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Why is it assumed that magnetic forces arising from magnetic fields do not do work on a current carrying conductor?

The reason that it is assumed that magnetic forces do no work is not really an assumption at all; it can be proven directly from Maxwell's equations. This is known as Poynting's theorem. My favorite derivation is found at section 11.2 here: http://web.mit.edu/6.013_book/www/book.html

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right)+ \mathbf E \cdot \frac{\partial}{\partial t}\mathbf P + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf H \cdot \frac{\partial}{\partial t} \mathbf M + \mathbf E \cdot \mathbf J$$

In your case, with just a conductor, we have $\mathbf P=0$ and $\mathbf M=0$ so the equation simplifies to a more commonly recognizable form

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right) + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf E \cdot \mathbf J$$

Where the term on the left is the flow of energy from one region of the field to another and the first two terms on the right are the change in the energy density of the electric and magnetic fields respectively. Those are purely field terms; the only term involving an interaction with matter is the last term $\mathbf E \cdot \mathbf J$ where $\mathbf J$ is the free current. This means that the only way that work can be done on a conductor is via the $\mathbf E$ field.

Note, this is a derivation based on the macroscopic Maxwell's equations. A similar derivation can be done based on the microscopic Maxwell's equations, and again the only term involving an interaction with the matter of a conductor is $\mathbf E \cdot \mathbf J$. So regardless of if you are talking about the macroscopic or microscopic Maxwell's equations, for a conductor, the conclusion is the same: all work is done by the $\mathbf E$ field and the amount of work is given by $\mathbf E \cdot \mathbf J$.

It is not an assumption, it is a theorem. It holds as long as Maxwell's equations hold.

As @Alfred Centauri described in his answer whenever you have a situation that looks like there is a magnetic field doing work, you can always "dig deeper" and find where it is actually the $\mathbf E$ field doing the work.

However, instead of digging deeper, let's say that we want to step back a bit. Is there anything else we can learn? The term $\mathbf E \cdot \mathbf J$ includes not only the mechanical work, but also the "Ohmic" work, i.e. the resistive work that basically just generates heat. Usually we want to maximize the mechanical work and we want to minimize the Ohmic work. So suppose, instead of trying to find where the $\mathbf E$ field is, we try to separate out the Ohmic work from the mechanical work.

To do so, we will transform to the rest-frame of the conductor, since in that frame there is no mechanical work. Assuming that $v<<c$ the transformation equations are: $$\mathbf E' = \mathbf E + \mathbf v \times \mathbf B$$ $$\mathbf J' = \mathbf J - \rho \mathbf v$$ where the primed quantities are quantities in the rest frame of the conductor. Substituting those into $\mathbf E \cdot \mathbf J$ and simplifying, we get: $$\mathbf E \cdot \mathbf J = \mathbf E' \cdot \mathbf J' + \mathbf v \cdot (\rho \mathbf E + \mathbf J \times \mathbf B)$$

So, that means that the $\mathbf E \cdot \mathbf J$ term itself contains within it the mechanical work due to the magnetic field that you are interested in. If you pull out the Ohmic work, then the mechanical work is exactly what you would expect including a term from the magnetic field.

This result may seem a little surprising or confusing, as it appears to contradict the above, but it does not. The thing is that all of the fields in electromagnetism are closely related to each other. You can often express the same thing in multiple ways, or dig out hidden dependencies. So although Poynting's theorem holds and although it clearly states that the total work is always $\mathbf E \cdot \mathbf J$, it is not a mere coincidence that formulas describing only the mechanical work correctly include the $\mathbf B$ field and show that it does mechanical work.

Why is it assumed that magnetic forces arising from magnetic fields do not do work on a current carrying conductor?

