Magnetic field and electromotive force

[Kernul]

The exercise says this:

A metal stick of mass [latex]m[/latex] and length [latex]L[/latex], initial still, can slide on two long horizontal binaries without friction. A uniform vertical magnetic field is present in the region in which the stick can move. The battery [latex]G[/latex] applies a constant electromotive force [latex]\epsilon[/latex] to the circuit formed by the binaries and the stick, letting flow a current which direction is showed in the figure. Demonstrate that the velocity of the stick approaches a constant value [latex]v[/latex] and determine the module and the direction of it. What is the current that flows in the stick when this value is reached?

 

So, we know there's an electromotive force produced by the battery G. Let's call it [latex]\epsilon_G[/latex]. There is then another electromotive force produced by the magnetic field in which the circuit is placed. Let's call it [latex]\epsilon_B[/latex].

This last one can be found with the Faraday-Neumann Law this way:

[latex]\epsilon_B = - \frac{d \Phi (\vec B)}{dt}[/latex]

where the [latex]\Phi (\vec B)[/latex] is the magnetic flux found this way:

[latex]\Phi (\vec B) = \vec B \cdot d\vec S[/latex]

After a couple of passages, I arrive at this:

[latex]\epsilon_B = -BLv[/latex]

Now here comes what I'm probably not sure anymore.

The electromotive force [latex]\epsilon_G[/latex] is simply [latex]IR[/latex]? The current circulating the circuit and the resistance of the stick? Does this mean that I have to sum the two electromotive forces?

Do I have to introduce the induced electromotive field [latex]\vec E_i[/latex]? Because I know that the electromotive force can also be found with this formula:

[latex]\epsilon = \oint_l \vec E_i \cdot d\vec l[/latex]

where the [latex]\vec E_i[/latex] can be written this way:

[latex]\vec E_i = \vec E + (\vec v_{drag} + \vec v_{drift}) \times \vec B[/latex]

Where [latex]\vec v_{drag}[/latex] is the drag velocity, and [latex]\vec v_{drift}[/latex] is the drift velocity. Knowing that the latter is null when we do the cross product with [latex]\vec B[/latex] (because this velocity is always parallel to the element of circuit [latex]d\vec l[/latex]), we simply have:

[latex]\vec E_i = \vec E + \vec v_{drag} \times \vec B[/latex]

and that velocity right there is the one I'm searching for, right? But I really don't understand how to arrive at finding that the velocity of the stick approaches a velocity [latex]v[/latex].

About the other point, the one related to the current, I basically need the velocity and then apply this formula:

[latex]i = \frac{\epsilon}{R} = \frac{BLv}{R}[/latex]

Am I right?

[zztop]

The exercise says this:

A metal stick of mass [latex]m[/latex] and length [latex]L[/latex], initial still, can slide on two long horizontal binaries without friction. A uniform vertical magnetic field is present in the region in which the stick can move. The battery [latex]G[/latex] applies a constant electromotive force [latex]\epsilon[/latex] to the circuit formed by the binaries and the stick, letting flow a current which direction is showed in the figure. Demonstrate that the velocity of the stick approaches a constant value [latex]v[/latex] and determine the module and the direction of it. What is the current that flows in the stick when this value is reached?

Exercise 3.jpg

 

So, we know there's an electromotive force produced by the battery G. Let's call it [latex]\epsilon_G[/latex]. There is then another electromotive force produced by the magnetic field in which the circuit is placed. Let's call it [latex]\epsilon_B[/latex].

This last one can be found with the Faraday-Neumann Law this way:

[latex]\epsilon_B = - \frac{d \Phi (\vec B)}{dt}[/latex]

where the [latex]\Phi (\vec B)[/latex] is the magnetic flux found this way:

[latex]\Phi (\vec B) = \vec B \cdot d\vec S[/latex]

After a couple of passages, I arrive at this:

[latex]\epsilon_B = -BLv[/latex]

Correct, so far.

Now, what causes the conductor to move and in what direction will it move? There is a force, called Ampere force....What is its direction, what is its expression?

There is another force that RESISTS the move of the conductor. It is called the Lorentz force. What is its direction and expression? Do the two forces cancel each other?

Edited February 19, 2017 by zztop

[Kernul]

Correct, so far.

