Where did this formula for velocity come from?

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Munky
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I am studying for an upcoming test and, while looking through my notes, came upon this formula:

v= sqrt((4/3)(IAB/M))

It solved my problem, but I don't know where I got it.

The problem is accelerating a rod with a current through it on rails across a magnetic field.

Can someone please help me understand why this works and where in my general physics book I may have found it?

Thanks.

munky
 
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Sounds like an induced EMF problem.

I'm guessing that the rod and rails are initially conducting a current I. The current in the rails produces a magnetic field, and the rod experiences a force (due to the magnetic field and current in the rod) which accelerates the rod along the rails.

However, the rod motion causes an increase in magnetic flux through the loop due to the increasing area. This creates an EMF that opposes the current. At some velocity, the induced EMF makes the current, and hence the magnetic field, go to zero.

At that velocity there is no force or acceleration, and may very well be the velocity given by your formula.

EDIT:
So in your textbook, look up Faraday's Law or Magnetic Induction or Induced EMF.
 
There is another possibility.
A rod is placed on a rails and a power supply is connected to the rails.
A magnetic field, making certain angle to the plane containing rails and rod, is applied. Its vertical component is perpendicular to the rod and horizontal component is parallel to the rod.
Current carrying rod experiences a force, which is given by F= BILsin(theta) where L is the length of the rod. Due to this force rod get accelerated. If its velocity is v when it displaces a distance x, work done on the rod = F*x. And that is equal to KE of the rod.
So, 1/2*M*v^2 = BILxsin(theta). Lx = A. In your problem sin (theta) may be 2/3. Put this value in the equation and find the expression for v.
 
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Thanks for the responses. I can see where the formula came from now (physics one was so long ago!), but the 4/3 is buggin me. The B field is perpendicular to the rod and the rails, pointed down. Would the fact that the rod is rolling as opposed to sliding have anything to do with it?

thanks again!