The reason that it is assumed that magnetic forces do no work is not really an assumption at all; it can be proven directly from Maxwell's equations. This is known as Poynting's theorem. My favorite derivation is found at section 11.2 here: http://web.mit.edu/6.013_book/www/book.html

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right)+ \mathbf E \cdot \frac{\partial}{\partial t}\mathbf P + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf H \cdot \frac{\partial}{\partial t} \mathbf M + \mathbf E \cdot \mathbf J$$

In your case, with just a conductor, we have $\mathbf P=0$ and $\mathbf M=0$ so the equation simplifies to a more commonly recognizable form

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right) + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf E \cdot \mathbf J$$

Where the term on the left is the flow of energy from one region of the field to another and the first two terms on the right are the change in the energy density of the electric and magnetic fields respectively. Those are purely field terms; the only term involving an interaction with matter is the last term $\mathbf E \cdot \mathbf J$ where $\mathbf J$ is the free current. This means that the only way that work can be done on a conductor is via the $\mathbf E$ field.

Note, this is a derivation based on the macroscopic Maxwell's equations. A similar derivation can be done based on the microscopic Maxwell's equations, and again the only term involving an interaction with the matter of a conductor is $\mathbf E \cdot \mathbf J$. So regardless of if you are talking about the macroscopic or microscopic Maxwell's equations, for a conductor, the conclusion is the same: all work is done by the $\mathbf E$ field and the amount of work is given by $\mathbf E \cdot \mathbf J$.

It is not an assumption, it is a theorem. It holds as long as Maxwell's equations hold.

As @Alfred Centauri described in his answer whenever you have a situation that looks like there is a magnetic field doing work, you can always "dig deeper" and find where it is actually the $\mathbf E$ field doing the work.

However, instead of digging deeper, let's say that we want to step back a bit. Is there anything else we can learn? The term $\mathbf E \cdot \mathbf J$ includes not only the mechanical work, but also the non-mechanical work. Usually we want to maximize the mechanical work and we want to minimize the non-mechanical work. So suppose, instead of trying to find where the $\mathbf E$ field is, we try to separate out the non-mechanical work from the mechanical work.

To do so, we will transform to the rest-frame of the conductor, since in that frame there is no mechanical work so all of the work in that frame is non-mechanical. Assuming that $v<<c$ the transformation equations are: $$\mathbf E' = \mathbf E + \mathbf v \times \mathbf B$$ $$\mathbf J' = \mathbf J - \rho \mathbf v$$ where the primed quantities are quantities in the rest frame of the conductor. Substituting those into $\mathbf E \cdot \mathbf J$ and simplifying, we get: $$\mathbf E \cdot \mathbf J = \mathbf E' \cdot \mathbf J' + \mathbf v \cdot (\rho \mathbf E + \mathbf J \times \mathbf B)$$

So, that means that the $\mathbf E \cdot \mathbf J$ term itself contains within it the mechanical work due to the magnetic field that you are interested in. If you pull out the non-mechanical work, then the mechanical work is exactly what you would expect including a term from the magnetic field.

This result may seem a little surprising or confusing, as it appears to contradict the above, but it does not. The thing is that all of the fields in electromagnetism are closely related to each other. You can often express the same thing in multiple ways, or dig out hidden dependencies. So although Poynting's theorem holds and although it clearly states that the total work is always $\mathbf E \cdot \mathbf J$, it is not a mere coincidence that formulas describing only the mechanical work correctly include the $\mathbf B$ field and show that it does mechanical work.

Why is it assumed that magnetic forces arising from magnetic fields do not do work on a current carrying conductor?

The reason that it is assumed that magnetic forces do no work is not really an assumption at all; it can be proven directly from Maxwell's equations. This is known as Poynting's theorem. My favorite derivation is found at section 11.2 here: http://web.mit.edu/6.013_book/www/book.html

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right)+ \mathbf E \cdot \frac{\partial}{\partial t}\mathbf P + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf H \cdot \frac{\partial}{\partial t} \mathbf M + \mathbf E \cdot \mathbf J$$

In your case, with just a conductor, we have $\mathbf P=0$ and $\mathbf M=0$ so the equation simplifies to a more commonly recognizable form

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right) + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf E \cdot \mathbf J$$

Where the term on the left is the flow of energy from one region of the field to another and the first two terms on the right are the change in the energy density of the electric and magnetic fields respectively. Those are purely field terms; the only term involving an interaction with matter is the last term $\mathbf E \cdot \mathbf J$ where $\mathbf J$ is the free current. This means that the only way that work can be done on a conductor is via the $\mathbf E$ field.

