What direction should the electric field point in a velocity selector?

AI Thread Summary
In a velocity selector, a charged particle moves through crossed electric and magnetic fields, with forces acting in opposite directions. The magnetic field is directed north, and the particle moves east, leading to the conclusion that the electric field should point out of the page if considering positive charges. However, the direction of the electric field can also be analyzed for negative charges, which would result in it pointing into the page. Clarification arises from the perspective of the coordinate system used, where East and North are treated as parallel to the ground. Ultimately, the correct direction of the electric field is determined by the charge of the particle and the chosen coordinate system.
cdymdcool
Messages
10
Reaction score
0
A velocity selector consists of a charged particle passing through crossed electirc magnetic fields. The forces exerted by these fields are in opposite directions, and only particles of a certain velocity will move in a straight line. In the following, disregard the magnitudes of the fields. In a velocity selector, the particles move toward the east and the magnetic field is directed to the north. What direction should the electric field point?

Use the right hand rule the B-field force is into the page, so the elctric field force should be out of page. But we don't know the particle's charge how do we determine the direction of the electric field then?
 
Physics news on Phys.org
cdymdcool said:
Use the right hand rule the B-field force is into the page, so the elctric field force should be out of page. But we don't know the particle's charge how do we determine the direction of the electric field then?
Since you don't know the charge, how did you determine the direction of the magnetic force?

Hint: Solve it for postive particles then for negative particles. Compare the two solutions.
 
So the B-field Force is going out of the page, the electric field should be the opposite, going into the page. If particle is positive or negative, the electric field point will both going into the page? But the answer key says down, is it from different perspective?
 
To avoid ambiguity, let's use the y-axis for North and the x-axis for East. Assuming East and North (the x-y plane) are parallel to the ground (like usual, on the earth), then the +z-axis would be up. (If your paper is the x-y plane, then the +z-axis would be out of the page.)
cdymdcool said:
So the B-field Force is going out of the page, the electric field should be the opposite, going into the page.
My point was that you cannot know the direction of the magnetic force, since it depends on the sign of the charge. But it doesn't matter! If the particle was positive, then the magnetic force would be in the +z direction and the electric field must point in the -z direction.
If particle is positive or negative, the electric field point will both going into the page?
Figure out what direction the electric field must point if the charge is negative.
But the answer key says down, is it from different perspective?
It looks like they are taking East and North as parallel to the ground, not in the plane of the paper. (See my comments above.)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top