Question about a rod moving through a magnetic field (ampere,faraday)

[vande060]

Homework Statement

Homework Equations

Since we are looking for force, this equation is what I thought we would use : F = qv X B

I need to find B, so I am guessing amperes law will be used

∫ B.dr

The Attempt at a Solution

If the sliding rod has motion, we can say F = ma = qv X B

computing amperes law ∫ B.dr = uI

here I get into a bind, because I am used to calculating amperes law for a closed circle of radius r, but here I believe the case is different because we have a square type structure to calculate the magnetic field for.

I remeber that our professor said that J can be worked into examples like these, where J = dI/dw, so the change in current as the rod moves through the field.

If anyone could give me a hand starting out, it would be much appreciated

 

[vela]

Homework Equations

Since we are looking for force, this equation is what I thought we would use : F = qv X B

This equation applies only to a point charge. You don't have a point charge in this problem, so it doesn't apply.

I need to find B, so I am guessing amperes law will be used

∫ B.dr

Why do you need to find B? It's a given in this problem.

The Attempt at a Solution

If the sliding rod has motion, we can say F = ma = qv X B

computing amperes law ∫ B.dr = uI

here I get into a bind, because I am used to calculating amperes law for a closed circle of radius r, but here I believe the case is different because we have a square type structure to calculate the magnetic field for.

I remeber that our professor said that J can be worked into examples like these, where J = dI/dw, so the change in current as the rod moves through the field.

If anyone could give me a hand starting out, it would be much appreciated

You need to step back and assess what's going on in the problem first before throwing equations at it.

First, you want the rod to move at constant speed. What does that tell you about its acceleration? What can then infer from applying Newton's second law?

Suppose the magnetic field isn't present. Newton's first law tells us that no force would be required to keep the rod moving at a constant speed v. When the magnetic field is present, a force is required. Why? What exactly is going on to introduce forces into this problem when B is not zero?

 

[vande060]

first, you want the rod to move at constant speed. What does that tell you about its acceleration? What can then infer from applying Newton's second law?

acceleration is zero, and we can infer f=ma=0

Suppose the magnetic field isn't present. Newton's first law tells us that no force would be required to keep the rod moving at a constant speed v. When the magnetic field is present, a force is required. Why? What exactly is going on to introduce forces into this problem when B is not zero

B must produce a force that is is the opposite direction of the velocity vector

 

[vela]

Good! So why does the magnetic field exert a force on the rod?

 

[vande060]

as the bar moves through the magnetic field the magnetic field exerts a force on the electrons within the bar.

 

[vela]

OK, so the charges initially have velocity v because they're in the bar and the bar moves with velocity v. Which way is the force on those charges?

 

[vande060]

OK, so the charges initially have velocity v because they're in the bar and the bar moves with velocity v. Which way is the force on those charges?

in the positive j direction

 

[vela]

Your answer is meaningless to me since you haven't said how you're aligning your axes. What direction is the +j direction on your diagram? To the left, to the right, up the page, down the page, out of the page, or into the page? Also, explain how you figured out the direction. If you just give me an answer that's incorrect without explaining how you reached that conclusion, it doesn't tell me anything other than you made a mistake somewhere.

 

[vande060]

Your answer is meaningless to me since you haven't said how you're aligning your axes. What direction is the +j direction on your diagram? To the left, to the right, up the page, down the page, out of the page, or into the page? Also, explain how you figured out the direction. If you just give me an answer that's incorrect without explaining how you reached that conclusion, it doesn't tell me anything other than you made a mistake somewhere.

sorry,

in taking x,y,z and the corresponding i,j,k the force on the charges in the bar are in the positive y, or positive j direction. the velocity vector of the bar is in the negative j direction, so a force to slow down the rod must be in the positive j direction

 

[vela]

What does the right-hand rule say about the direction of the force?

 

[vande060]

What does the right-hand rule say about the direction of the force?

the negative i direction?

 

[vela]

Yup, the movement of the rod causes the (positive) charges to move to the left. In other words, there's a current that flows to the left in the rod and clockwise around the loop. The movement of the rod induces this current in the loop. This particular force is not the one that tries to slow the rod down.

But now the rod carries a current, and this current is in a magnetic field. What does that mean?

 

[vande060]

Yup, the movement of the rod causes the (positive) charges to move to the left. In other words, there's a current that flows to the left in the rod and clockwise around the loop. The movement of the rod induces this current in the loop. This particular force is not the one that tries to slow the rod down.

But now the rod carries a current, and this current is in a magnetic field. What does that mean?

current loop + magnetic field => torque

 

[vela]

Only one side of the loop, the rod, can move, so the torque on the loop really doesn't matter here. In which direction is the force on the rod due to the current flowing through it?

 

[vande060]

Only one side of the loop, the rod, can move, so the torque on the loop really doesn't matter here. In which direction is the force on the rod due to the current flowing through it?

We covered this variety in more detail in class today, I know how it is done now. Your help was a great advantage though thank you