# Conducting bar in a uniform magnetic field

I'm studying for the MCAT and this problem came up. The correct answer is D. However the explanation was very confusing.

The explanation is verbatim:
" Let's start by considering A. Using the right-hand rule on a current running through the wire in figure 2 shows you that the rod is pushed either left or right (depending on the direction of the current). This invalidates choice A. Magnetic flux does not tend to change. That is why as stated int he passage the current generated in a wire loop will orient itself so that its magnetic field opposes the induced change in flux--a condition described by Lenz's law. This invalidates choices B, C, and favors D. The best answer is choice D."

I cannot for the life of me understand how the right hand rule on the current through the wire in figure 2 shows you that the rod is either pushed left or right :(

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Andrew Mason
Homework Helper
I cannot for the life of me understand how the right hand rule on the current through the wire in figure 2 shows you that the rod is either pushed left or right :(

The right hand rule applied to the conducting rod with a current going from the bottom to top in the diagram, gives you a magnetic field going into the paper on the right side and coming out from the paper on the left side. So it adds to the field on the right and reduces the field on the left, so it moves to the left. If you reverse the current, the motion is to the right.

You can just apply conservation of energy to see that the bar will not move. There is no change in the magnetic field strength, no applied emf so no current running through the bar and no rate of change of flux. So there is no source of energy. If you gave it a little push, the current would flow in such a way to oppose the motion (Lenz' law) so it would keep slowing down, its kinetic energy being used to generate heat in the resisitor.

AM

TSny
Homework Helper
Gold Member
[Edit: I see now that Andrew has already posted. Below is just another way to think about it in terms of using a right hand rule.]

For choice A we imagine that somehow there is a current in the resistor. (How the current got there is left as a mystery.) Suppose the current is upward as shown in the attachment. Are you familiar with the right-hand-rule for determining the direction of the magnetic force on a moving charge?

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Dope! I still don't get it!!!!

I followed you up until a certain point. I understood that the B field goes into the paper on the right side and comes out from the paper on the left side and that it adds to the field on the right and reduces the left BUT how did you make the connection that it is pushed to the left?!

What is pushing it to the left? I am so sorry but I have always had trouble with electrostatics and electromagnetism!!! It is so mind-boggling :/

Hello Tsny thank you for your reply. Yes I am familiar with it. Thumb is velocity, fingers are the B field, and palm shows you the direction of force on a positive charge.

DOHHH ! I think I get it! Please hold on and let me reformulate my question so that I can ask you to see if I am correct!

TSny
Homework Helper
Gold Member
Hello Tsny thank you for your reply. Yes I am familiar with it. Thumb is velocity, fingers are the B field, and palm shows you the direction of force on a positive charge.

OK. For the current, the velocity is the same as the direction of the current.

Mr. Andrew Mason and Tsny, could you please let me know if my thoughts are correct? They are as follows:

The external magnetic field is pointing INTO The plane of the computer screen and the velocity of the bar is from left to right. Using right hand rule, the positive charge moves to the top of the bar, so the induced current is counterclockwise.

As you pointed out, the bar itself creates a magnetic field. Using the right hand rule, we see that the induced magnetic field of the bar goes into the screen on the right, and out of the screen on the left.

Thus the magnetic field on the right is increased while the magnetic field on the left of the bar is decreased....

......yeah nevermind wow I don't get still....*banging head on desk*

OK. For the current, the velocity is the same as the direction of the current.

Yes I understand that, but I don't see how that relates how the bar is moved right or left

is it as simple as i think it is? the current goes from bottom to the top of the bar...so as Tsny said current is the same as velocity. SO using the right hand rule it's pushed left?? Is that all there is to it?

Then I don't see how Andrew Mason's explanation fit in. The current in the rod adds to the external magnetic field on the right but reduces it on the left....? How does that fit in...?

TSny
Homework Helper
Gold Member
is it as simple as i think it is? the current goes from bottom to the top of the bar...so as Tsny said current is the same as velocity. SO using the right hand rule it's pushed left?? Is that all there is to it?

Yes. The current is just a bunch of moving charged particles. For convenience you may think of the particles as positively charged and moving in the same direction as the current. The magnetic field produces a magnetic force on the moving particles in a direction given by the right hand rule. This force is communicated to the rod as a whole.

Thank you tsny !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! )))))))))))))))))))))))))

Andrew Mason
Homework Helper
is it as simple as i think it is? the current goes from bottom to the top of the bar...so as Tsny said current is the same as velocity. SO using the right hand rule it's pushed left?? Is that all there is to it?

Then I don't see how Andrew Mason's explanation fit in. The current in the rod adds to the external magnetic field on the right but reduces it on the left....? How does that fit in...?
I was trying to give you a way of understanding the force in terms of the interaction of the magnetic fields (the external B field and that produced by a current-carrying conductor).

The reason the right hand rule works has to do with cross products. The cross product convention is given by the right hand rule. The direction of the current is the direction that positive charges flow (actually, current direction is opposite to the direction that the negative charges - electrons - flow but the convention for current direction was established before we knew that). The Lorentz force is F = q v x B = q ds/dt x B. For current it becomes: F = dq/dt L x B = IL x B.

So when you apply the right hand rule, you are really finding the direction of the cross product of the vectors for charge velocity (current) and the magnetic field, which gives you the direction of the force.

AM

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Thank you for clearing that up! It all makes so much sense now! :)