows:
[tex]\vec F_{Lorentz}=0[/tex]
[tex]\vec E = qvB \hat y[/tex]
[tex]\vec J = \frac{\vec E}{\rho}[/tex]
[tex]I=\vec J \cdot \vec A = \frac{E}{\rho}Lh[/tex]
Now all this is still in the [tex]+\hat y[/tex] direction.
I feel like I should somehow acknowledge the fact that the circuit is shorted out outside the material, and that that's the reason the current in the material goes in the opposite direction, but that doesn't seem right.
Any advice on what I've missed would be greatly appreciated. :)