Am I belitting this question? Magnetic force

[flyingpig]

Homework Statement

http://img710.imageshack.us/img710/9295/roachj.th.png

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http://img146.imageshack.us/img146/6360/61565460.th.png

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The Attempt at a Solution

The vertical conductors does nothing because their forces are only horizontal. The horizontal conductor will produce a force (up or down).

Now the problem is that I got a B-field going into the page at the top of the horizontal wire and a B-field going out.

I chose the B-field going out so that the magnetic force would be up and hence opposite to mg because I feel like that is what I should get at the end...

Anyways here is the math

$$F_b = F_g$$

$$IlB=mg$$

$$B = \frac{mg}{IL}$$

Now the problem is that this is too simple...I feel like I overlooked something big. For one I am not 100% sure why must the B-field be going in. I mean it works, but I don't know why I must choose that one

 

[flyingpig]

Anyone...? Just anything is okay...

 

[gneill]

The problem states that there is a separate uniform magnetic field that is perpendicular to the page. Think of the current in the wire as moving electric charges. The direction of the force that results on a moving charge in a magnetic field is determined by a cross product (v x B).

So, if v is in the direction of the current, and you want the resultant of the cross product to be upwards, what direction does B have to be pointing? Once again a right hand rule can "point the way".

Alternatively, if you consider the directions of the field surrounding the wire due to the current and its interaction with the external uniform field, the wire will want to move in the direction that the field is weakest and away from where it's strongest. So the external field should be in a direction that will oppose the wire's field on top of the wire (for mutual cancellation) and reinforce it on the bottom.

 

[flyingpig]

The problem states that there is a separate uniform magnetic field that is perpendicular to the page. Think of the current in the wire as moving electric charges. The direction of the force that results on a moving charge in a magnetic field is determined by a cross product (v x B).

My redrawn field

[PLAIN]http://img863.imageshack.us/img863/2368/97467875.png

I am guessing the vertical bars still have no effect since the net force (into the page and out to the page) cancels.

So, if v is in the direction of the current, and you want the resultant of the cross product to be upwards, what direction does B have to be pointing? Once again a right hand rule can "point the way".

I chose initially it is out of the page. Even though it worked, I don't understand why it couldn't be into the page. As shown in the picture it goes into the page at the top of the bar and goes out of the page below it

Alternatively, if you consider the directions of the field surrounding the wire due to the current and its interaction with the external uniform field, the wire will want to move in the direction that the field is weakest and away from where it's strongest. So the external field should be in a direction that will oppose the wire's field on top of the wire (for mutual cancellation) and reinforce it on the bottom.

But the external field and the field at the top (going into the page) are in different directions

 

[flyingpig]

By the way, did I get it right though? Because that's important...

 

[gneill]

The external field comes out of the page. What you've drawn seems to be the field due to the horizontal wire.

Note that the vertical wires will contribute some to the net magnetic field between the wires -- both of them contribute a field that comes out of the page between the wires. I haven't done a calculation to see if it's significant compared to the required strength of the external uniform field.

 

[flyingpig]

But it says perpendicular to the page? How do you know the external field is out of the page?

Yes the field with the dots and crosses are for the horizontal wire.

So that means my initial drawing was right? My problem is that I don't know which B-field to cross it with for the horizontal bar

 

[gneill]

But it says perpendicular to the page? How do you know the external field is out of the page?

Yes the field with the dots and crosses are for the horizontal wire.

So that means my initial drawing was right? My problem is that I don't know which B-field to cross it with for the horizontal bar

The problem statement says that there is a uniform field perpendicular to the page. That is the field that the wire's field is interacting with in order to create the "push" that moves the wire. Or, if you're looking at it in terms of the charges moving in the wire, that is the field that you are crossing with the velocity vector.

 

[flyingpig]

Or, if you're looking at it in terms of the charges moving in the wire, that is the field that you are crossing with the velocity vector.

But this part is where I am having uncertainties about

 

[gneill]

But this part is where I am having uncertainties about

How so? You have a field that is either going into or coming out of the page. You have a velocity vector determined by the direction of the current flowing in the horizontal wire. Determine the direction of the force that results for each case by the right hand rule for cross products (or whatever method you prefer).

 

[flyingpig]

How so? You have a field that is either going into or coming out of the page. You have a velocity vector determined by the direction of the current flowing in the horizontal wire. Determine the direction of the force that results for each case by the right hand rule for cross products (or whatever method you prefer).

Yes that's the part, which B-field should I cross with, that's the dilemma I have been problems with.

The problem sort of hints that I use the on going out so that Fb = Fg, but I don't understand why I must cross with the one coming out of page

 

[gneill]

Yes that's the part, which B-field should I cross with, that's the dilemma I have been problems with.

The problem sort of hints that I use the on going out so that Fb = Fg, but I don't understand why I must cross with the one coming out of page

Because the external field is the one that provides the "traction" for applying a force to the wire. In the absence of an external field, a current carrying wire does not experience any force; it's own field is symmetrical about the wire, so there is no net force in any direction.

 

[flyingpig]

Because the external field is the one that provides the "traction" for applying a force to the wire. In the absence of an external field, a current carrying wire does not experience any force; it's own field is symmetrical about the wire, so there is no net force in any direction.

Oh okay, so that means that the bottom B-field that's coming out of the page is the sum of the rod's own B-field and the vertical bars and hence we cross it with that!

Wow that makes so much more sense now

thanks lol

 

[gneill]

F = I*L x B, where B is the external magnetic field, I is the current in the wire, L is the length of the wire. You don't include the wire's own field in the calculation; that's taken care of by the I*L term.