Magnetic Field & Potential: Northward Motion of a Horizontal Rod

[leolaw]

A horizontal rod is moved northward at a constant velocity throough a magnetic field that points straight down. Which statement is true?

a) The West end of the rod is at higher potential than the east end.
b) The East end of the rod is at higher potential than the West end.
c) The top surface of the rod is at higher potential than the bottom surface.
d) The bottom surface of the rod is at higher potential than top surface.

By applying right hand rule #3, I have my finger pointing to the north, then curve my fingers downward since that the magnetic field is pointing downward. My thumb ends up pointing to the west. Since that this rule assuming that the particle is a positive charge, so I conclude that the west end of the rod is at a higher potential than the east.

Am I right?

 

[JFo]

yep... assuming the directions North, West, South, lie in a plane, and a component of B is perpendicular to that plane.

 

[Andrew Mason]

A horizontal rod is moved northward at a constant velocity throough a magnetic field that points straight down. Which statement is true?

a) The West end of the rod is at higher potential than the east end.
b) The East end of the rod is at higher potential than the West end.
c) The top surface of the rod is at higher potential than the bottom surface.
d) The bottom surface of the rod is at higher potential than top surface.

By applying right hand rule #3, I have my finger pointing to the north, then curve my fingers downward since that the magnetic field is pointing downward. My thumb ends up pointing to the west. Since that this rule assuming that the particle is a positive charge, so I conclude that the west end of the rod is at a higher potential than the east.

Am I right?

Applying the right hand rule, I get current flowing east to west, so the east is positive and the west negative (if the rod is aligned in east west direction). Thumb points north (motion), index finger down (field), second finger (current) points west.

AM

 

[JFo]

Applying the right hand rule, I get current flowing east to west, so the east is positive and the west negative (if the rod is aligned in east west direction). Thumb points north (motion), index finger down (field), second finger (current) points west.

AM

Yes but positive or negative current?

The direction of magnetic force on a positive charge is given by v X B so using the right-hand rule, the positive charges feel a force towards the west, and hence the negatives feel a force towards the east.

Hence positive current is from east to west, leaving the west end with a net positive charge, so the west hend is at a highter potential.

 

[Andrew Mason]

Yes but positive or negative current?

The direction of magnetic force on a positive charge is given by v X B so using the right-hand rule, the positive charges feel a force towards the west, and hence the negatives feel a force towards the east.

Hence positive current is from east to west, leaving the west end with a net positive charge, so the west hend is at a highter potential.

I never could get those conventions right.

AM

 

[Gamma]

What would be the answer if this is a closed loop?



Since current flows from east to west, than east is at a higer potantial than the west since current flows in the direction of the potantial drop??

 

[leolaw]

I wouldn't assume that it is a closed loop because it is a horizontal rod, which I think that it looks like the one we use for the compass.

 

[JFo]

gamma: for the horizontal rod, "current" really only flows for a fraction of a second before the the charge builds up on each end, and all forces cancel. Once this equillibrium is reached (assuming constant v and B) then there is no current, just a potential difference across the rod due to the charge separation.

I think for a closed loop, there would still be a charge separation allong the "width of the loop, just like the rod, but unless the B was changing as the loop moved, there would be no current since there is no change in flux, according to Faraday's law.