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Hey,

I just stumbled apon this question in a past paper that im having trouble understanding,

http://img190.imageshack.us/img190/6266/78199113.jpg [Broken]

For a) its just straight forward isnt it, C = Aε/d even though they are moving.

b and c im having trouble imagining the direction of the current. If the charges are moving to the right at v, then that would also be the direction the current right? But each plate has a different charge, so the since the top plate is positive the current is to the right with v and as the bottom plate is negative the current is to the left.

Is that the correct way to see it?

Then would you take ampereian loops through the centre of each plate enclosing a section of the plate, so the v vector / I will pass through the loop. One way to explain it is that the area of the loop has elements which point in the y direction, to the right.

I'm getting a little stuck here, i keep thinking what i thought was wrong, but using the ampereian loop doesnt help because the top part of the loop will be outside, so itll get Bl=μI

Similarly for the bottom plate as well. And I don't really know what Ienc is, so i cant really say what B is because it doesnt seem right to leave the answer with Ienc.

In the end, adding the field inside gives B=-2uI/l and that would be in the -x direction.

Is that any where near correct?

for c)

q/A (vxB)= q/A vB and that would be in the positive z direction,

that seemed to easy so i must have missed somthing

d)

qE=q(vxB)

E=vxB

v= E/B?

e)

magnetic fields do no work, the work was done by electrostatic forces

I think b is the worst so far, but im sure i missed somthing in part c), 5 marks is way to many marks for such little work.

Thanks in advance for any help =]

I just stumbled apon this question in a past paper that im having trouble understanding,

http://img190.imageshack.us/img190/6266/78199113.jpg [Broken]

For a) its just straight forward isnt it, C = Aε/d even though they are moving.

b and c im having trouble imagining the direction of the current. If the charges are moving to the right at v, then that would also be the direction the current right? But each plate has a different charge, so the since the top plate is positive the current is to the right with v and as the bottom plate is negative the current is to the left.

Is that the correct way to see it?

Then would you take ampereian loops through the centre of each plate enclosing a section of the plate, so the v vector / I will pass through the loop. One way to explain it is that the area of the loop has elements which point in the y direction, to the right.

I'm getting a little stuck here, i keep thinking what i thought was wrong, but using the ampereian loop doesnt help because the top part of the loop will be outside, so itll get Bl=μI

_{enc}then inside itll get negative that because the dl is facing the opposite direction, and the sides of the loop will give zero since their dl is perpendicular to B.Similarly for the bottom plate as well. And I don't really know what Ienc is, so i cant really say what B is because it doesnt seem right to leave the answer with Ienc.

In the end, adding the field inside gives B=-2uI/l and that would be in the -x direction.

Is that any where near correct?

for c)

q/A (vxB)= q/A vB and that would be in the positive z direction,

that seemed to easy so i must have missed somthing

d)

qE=q(vxB)

E=vxB

v= E/B?

e)

magnetic fields do no work, the work was done by electrostatic forces

I think b is the worst so far, but im sure i missed somthing in part c), 5 marks is way to many marks for such little work.

Thanks in advance for any help =]

## Homework Statement

## Homework Equations

## The Attempt at a Solution

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