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

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.

 

[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.

 

[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

 

[asteriatic]

I would like to commend you on your attempt at solving this problem using basic principles of electromagnetism. Your reasoning and calculations are mostly correct, but there are a few things that need clarification.

Firstly, the direction of the induced current in the loop will indeed be counter-clockwise, as you correctly stated. This is in accordance with Lenz's law, which states that the induced current will always oppose the change in magnetic flux.

Secondly, your calculation of the induced EMF (E) is correct, but your interpretation of it is not entirely accurate. The induced EMF is not equal to the applied voltage (V), but rather it is equal to the rate of change of magnetic flux through the loop. In this case, the induced EMF is equal to BLv, where v is the velocity of the bar.

Next, your calculation of the current in the loop is also correct, but again, your interpretation is not completely accurate. The current in the loop is not a constant value, but rather it will change as the bar moves due to the changing magnetic flux. So, the correct expression for the current would be I = -BLv/R.

Finally, your calculation of the Lorentz force is also correct, but again, your interpretation needs some clarification. The force on the bar due to the magnetic field is not simply equal to ILB, but rather it is equal to the product of the current and the length of the bar (L). So, the correct expression for the force would be F = -BL^2v/R.

In summary, the force required to move the bar at a constant speed v would be F = -BL^2v/R. This force is needed to overcome the opposing force due to the induced current in the loop, which is also equal to -BL^2v/R. So, the net force on the bar would be zero, and it would continue to move at a constant speed v.

I hope this clarifies any confusion and helps you to better understand the physics behind this problem. Keep up the good work!