# Magnetic Force/Field, Current, Inclined Plane Problem

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1. Oct 16, 2015

### Callix

Hello! I just need someone to check my work here. I already turned in the homework with my hunch, but now I just want to check my work because it is killing me not knowing 100% :) I am having strong conflicting thoughts on particular parts and thought I'd post it here for clarification. I will post the question and my solution and then ask the question that needs clarification.

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

A metal bar with a length L has a mass of m. Initially it is placed on an incline θ above the horizontal. There is a magnetic field T that is directed upward around the bar and the metal contacts along the incline allow a current through the bar. If the coefficient of static friction is μ,

a). What are the minimum magnitude and the direction of the current that must pass through the bar to keep it from moving once released?

b). Is there a maximum current that can be used in that same direction that will also hold the bar in place?

2. Relevant equations
Presented in the solution

3. The attempt at a solution
The first thing I notice is that B is perpendicular to I. Following the right hand rule, I can determine that F is either into the inclined plane (not up it, but into) or away from the inclined plane depending on the direction of the current. If the current is into the page, then the magnetic force is into the inclined plane. If it is out of the page, then away. I chose the convention that the magnetic force is into the inclined plane. Therefore I can create the following FBD:

*Everything in blue is related to the magnetic field, force, and current.

I can check to see if the current, field, and force are correct by doing the right hand rule. And due to the fact that I already know that the current and magnetic field are perpendicular.

Now, I can set up the following equations to solve for the minimum current necessary to create a magnetic force that will keep the bar static.
$$\sum F= F\sin(\theta)-F_f-mg\cos(\theta)=0$$
$$F_{magnetic}=BIL\sin(\phi) \rightarrow \phi=90^o \rightarrow F_{magnetic}=BIL$$
$$F_{friction}=\mu mg\cos(\theta)$$
$$\sum F=(BIL)\sin(\theta)-\mu mg\cos(\theta)-mg\sin(\theta)=0$$
$$\rightarrow BIL\sin(\theta)=\mu mg\cos(\theta)+mg\sin(\theta)$$
$$\rightarrow I=\frac{\mu mg\cos(\theta)+mg\sin(\theta)}{BL\sin(\theta)}$$

Now my question is: Is this equation correct and is it true that the component UP the incline is $F\sin(\theta)$? Because if it wasn't and it was just F, that would mean that B would have to be perpendicular to the surface of the inclined plane using the right hand rule. Given the diagram in the initial problem, this is not the case. Therefore the force must be either into the plane or away from it depending on the direction of the current.

**Also, for B I'm a bit confused. The maximum value the current would be, would be the force that JUST ALMOST causes the bar to move. As soon as it moves, it would experience kinetic friction. So I could set up the equation using kinetic friction but I don't have kinetic friction. How would I go about determine this?

Thanks for your time and any help would be greatly appreciated!

-Callix

2. Oct 16, 2015

### Staff: Mentor

I don't see where you've accounted for the component of the magnetic force that will increase the friction (Think of it as another "gravity" that's working horizontally instead of vertically. There will be a component along the slope and one normal to the slope). And I think you should check your direction for the friction force. The static friction will oppose the forces that are trying to make the block slip. So if the block is just on the verge of slipping downwards on the wedge, what direction will the static friction force point?

As the current increases, the magnetic "gravity" will increase. So you'd expect any frictional force to increase, too. The question is, will the increase in the "upslope" component of the magnetic force outpace the increased normal force and hence the friction. Which way will the friction vector be pointing when the block is poised to move upslope?

3. Oct 17, 2015

### Callix

Ah yes you're right about the sign for the friction component. Hmm, I don't understand what you mean about the component of the magnetic force. In my diagram I chose for the magnetic force to go into the inclined plane. So the component that would go up the inclined plane would be $F\sin(\theta)$ wouldn't it? Actually, wouldn't it be a cos instead?

So the second case is actually what I had for the first case then, right? Because when it's right about to move, friction is about to change to down the slope rather than up the slope. So it should be opposite of the force where as in the first case it should be up the slope with the magnetic force.

4. Oct 17, 2015

### Staff: Mentor

The magnetic force is horizontal, so there will be vector components of that force which are parallel to the slope and perpendicular to the slope ("normal" force"). We do the same thing with the gravitational force.

You should be able to work out where to use sine or cosine from the figure above.
I think you've got it. Essentially static friction opposes the net resultant of any other forces that want to make the object move along the surface.

You can imagine a situation where the gravitational and magnetic forces are balanced in such a way that there is no net upslope or downslope force. The system would be in equilibrium with or without friction. In such an equilibrium state the static friction would be zero, since the object isn't "trying" to move. Now suppose the magnetic force were to increase slightly so there's a net upslope force force resulting from the sum of the two field forces (gravitational and magnetic fields). The static friction would then operate to prevent movement, so it would be directed downslope. In a similar manner, if the magnetic force were to decrease a tad and there was a net downslope force, the static friction would point upslope to oppose it and thus prevent motion. The object actually moves when the net upslope or downslope force exceeds the maximum static friction force.

Of course, any frictional forces will vary with the size of the net normal force. So be sure to take into account the normal forces due to both field forces.

5. Oct 20, 2015

### Callix

Sorry for the late reply, but thank you very much! This makes complete sense now! :)