# Finding the direction of the force (right hand rule)

1. May 11, 2007

### t_n_p

1. The problem statement, all variables and given/known data

http://img329.imageshack.us/img329/9304/righthandcq2.jpg [Broken]

3. The attempt at a solution

For this method I was taught that the direction of the fingers points in the direction of velocity, the thumb in the direction of the magnetic field, and the force would shoot out of your palm. For this question I noticed it wants us to find the direction of the force on a negative charge, does that mean I would go about the situation any differently.

If I use the rule I have stated above, I know some of the answers are conflicting in consistancy. Would somebody be able to help me straighten this out?

Thanks

edit: velocity is the green color, the direction of the magnetic field is in blue and my suggested answer for direction of force is in black/greylead

Last edited by a moderator: May 2, 2017
2. May 11, 2007

### Hootenanny

Staff Emeritus
The right hand rule is a handy () rule for determining the direction of crossed vectors and follows directly from the definition of the cross product. So, a x b produces a vector that is orthogonal to both a and b. Lets look at the magnetic component of the Lorentz force;

$$\vec{F} = q\cdot\left( \vec{v}\times \vec{B}\right)$$

So, your right hand rule, gives you the relative direction of the force acting on a positive particle (i.e. q >0) travelling at velocity v in a magnetic field B. So now, if we have a negatively charged particle we can write;

$$\vec{F_{-}} = -|q|\cdot\left( \vec{v}\times \vec{B}\right)\hspace{1cm}\left\{q: q\leq0\right\}$$

This negative sign implies that the force acts in the opposite direction to a positive charge. So using your right hand rule, for a negative charge, the force would come out of the back of your hand.

I hope that makes sense

Last edited: May 11, 2007
3. May 11, 2007

### t_n_p

that makes alot of sense now! So with that in mind, here are my corrections
a) right
b) right
c) down
d) out of page
e) zero
f) up

correct?!

4. May 11, 2007

### Hootenanny

Staff Emeritus
I'm not quite sure of your diagrams, if you could write them out something like this;

(a)
Direction of magnetic field: In/out of the page
Direction of Velocity: Left / right / up /down
Direction of Force: Left / right / up /down

It may be a little easier to follow.

5. May 11, 2007

### t_n_p

no worries!
a) magnetic field is down, velocity is out of page, force is right
b) magnetic field is into page, velocity is down, force is right
c) magnetic field right, velocity is into of page, force is down
d) magnetic field is up, velocity is right, force is out of page
e) magnetic field is right, velocity is left, force is zero
f) magnetic field is out of page, velocity is left, force is upwards

6. May 11, 2007

### t_n_p

thanks for the speedy replies. I've got another similar question, but the wording on this one is a bit different.

"Determine the direction of B for each case where F represents the maximum magnetic force on a positively charged particle moving with velocity V"

I'm weary of the term "positively charged particle". I'm assuming I have to change the direction of the velocity to the opposite side? (That is if it is originally left, the velocity will then be to the right)

The original velocities are used in the following:
a) F is out of page, V is left, B is upwards
b) F is upwards, V is right, B is into the page
c) F is dowards, V is into the page, B is left.

7. May 11, 2007

### Hootenanny

Staff Emeritus
Correction, sorry I was working in postive charges. All your answers would be correct if we were working with a positve charge. However, you need to reverse the direction of all your forces, as we are working with a negative charge, sorry for the confusion!

8. May 11, 2007

### t_n_p

so the arrow comes out the front of my palm when working with a negative charge and the back of my palm when working with a positive charge. point taken.

9. May 11, 2007

### Hootenanny

Staff Emeritus
Sorry, your either using a different right hand rule to me, or using it wrong. Which on of these are you using http://www.physics.brocku.ca/faculty/sternin/120/slides/rh-rule.html [Broken] ?

Last edited by a moderator: May 2, 2017
10. May 11, 2007

### t_n_p

the third one, however i think i may have been using b as thumb and fingers as V instead. But using this method, it seems a) should be right etc...

This is so confusing!

11. May 11, 2007

### Hootenanny

Staff Emeritus
That is your problem. So, now using your thumb as the velocity, your fingers as the magnetic field, the force on a positive particle comes out of you palm; you should reverse the direction for a negative charge.

This can be quite confusing, so I suggest you chose one rule and stick with it. I myself use the RH screw rule. Imagine turning a RH (normal) screw from the velocity vector, to the magnetic field vector. The direction the screw would move, will be the direction of the force on a positive particle. It is important however, to find a method that works for you and stick with it.

12. May 11, 2007

### t_n_p

Phew! It all makes so much sense now! I'll definately stick to the thumb/fingers/palm method, the "screw" method really makes me :yuck: :surprised

Could you now help me with the "positively charged" question? positive charge means coming out of my palm?

13. May 11, 2007

### Hootenanny

Staff Emeritus
I'm not entirely sure what the question means by "maximum force" since, the magnitude of F will always maximal provided $\vec{v}\bot \vec{B}$.

14. May 12, 2007

### t_n_p

I suppose maximum force is just force then!

15. May 12, 2007

### t_n_p

*bump*
just need to finally clear this one up.
for a force of a positively charged particle will the force be coming out of my palm if im using the thumb as v, fingers as B, palm as force?

16. May 12, 2007

### Hootenanny

Staff Emeritus
Correct

See the bottom diagram in the link I gave you