Right-hand rule regarding magnetic fields

[clairez93]

Homework Statement

Determine the initial direction of the deflection of charged particles as they enter the magnetic fields as shown in Figure P29.1 (attached).

Homework Equations

I don't think there are any.

The Attempt at a Solution

I believe this problem will need to make use of the right-hand rule, if I'm not mistaken. However I don't understand how to use the right-hand rule. Could someone please help me to understand how to use it? Thank you.

 

[kreil]

This site ought to help:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html#c3

 

[berkeman]

This site ought to help:

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html#c3

Very helpful webpage, kreil. Thanks.

 

[clairez93]

I'm still having trouble grasping it. In my problem, for part b, I would put my finger sin the direction of upwards, and then curl it where?

 

[kreil]

Point your thumb in the direction of particle velocity and your fingers in the direction of the field. Then, for (b) you would get the force on the particle as into the page (direction that your palm faces).

 

[clairez93]

That doesn't quite work for part (c), however. Because if I can't point my thumb in the direction of the velocity of the particle and have my fingers poniting the opposite way...

 

[berkeman]

Point your thumb in the direction of particle velocity and your fingers in the direction of the field. Then, for (b) you would get the force on the particle as into the page (direction that your palm faces).

Not in the direction of the particle velocity, necessarily. It's in the direction of qV. So if the particle has a negative charge, you would point your thumb in which direction?

 

[berkeman]

That doesn't quite work for part (c), however. Because if I can't point my thumb in the direction of the velocity of the particle and have my fingers poniting the opposite way...

What's the cross product between two parallel (or antiparallel) vectors?

 

[clairez93]

What's the cross product between two parallel (or antiparallel) vectors?

I think it is 0.

 

[cryptoguy]

Right. so the force in part C is 0

 

[clairez93]

Oh, I see.
Then what about part d? How would that one work?
I'm still confused.

 

[cryptoguy]

well is there at a component of the magnetic field that is perpendicular to the motion of the proton? If so, you can use the right hand rule and find out the force

 

[clairez93]

The x component would be perpendicular to the motion. So do my fingers go in the direction of the x component and then my thumb with the red arrow from the particle?

 

[cryptoguy]

yessir