Note, this is a derivation based on the macroscopic Maxwell's equations. A similar derivation can be done based on the microscopic Maxwell's equations, and again the only term involving an interaction with the matter of a conductor is $\mathbf E \cdot \mathbf J$. So regardless of if you are talking about the macroscopic or microscopic Maxwell's equations, for a conductor, the conclusion is the same: all work is done by the $\mathbf E$ field and the amount of work is given by $\mathbf E \cdot \mathbf J$.

It is not an assumption, it is a theorem. It holds as long as Maxwell's equations hold.

As @Alfred Centauri described in his answer whenever you have a situation that looks like there is a magnetic field doing work, you can always "dig deeper" and find where it is actually the $\mathbf E$ field doing the work.

However, instead of digging deeper, let's say that we want to step back a bit. Is there anything else we can learn? The term $\mathbf E \cdot \mathbf J$ includes not only the mechanical work, but also the "Ohmic" work, i.e. the resistive work that basically just generates heat. Usually we want to maximize the mechanical work and we want to minimize the Ohmic work. So suppose, instead of trying to find where the $\mathbf E$ field is, we try to separate out the Ohmic work from the mechanical work.

To do so, we will transform to the rest-frame of the conductor, since in that frame there is no mechanical work. Assuming that $v<<c$ the transformation equations are: $$\mathbf E' = \mathbf E + \mathbf v \times \mathbf B$$ $$\mathbf J' = \mathbf J - \rho \mathbf v$$ where the primed quantities are quantities in the rest frame of the conductor. Substituting those into $\mathbf E \cdot \mathbf J$ and simplifying, we get: $$\mathbf E \cdot \mathbf J = \mathbf E' \cdot \mathbf J' + \mathbf v \cdot (\rho \mathbf E + \mathbf J \times \mathbf B)$$

So, that means that the $\mathbf E \cdot \mathbf J$ term itself contains within it the mechanical work due to the magnetic field that you are interested in. If you pull out the Ohmic work, then the mechanical work is exactly what you would expect including a term from the magnetic field.

This result may seem a little surprising or confusing, as it appears to contradict the above, but it does not. The thing is that all of the fields in electromagnetism are closely related to each other. You can often express the same thing in multiple ways, or dig out hidden dependencies. So although Poynting's theorem holds and although it clearly states that the total work is always $\mathbf E \cdot \mathbf J$, it is not a mere coincidence that formulas describing only the mechanical work correctly include the $\mathbf B$ field as doing mechanical work.

Why is it assumed that magnetic forces arising from magnetic fields do not do work on a current carrying conductor?

The reason that it is assumed that magnetic forces do no work is not really an assumption at all; it can be proven directly from Maxwell's equations. This is known as Poynting's theorem. My favorite derivation is found at section 11.2 here: http://web.mit.edu/6.013_book/www/book.html

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right)+ \mathbf E \cdot \frac{\partial}{\partial t}\mathbf P + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf H \cdot \frac{\partial}{\partial t} \mathbf M + \mathbf E \cdot \mathbf J$$

In your case, with just a conductor, we have $\mathbf P=0$ and $\mathbf M=0$ so the equation simplifies to a more commonly recognizable form

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right) + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf E \cdot \mathbf J$$

Where the term on the left is the flow of energy from one region of the field to another and the first two terms on the right are the change in the energy density of the electric and magnetic fields respectively. Those are purely field terms; the only term involving an interaction with matter is the last term $\mathbf E \cdot \mathbf J$ where $\mathbf J$ is the free current. This means that the only way that work can be done on a conductor is via the $\mathbf E$ field.

