Derivation of induced voltage in loop

[annamal]

$\phi_m = Blwcos\theta$
voltage = $\frac{-d\phi_m}{dt} = Blwsin\theta\frac{d\theta}{dt}$ where l is the length of the loop and w is the width of the loop

The top and bottom of the loop have magnetic forces perpendicular to the loop. My book says that means that there is no current through it. I do not understand why there is no current through it if the magnetic force is perpendicular to it? $d\vec F = I\vec dl \times B$

 

[alan123hk]

Is it because $~sin\theta~~$ is equal to zero ?

 

[Delta2]

I can't understand either, can you please post an image with the quoted text from the book

 

[annamal]

I can't understand either, can you please post an image with the quoted text from the book

 

[annamal]

Is it because $~sin\theta~~$ is equal to zero ?

I don't see how it is $sin\theta$ = 0. What is $\theta$ the angle between of?

 

[Delta2]

Yes well, it is one of those passages in a book that can be interpreted totally wrong by a student. Hold on while I try to explain to you what your book really means.

 

[Delta2]

Your book calls "current" the motion of the electrons inside the wires under the influence of the Lorentz Force $F_L=(\vec{B}\times \vec{v})q_e$, where $\vec{v}$ the velocity that the electrons have due to the rotation of the wires (they rotate together with the wire, I guess that stating the obvious but anyway).
Now we usually call current the motion of the electrons under the electric field due to the force $F_E=\vec{E}q_e$ and that's the main problem that causes the misinterpretation here.

In the side wires the Lorentz force is parallel to the side wires, hence it moves the rotating electrons towards the top of the wire, where they coalescing while in the bottom positive ions are coalescing. This coalescing of charges negative in the top of the side wire, and positive in the bottom of the side wire, creates a potential difference between the bottom and the top of the side wire.

In the top and bottom wires though, the Lorentz force remains parallel to the side wires (or normal to the top and bottom wires) hence it tends to move the electrons up too, but the electrons have no space to move towards up because the top and bottom wires we consider them to be one dimensional, the electrons can move left or right there but not up or down. So the "current" there is zero and the potential difference is zero there too.

I know I have to make my own figure to explain this in full detail but sorry i am bad in drawing figures.

 

[vanhees71]

That's of course a problem, because it's wrong. Any moving charge distribution implies that there is a current (density), $\vec{j}=\rho \vec{v}$, where $\vec{v}$ is the flow field of the charges.

In this case you consider the conduction current in a wire. Here the wire (i.e., the ion lattice of the metal) rotates in a magnetic field, and since there's a force on the conduction electrons due to the magnetic Lorentz force, you have a current since these conduction electrons can moving against the ion lattice of the wire, which is due to the complete (!) form of Ohm's Law, [corrected from a previous version]
$$\vec{j}=\sigma \left (\vec{E} + \frac{1}{\rho} \vec{j} \times \vec{B} \right),$$
where $\sigma$ is the electric conductivity of the wire.

Another way to calculate this is to use Faraday's Law in integral form, choosing the integration path along the wire loop, $\partial A$ (where $A$ is the area encircled by this integration path) and taking into account that this integration path moves, i.e., is not time-independent. Then Faraday's Law reads
$$\int_{\partial A} \mathrm{d}r (\vec{E}+\vec{u} \times \vec{B}) = -\frac{\mathrm{d}}{\mathrm{d} t} \int_{A} \mathrm{d}^2 \vec{f} \cdot \vec{B}=-\dot{\Phi}_{\vec{B},A}(t).$$
Here $\vec{u}(t,\vec{x})$ is the velocity of the wire (making up the integration path, $\partial A$).

 

[Delta2]

@vanhees71 can you please state again the complete form of Ohm's law cause I think you have a typo, you have $\vec{j}$ both in the left and right side of the equation, i think the second $\vec{j}$ should be $\vec{v}$.

Also when we say the magnetic force in a current element $I\vec{dl}$ is $dF=I\vec{dl}\times \vec{B}$ we take the current $I$ to be from a current density of the complete Ohm's law or from the usual Ohm's law $\vec{J}=\sigma \vec{E}$?

 

[vanhees71]

Ohm's Law in this form is correct. It's a constitutive equation. It follows from the simple Drude model of electric conductivity. You have to take into account not only the electric but also the magnetic part of the Lorentz force on the charge carriers. For wires at rest under household conditions you can safely neglect the magnetic part but not when the wires are in motion (except if you want to do the current-carrying wire relativistically correctly, and it's done almost everywhere wrong, including the Feynman lectures and Berkeley physics course by Purcell [*]).

[*] https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf

Another related problem is the homopolar generator, which also can only be understood only relativistically, even when all motions are with speeds $\ll c$:

https://itp.uni-frankfurt.de/~hees/pf-faq/homopolar.pdf

 

[Delta2]

Ohm's Law in this form is correct

Ohm's Law seems like an equation we have to solve for $\vec{j}$ because it appears both in left and right side of the equation.

