Force to move a bar in a magnetic field

1. Apr 15, 2012

QuarkCharmer

1. The problem statement, all variables and given/known data
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:

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

3. 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,
$$\Phi = B \dot A\\ A = Lx\\ \Phi = BLxcos(\theta)\\$$

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

$$\Phi = BLx$$

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:

$$\Phi = BLx\\ \frac{d\Phi}{dt} = BL\\$$

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:
$$V=IR\\ I=\frac{V}{R}\\ I = \frac{-BL}{R}\\$$

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

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.

Last edited: Apr 15, 2012
2. Apr 15, 2012

QuarkCharmer

I forgot this part:

Since:
$$I = \frac{-BL}{R}\\$$

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.

3. Apr 15, 2012

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.

4. Apr 15, 2012

QuarkCharmer

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

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