Note, this is a derivation based on the macroscopic Maxwell's equations. A similar derivation can be done based on the microscopic Maxwell's equations, and again the only term involving an interaction with the matter of a conductor is $\mathbf E \cdot \mathbf J$. So regardless of if you are talking about the macroscopic or microscopic Maxwell's equations, for a conductor, the conclusion is the same: all work is done by the $\mathbf E$ field and the amount of work is given by $\mathbf E \cdot \mathbf J$.

It is not an assumption, it is a theorem. It holds as long as Maxwell's equations hold.

As @Alfred Centauri described in his answer whenever you have a situation that looks like there is a magnetic field doing work, you can always "dig deeper" and find where it is actually the $\mathbf E$ field doing the work.

However, instead of digging deeper, let's say that we want to step back a bit. Is there anything else we can learn? The term $\mathbf E \cdot \mathbf J$ includes not only the mechanical work, but also the "Ohmic" work, i.e. the resistive work that basically just generates heat. Usually we want to maximize the mechanical work and we want to minimize the Ohmic work. So suppose, instead of trying to find where the $\mathbf E$ field is, we try to separate out the Ohmic work from the mechanical work.

To do so, we will transform to the rest-frame of the conductor, since in that frame there is no mechanical work. Assuming that $v<<c$ the transformation equations are: $$\mathbf E' = \mathbf E + \mathbf v \times \mathbf B$$ $$\mathbf J' = \mathbf J - \rho \mathbf v$$ where the primed quantities are quantities in the rest frame of the conductor. Substituting those into $\mathbf E \cdot \mathbf J$ and simplifying, we get: $$\mathbf E \cdot \mathbf J = \mathbf E' \cdot \mathbf J' + \mathbf v \cdot (\rho \mathbf E + \mathbf J \times \mathbf B)$$

So, that means that the $\mathbf E \cdot \mathbf J$ term itself contains within it the mechanical work due to the magnetic field that you are interested in. If you pull out the Ohmic work, then the mechanical work is exactly what you would expect including a term from the magnetic field.

This result may seem a little surprising or confusing, as it appears to contradict the above, but it does not. The thing is that all of the fields in electromagnetism are closely related to each other. You can often express the same thing in multiple ways, or dig out hidden dependencies. So although Poynting's theorem holds and although it clearly states that the total work is always $\mathbf E \cdot \mathbf J$, it is not a mere coincidence that formulas describing only the mechanical work correctly include the $\mathbf B$ field and show that it does mechanical work.

Why is it assumed that magnetic forces arising from magnetic fields do not do work on a current carrying conductor?

The reason that it is assumed that magnetic forces do no work is not really an assumption at all; it can be proven directly from Maxwell's equations. This is known as Poynting's theorem. My favorite derivation is found at section 11.2 here: http://web.mit.edu/6.013_book/www/book.html

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right)+ \mathbf E \cdot \frac{\partial}{\partial t}\mathbf P + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf H \cdot \frac{\partial}{\partial t} \mathbf M + \mathbf E \cdot \mathbf J$$

In your case, with just a conductor, we have $\mathbf P=0$ and $\mathbf M=0$ so the equation simplifies to a more commonly recognizable form

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right) + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf E \cdot \mathbf J$$

Where the term on the left is the flow of energy from one region of the field to another and the first two terms on the right are the change in the energy density of the electric and magnetic fields respectively. Those are purely field terms; the only term involving an interaction with matter is the last term $\mathbf E \cdot \mathbf J$ where $\mathbf J$ is the free current. This means that the only way that work can be done on a conductor is via the $\mathbf E$ field.

Note, this is a derivation based on the macroscopic Maxwell's equations. A similar derivation can be done based on the microscopic Maxwell's equations, and again the only term involving an interaction with the matter of a conductor is $\mathbf E \cdot \mathbf J$. So regardless of if you are talking about the macroscopic or microscopic Maxwell's equations, for a conductor, the conclusion is the same: all work is done by the $\mathbf E$ field and the amount of work is given by $\mathbf E \cdot \mathbf J$.

It is not an assumption, it is a theorem. It holds as long as Maxwell's equations hold.