 

[vanhees71]

Yes, indeed! It's a "constitutive equation". However, in this case it simplifies a bit, because you can neglect the current density along the wire against the contribution from the motion of the wire as a whole, i.e.,
$$\vec{j}=\sigma \left (\vec{E} + \frac{1}{\rho} \vec{j} \times \vec{B} \right) \simeq \sigma (\vec{E} + \vec{u} \times \vec{B}).$$

 

[Delta2]

Yes, indeed! It's a "constitutive equation".

May I ask how we solve it for $\vec{j}$?

 

[vanhees71]

It's usually not that easy. Only for very simple cases you have a simple solution of the complete magnetostatic equations. One example is the complete relativistic treatment of the infinitely long cylindrical wire with a DC current:

https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf

 

[alan123hk]

Most likely I don't get it, but I feel like it's two different currents, $~\vec{j_c}=\sigma (\vec{E} + \vec{j_d} \times \vec{B})$

Also when we say the magnetic force in a current element Idl→ is dF=Idl→×B→ we take the current I to be from a current density of the complete Ohm's law or from the usual Ohm's law J→=σE→?

Since this is a complete form of Ohm's law, it appears that it should also apply to the current in equation $~dF=I\vec{dl}\times \vec{B}$.

 

[bob012345]

The top and bottom of the loop have magnetic forces perpendicular to the loop. My book says that means that there is no current through it. I do not understand why there is no current through it if the magnetic force is perpendicular to it? $d\vec F = I\vec dl \times B$

Your book only says the forces on the top and bottom are perpendicular and thus cause no current meaning it doesn't add to the current already flowing. The current is the same around the whole loop but some segments of the loop are driving the current to increase just like a battery was in the side segments but not the top and bottom.

 

[nasu]

That's of course a problem, because it's wrong. Any moving charge distribution implies that there is a current (density), $\vec{j}=\rho \vec{v}$, where $\vec{v}$ is the flow field of the charges.

In this case you consider the conduction current in a wire. Here the wire (i.e., the ion lattice of the metal) rotates in a magnetic field, and since there's a force on the conduction electrons due to the magnetic Lorentz force, you have a current since these conduction electrons can moving against the ion lattice of the wire, which is due to the complete (!) form of Ohm's Law, [corrected from a previous version]
$$\vec{j}=\sigma \left (\vec{E} + \frac{1}{\rho} \vec{j} \times \vec{B} \right),$$
where $\sigma$ is the electric conductivity of the wire.

Another way to calculate this is to use Faraday's Law in integral form, choosing the integration path along the wire loop, $\partial A$ (where $A$ is the area encircled by this integration path) and taking into account that this integration path moves, i.e., is not time-independent. Then Faraday's Law reads
$$\int_{\partial A} \mathrm{d}r (\vec{E}+\vec{u} \times \vec{B}) = -\frac{\mathrm{d}}{\mathrm{d} t} \int_{A} \mathrm{d}^2 \vec{f} \cdot \vec{B}=-\dot{\Phi}_{\vec{B},A}(t).$$
Here $\vec{u}(t,\vec{x})$ is the velocity of the wire (making up the integration path, $\partial A$).

Do you really believe that this will help at the level of the OP poster? He is strugling with the very basics.

 

[alan123hk]

I also thought for a long time, I tried my best, and still don't understand why there is a magnetic force perpendicular to the loop at the top and bottom of the loop which means no current is flowing.

 

[nasu]

They mean the force on the free electrons is perpendicular to the wire. This is what is said in the text shown. Not perpendicular to the loop. The electrons in the top side are moving around horizontal circles due to the rotation of the loop. So the component of velocity due to the rotation of the loop is always horizontal. The magnetic field is horizontal too. The magnetic force will be perpendicluar to the plane of these two vectors (horizontal at any position) so it can be only vertical. This will never have a component along the wire (in the horizotal plane). It would just push the electrons towards one lateral side of the conductor, like in Hall effect.

 

[bob012345]

I also thought for a long time, I tried my best, and still don't understand why there is a magnetic force perpendicular to the loop at the top and bottom of the loop which means no current is flowing.

The statement "which means no current is flowing" is wrong and was a misquote of the book. Current is flowing as I stated above but no emf is induced in the top and bottom segments.

 

[Delta2]

The generator seems open circuited so no conventional current ($\vec{J}=\sigma \vec{E}$) is flowing. It is just that the book calls current the term $\sigma (\vec{v}\times\vec{B})$ which causes the misinterpretation by student.

 

[annamal]

Your book calls "current" the motion of the electrons inside the wires under the influence of the Lorentz Force $F_L=(\vec{B}\times \vec{v})q_e$, where $\vec{v}$ the velocity that the electrons have due to the rotation of the wires (they rotate together with the wire, I guess that stating the obvious but anyway).
Now we usually call current the motion of the electrons under the electric field due to the force $F_E=\vec{E}q_e$ and that's the main problem that causes the misinterpretation here.

