Force to move a bar in a magnetic field

[QuarkCharmer]

Homework Statement

I made this image to illustrate:

A rod with resistance R lies across frictionless conducting rails in a constant uniform magnetic field B. Assume the rails have negligible resistance. The magnitude of the force that must be applied by a person to pull the rod to the right at a constant speed v is:

Homework Equations

E = -\frac{d \phi}{dt}
Plus any of maxwells equations et al.

The Attempt at a Solution

From what I figure, as the bar moves to the right, the magnetic flux will be increasing, and so, the current in the loop will produce a field opposing the change. This B field would be coming out of the screen/page, so the current must be going counter clockwise.

Now,
<br /> \Phi = B \dot A\\<br /> A = Lx\\<br /> \Phi = BLxcos(\theta)\\<br />

But theta (angle between area vector and B is 0:

<br /> \Phi = BLx<br />

The velocity of the bar is basically in x per seconds, so I think I can say that basically x is time. That way, I can differentiate the flux as so:

<br /> \Phi = BLx\\<br /> \frac{d\Phi}{dt} = BL\\<br />

Now, E in the loop will equal negative BL since:
E = -\frac{d\Phi}{dt}

and so, via Ohms law, the current in the loop is:
<br /> V=IR\\<br /> I=\frac{V}{R}\\<br /> I = \frac{-BL}{R}\\<br />

Then, by the equation of the Lorentz force:
<br /> F_{B} =I(L×B)\\<br /> F_{B} = ILBsin(\theta)\\<br />

But theta here is 90 degrees, and sin(90) = 1,

So I think the "opposing" force on the bar due to it moving and increasing the flux through the loop is equal to ILB.

So at least a force of ILB is required to move the bar to the right, and with a force of ILB, the bar will remain still. So now I am lost.

What's the next step? I know the solution but I need to figure this out myself.

 

[QuarkCharmer]

I forgot this part:

Since:
I = \frac{-BL}{R}\\<br />

and

F = ILB

Then:
F = -\frac{B^{2}L^{2}}{R}

Which is almost the answer I need, it's definitely the force to the left. So to pull it to the right I need to put that force.

There should be a v in that equation though, and I don't see how.

 

[WannabeNewton]

Well you have \Phi = BLx so \frac{\mathrm{d}\Phi }{\mathrm{d} t} = BL\frac{\mathrm{d} x}{\mathrm{d} t} = Blv. So there's your v.

 

[QuarkCharmer]

Well you have \Phi = BLx so \frac{\mathrm{d}\Phi }{\mathrm{d} t} = BL\frac{\mathrm{d} x}{\mathrm{d} t} = Blv as per the chain rule. So there's your v.

Ah, I see. Idk what I was thinking with this sentence:

"The velocity of the bar is basically in x per seconds, so I think I can say that basically x is time. That way, I can differentiate the flux as so:"Thanks