Self-Induced Magnetic Field in a moving rod with current

[shespuzzling]

Hi,

In my book, they use an example to explain motional emf. The example consists of a rectangular circuit with a resistor on the left side and a moving bar on the right side. There already exists a magnetic field pointing into the paper and the bar is given an initial velocity to the right. I'm having some trouble determining exactly what is the force on the movable bar. I will explain what I think is happening, and then hopefully somebody can help me out!

1. Initially there is no current. As the bar moves to the right, the magnetic flux is changing, and thus an emf and a current is induced in the circuit. This current is in the counter clockwise direction.

2. Now that there is current in the circuit, a magnetic field is induced due to the current. My understanding is that the magnetic field will encircle the wire in the circuit and will come out of the page on the left and into the page on the right (**is this correct? the book says that the induced magnetic field is simply "out" of the page, but if it is a circle it has to go both ways, right?)

3. My understanding is that there are now two magnetic fields that come into play. There is the uniform magnetic field initially present that is directed into the page, and there is now the induced magnetic field due to the induced current that encircles the circuit. To me, this means that to the left of the movable bar, the induced magnetic field goes against the original magnetic field, but to the right of the bar, the induced field actually goes in the same direction as the original field. Thus, to the left the fields subtract and to the right they add. (**something tells me I'm wrong about this but I can't figure out why)

4. If there are indeed the two fields that come into play, then the total force on the bar from the magnetic fields is in the left direction (assuming that the original B field is much greater than that induced by the current). However, if you calculate it individually, to the left of the bar you have F=ILx(B1-B2) where B1 is the original field and B2 is the induced field. To the right you have F=ILx(B1+B2). Then, adding these forces you get the total force to equal 2*IL(B1).

If anybody can shed some light on this I'd be really appreciative! I'm trying to teach myself physics and it can be pretty confusing at times.

Thanks!

 

[kuruman]

1. Initially there is no current. As the bar moves to the right, the magnetic flux is changing, and thus an emf and a current is induced in the circuit. This current is in the counter clockwise direction.

Good.

2. Now that there is current in the circuit, a magnetic field is induced due to the current. My understanding is that the magnetic field will encircle the wire in the circuit and will come out of the page on the left and into the page on the right (**is this correct? the book says that the induced magnetic field is simply "out" of the page, but if it is a circle it has to go both ways, right?)

Not good. The induced current in the loop is no different from a current produced by a battery. This current generates a magnetic field that is perpendicular to the plane of the loop and out of the page (or screen) according to the right hand rule. Yes, magnetic field lines form closed loops, but here the closed loops cross the plane of the current loop perpendicularly over the entire area enclosed by the current loop.

3. My understanding is that there are now two magnetic fields that come into play. There is the uniform magnetic field initially present that is directed into the page, and there is now the induced magnetic field due to the induced current that encircles the circuit.

That is correct. However, the induced magnetic field extends through the entire area of the moving loop whereas the uniform magnetic field goes through the part of the loop that has already entered the field. The rest of the loop is outside the uniform field area and has no uniform magnetic flux through it. In this particular case the induced field is out of the page and the uniform field is in the opposite direction, into the page.

4. If there are indeed the two fields that come into play, then the total force on the bar from the magnetic fields is in the left direction (assuming that the original B field is much greater than that induced by the current). However, if you calculate it individually, to the left of the bar you have F=ILx(B1-B2) where B1 is the original field and B2 is the induced field. To the right you have F=ILx(B1+B2). Then, adding these forces you get the total force to equal 2*IL(B1).

The bar on the left is out of the uniform field region so there is no force on it due to B1. When you add the forces together, you get I L B1. I should add that when the left bar crosses into the field region, there is no more induced current and no more force on the loop. That's because the motional emfs from the two bars cancel each other.