In the side wires the Lorentz force is parallel to the side wires, hence it moves the rotating electrons towards the top of the wire, where they coalescing while in the bottom positive ions are coalescing. This coalescing of charges negative in the top of the side wire, and positive in the bottom of the side wire, creates a potential difference between the bottom and the top of the side wire.

In the top and bottom wires though, the Lorentz force remains parallel to the side wires (or normal to the top and bottom wires) hence it tends to move the electrons up too, but the electrons have no space to move towards up because the top and bottom wires we consider them to be one dimensional, the electrons can move left or right there but not up or down. So the "current" there is zero and the potential difference is zero there too.

I know I have to make my own figure to explain this in full detail but sorry i am bad in drawing figures.

What about the other side wire? Is the current in the opposite direction of the original side wire?

 

[Delta2]

What about the other side wire? Is the current in the opposite direction of the original side wire?

Yes, the "current" (i put it inside marks cause it is the current due to magnetic field B and the rotational velocity, that is due to the term $\vec{v}\times\vec{B}$) is in the opposite direction there. That is because $\vec{v}$ is in the opposite direction there.

 

[rude man]

My book says that means that there is no current through it. I do not understand why there is no current through it if the magnetic force is perpendicular to it?

Your book does not say that. It says no current is induced in the horizontal sections but there is current induced in the vertical sections. If there is current in the vertical sections that same current must also flow in the horizontal sections. Ohm's current law.

PS that assumes there is a load for the generator, obviously.
If no load then no current but the emf is still being generated and is generated in the vertical sections only.

 

[Delta2]

PS that assumes there is a load for the generator, obviously.

yes this point would help clear some of the confusion regarding what exactly the book means by current there. If it means the current due to load in the generator or the current due to the term $\vec{v}\times\vec{B}$.

 

[annamal]

Yes, the "current" (i put it inside marks cause it is the current due to magnetic field B and the rotational velocity, that is due to the term $\vec{v}\times\vec{B}$) is in the opposite direction there. That is because $\vec{v}$ is in the opposite direction there.

But because the current is flowing in the opposite direction isn't the voltage of one side equal to the negative of the other side's voltage thus causing the net voltage to be 0?

 

[Delta2]

But because the current is flowing in the opposite direction isn't the voltage of one side equal to the negative of the other side's voltage thus causing the net voltage to be 0?

Yes the + of one side meets the - of the other side but this configuration is known as voltage sources in series, and the net voltage equals the sum (and not the difference) of the separate voltages.
The following youtube video might explain it better.

 

[vanhees71]

My book says that means that there is no current through it. I do not understand why there is no current through it if the magnetic force is perpendicular to it?

Your book does not say that. It says no current is induced in the horizontal sections but there is current induced in the vertical sections. If there is current in the vertical sections that same current must also flow in the horizontal sections. Ohm's current law.

PS that assumes there is a load for the generator, obviously.
If no load then no current but the emf is still being generated and is generated in the vertical sections only.

Of course there is a current through it, because there is an electromotive force. Note that indeed in the (local) restframe of the wire you have Ohm's Law in the usual form $\vec{j}'=\sigma \vec{E}'$. So you have a current through the wire given by $i=\mathcal{E}/R$, where $R$ is the resistance of the wire. That's how generators work!

 

[alan123hk]

I feel like I's starting to understand the meaning of $\vec{j}=\sigma \left (\vec{E} + \frac{1}{\rho} \vec{j} \times \vec{B} \right)$. Although I don't know how to apply it in a practical situation, because the math seems to get very complicated, e.g. the motion trajectory of electric charge is not restricted to very thin wires, but can move freely in 3D space. But anyway, it's a really cool expression.

On the other hand, I agree that the Lorentz Force $~\vec{F_L}=(\vec{B}\times \vec{v})q_e$ should not be interpreted as an electric field force $\vec{F_E}=\vec{E}q_e$ because it was zero unless the charge are moving. But the electric field created by magnetic EMF $(2Blv~sin~\theta )~$is of course real, because this electric field is created by the charge that accumulates on the conductor after being moved by the force $~\vec{F_L}=(\vec{B}\times \vec{v})q_e$.

 

[rude man]

I feel like I's starting to understand the meaning of $\vec{j}=\sigma \left (\vec{E} + \frac{1}{\rho} \vec{j} \times \vec{B} \right)$. Although I don't know how to apply it in a practical situation, because the math seems to get very complicated, e.g. the motion trajectory of electric charge is not restricted to very thin wires, but can move freely in 3D space. But anyway, it's a really cool expression.

If I were you I would forget all about that equation. It's not significant in any introductory physics course. Nor for that matter in an advanced undergraduate course.