Now, what causes the conductor to move and in what direction will it move? There is a force, called Ampere force....What is its direction, what is its expression?

There is another force that RESISTS the move of the conductor. It is called the Lorentz force. What is its direction and expression? Do the two forces cancel each other?

The magnetic field [latex]B[/latex] is the one that causes the conductor to move, right? The Ampère's force law is, in general, given by this expression:

[latex]\vec F = \frac{\mu_0 I_1 I_2}{2 \pi r} \vec r[/latex]

But here I have only one current. How does this work?

The Lorentz force is given by [latex]\vec F = q \vec v \times \vec B[/latex] and it's direction should be perpendicular to the magnetic field [latex]B[/latex], so moving along the binaries like the arrow shows in the picture. Am I right?

[zztop]

The magnetic field [latex]B[/latex] is the one that causes the conductor to move, right? The Ampère's force law is, in general, given by this expression:

[latex]\vec F = \frac{\mu_0 I_1 I_2}{2 \pi r} \vec r[/latex]

But here I have only one current. How does this work?

The Lorentz force is given by [latex]\vec F = q \vec v \times \vec B[/latex] and it's direction should be perpendicular to the magnetic field [latex]B[/latex], so moving along the binaries like the arrow shows in the picture. Am I right?

You have TWO currents , of equal absolute value and OPPOSITE senses.

[Kernul]

You have TWO currents , of equal absolute value and OPPOSITE senses.

Oh right! Because the current on stick is the same but flowing on the opposite direction of the one where there is the battery. Because of this, they push each other, but only the stick can move, so it moves far from the battery, right? Do I have to sum this one with the Lorentz force? Does this mean that the direction of the Lorentz force is contrary to the Ampère force? And does this mean that the Ampère force gradually decrease until it equals the Lorentz force? Is it a that point that the velocity is constant? And it's a that point that I have to calculate the current with the formula I gave in the initial post, right?

Edited February 19, 2017 by Kernul

[zztop]

Oh right! Because the current on stick is the same but flowing on the opposite direction of the one where there is the battery. Because of this, they push each other, but only the stick can move, so it moves far from the battery, right?

Correct.

 

 

 

Do I have to sum this one with the Lorentz force? Does this mean that the direction of the Lorentz force is contrary to the Ampère force?

 

Yes, the Lorentz force is [latex]\vec{F_L}=L\vec{I}X\vec{B}[/latex] and it opposes the Ampere force. The absolute value of the Lorentz force is [latex]F_L=LIB[/latex]

 

 

 

And does this mean that the Ampère force gradually decrease until it equals the Lorentz force? Is it a that point that the velocity is constant? And it's a that point that I have to calculate the current with the formula I gave in the initial post, right?

 

Yes.

Edited February 19, 2017 by zztop

[Kernul]

Thank you very much for the help!

[zztop]

Thank you very much for the help!

There is one more thing to consider: at equilibrium, the em induced voltage [latex]-BLv[/latex] will cancel out the battery voltage [latex]G[/latex] so, the speed [latex]v[/latex] is ....? And the current [latex]I[/latex] in the wires becomes....? And the Ampere and Lorentz forces become equal to...?

Edited February 19, 2017 by zztop

[Kernul]

There is one more thing to consider: at equilibrium, the em induced voltage [latex]-BLv[/latex] will cancel out the battery voltage [latex]G[/latex] so, the speed [latex]v[/latex] is ....? And the current [latex]I[/latex] in the wires becomes....? And the Ampere and Lorentz forces become equal to...?

This means that [latex]IR - BLv = 0[/latex], right? So the velocity would be [latex]v = \frac{IR}{BL}[/latex]?

For the current I just need to replace the [latex]v[/latex] I just found in [latex]i = \frac{BLv}{R}[/latex].

The sum of the two forces is zero at equilibrium, right?

[zztop]

This means that [latex]IR - BLv = 0[/latex], right? So the velocity would be [latex]v = \frac{IR}{BL}[/latex]?

For the current I just need to replace the [latex]v[/latex] I just found in [latex]i = \frac{BLv}{R}[/latex].

 

Not exactly: [latex]G - BLv = 0[/latex]

 

 

 

The sum of the two forces is zero at equilibrium, right?

 

Yes. Meaning that [latex]i[/latex] must be 0.

Edited February 19, 2017 by zztop