# Work done on a moving square wire loop in a B-field

• feynman1
In summary, the original post was edited to correct a mistake, but it is important to keep the original post for the sake of continuity and future reference. Instead, a new post should be created to address any corrections or updates.
feynman1

There’s a current i in the loop in the figure. The Ampere’s force iLB on a wire of length L exerts on charges in the wire but it does no work on the charges. The charges would go in circular motion if there were no wire. Then the wire exerts exactly iLB on those charges to keep the charges moving at constant speed. Then likewise the charges exert iLB on the wire. Then it is the charges that are doing work iLBv on the wire?

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As the loop moves to the right on both the right and left sides there is a large ##\partial \vec B/\partial t## and therefore by Faraday’s law we get a non zero ##\vec E## field.

Dale said:
As the loop moves to the right on both the right and left sides there is a large ##\partial \vec B/\partial t## and therefore by Faraday’s law we get a non zero ##\vec E## field.
i changed that pic so that there's no change in B

feynman1 said:
i changed that pic so that there's no change in B
You can change the pic, but now your pic doesn’t represent reality. If there is no change in the magnetic field then the wire cannot be carrying current.

Vanadium 50 and berkeman
I don't know, whether I read your drawing correctly: If the entire loop moves in a homogeneous magnetic field there's indeed no work done. It just moves with constant velocity.

The usual setup is shown here in the figure with the resistor (the explanations on the webpage are not optimal, to say it friendly):

http://labman.phys.utk.edu/phys222core/modules/m5/motional_emf.html

Take the situation in this figure as the initial condition and let the wire just run without any external force. Then we can make a simple analysis to see how work is done here. We use the quasistationary approximation.

The motion of the wire induces an electromotive force according to Faraday's Law (in SI units)
$$\mathcal{E}=-\dot{\Phi}=v B d.$$
According to Ohm's law this induces a current going upwards in the wire
$$i=\mathcal{E}/R=v B d/R.$$
Now due to this current there acts a force to left on the wire, i.e.,
$$m \dot{v}=-B i d=-\frac{B^2 d^2}{R} v.$$
The power is
$$P=m \dot{v} v = -\frac{B^2 d^2}{R} v^2.$$
This is precisely the heat due to the ohmic losses in the wire,
$$P_\text{ohm}=R i^2=\frac{B^2 d^2}{R} v^2.$$
So the work done here is due to friction of the charges in the wire. It's not the magnetic field directly doing any work. It cannot do work, because the force on each charge is always perpendicular to its velocity.

It's pretty obnoxious to ask a question, get an answer, and then change the question by editing it. That just confuses everybody.

As pointed out, there's no work done. If you draw the system at two different times, nothing changes: you could take two snapshots and overlay them so the loops are in the same place and there is otherwise no difference.

weirdoguy
feynman1 said:
i changed that pic so that there's no change in B

It's pretty obnoxious to ask a question, get an answer, and then change the question by editing it. That just confuses everybody.
@Vanadium 50 is right. @feynman1, the correct way to do it is to add a new post with the correction and not edit the original, especially after getting an answer.

how to delete the original post

feynman1 said:
how to delete the original post
You don't. I don't think you see the point here.

The original post #1 asks a question. An answer #2 is provided. If you then edit or delete #1, then #2 looks silly and hard to explain. It would be unfair to the author of #2.

The original post #1 asks a question. An answer #2 is provided. You post a modified question #3. Someone provides an answer #4. Now the whole thread #1, #2, #3, #4 can be read in sequence and understood. It becomes part of our archives for other students to find and to read in the future.

vanhees71
anorlunda said:
You don't. I don't think you see the point here.

The original post #1 asks a question. An answer #2 is provided. If you then edit or delete #1, then #2 looks silly and hard to explain. It would be unfair to the author of #2.

The original post #1 asks a question. An answer #2 is provided. You post a modified question #3. Someone provides an answer #4. Now the whole thread #1, #2, #3, #4 can be read in sequence and understood. It becomes part of our archives for other students to find and to read in the future.
i made a mistake in the original post, so don't see any sense keeping it. but if you want, i'll create another one.

## 1. What is work done on a moving square wire loop in a B-field?

The work done on a moving square wire loop in a B-field is the amount of energy transferred to the loop as it moves through the magnetic field. This work is dependent on the strength of the magnetic field, the speed of the loop, and the orientation of the loop with respect to the magnetic field.

## 2. How is work calculated for a moving square wire loop in a B-field?

The work done on a moving square wire loop in a B-field can be calculated using the formula W = BILcosθ, where B is the magnetic field strength, I is the current flowing through the loop, L is the length of the wire, and θ is the angle between the direction of the magnetic field and the direction of motion of the loop.

## 3. What is the significance of work done on a moving square wire loop in a B-field?

The work done on a moving square wire loop in a B-field is significant because it is a measure of the energy transferred to the loop, which can be used to power devices or do other work. It also helps in understanding the behavior of electromagnetic systems and how they interact with magnetic fields.

## 4. How does the orientation of the square wire loop affect the work done in a B-field?

The orientation of the square wire loop with respect to the magnetic field affects the work done in a B-field because it determines the angle θ in the formula W = BILcosθ. When the loop is perpendicular to the magnetic field, θ is 90 degrees and the work done is maximum. When the loop is parallel to the magnetic field, θ is 0 degrees and the work done is zero.

## 5. Can work be done on a stationary square wire loop in a B-field?

No, work cannot be done on a stationary square wire loop in a B-field because there is no motion to transfer energy to the loop. The work done is only applicable when there is relative motion between the loop and the magnetic field.

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