As @Alfred Centauri described in his answer whenever you have a situation that looks like there is a magnetic field doing work, you can always "dig deeper" and find where it is actually the $\mathbf E$ field doing the work.

However, instead of digging deeper, let's say that we want to step back a bit. Is there anything else we can learn? The term $\mathbf E \cdot \mathbf J$ includes not only the mechanical work, but also the "Ohmic" work, i.e. the resistive work that basically just generates heat. Usually we want to maximize the mechanical work and we want to minimize the Ohmic work. So suppose, instead of trying to find where the $\mathbf E$ field is, we try to separate out the Ohmic work from the mechanical work.

To do so, we will transform to the rest-frame of the conductor, since in that frame there is no mechanical work. Assuming that $v<<c$ the transformation equations are: $$\mathbf E' = \mathbf E + \mathbf v \times \mathbf B$$ $$\mathbf J' = \mathbf J - \rho \mathbf v$$ where the primed quantities are quantities in the rest frame of the conductor. Substituting those into $\mathbf E \cdot \mathbf J$ and simplifying, we get: $$\mathbf E \cdot \mathbf J = \mathbf E' \cdot \mathbf J' + \mathbf v \cdot (\rho \mathbf E + \mathbf J \times \mathbf B)$$

So, that means that the $\mathbf E \cdot \mathbf J$ term itself contains within it the mechanical work due to the magnetic field that you are interested in. If you pull out the Ohmic work, then the mechanical work is exactly what you would expect including a term from the magnetic field.

This result may seem a little surprising or confusing, as it appears to contradict the above, but it does not. The thing is that all of the fields in electromagnetism are closely related to each other. You can often express the same thing in multiple ways, or dig out hidden dependencies. So although Poynting's theorem holds and although it clearly states that the work is $\mathbf E \cdot \mathbf J$, it is not a mere coincidence that formulas describing the mechanical work only often correctly include the $\mathbf B$ field as doing work.

Why is it assumed that magnetic forces arising from magnetic fields do not do work on a current carrying conductor?

The reason that it is assumed that magnetic forces do no work is not really an assumption at all; it can be proven directly from Maxwell's equations. This is known as Poynting's theorem. My favorite derivation is found at section 11.2 here: http://web.mit.edu/6.013_book/www/book.html

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right)+ \mathbf E \cdot \frac{\partial}{\partial t}\mathbf P + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf H \cdot \frac{\partial}{\partial t} \mathbf M + \mathbf E \cdot \mathbf J$$

In your case, with just a conductor, we have $\mathbf P=0$ and $\mathbf M=0$ so the equation simplifies to a more commonly recognizable form

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right) + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf E \cdot \mathbf J$$

Where the term on the left is the flow of energy from one region of the field to another and the first two terms on the right are the change in the energy density of the electric and magnetic fields respectively. Those are purely field terms; the only term involving an interaction with matter is the last term $\mathbf E \cdot \mathbf J$ where $\mathbf J$ is the free current. This means that the only way that work can be done on a conductor is via the $\mathbf E$ field.

Note, this is a derivation based on the macroscopic Maxwell's equations. A similar derivation can be done based on the microscopic Maxwell's equations, and again the only term involving an interaction with the matter of a conductor is $\mathbf E \cdot \mathbf J$. So regardless of if you are talking about the macroscopic or microscopic Maxwell's equations, for a conductor, the conclusion is the same: all work is done by the $\mathbf E$ field and the amount of work is given by $\mathbf E \cdot \mathbf J$.

It is not an assumption, it is a theorem. It holds as long as Maxwell's equations hold.

As @Alfred Centauri described in his answer whenever you have a situation that looks like there is a magnetic field doing work, you can always "dig deeper" and find where it is actually the $\mathbf E$ field doing the work.

However, instead of digging deeper, let's say that we want to step back a bit. Is there anything else we can learn? The term $\mathbf E \cdot \mathbf J$ includes not only the mechanical work, but also the "Ohmic" work, i.e. the resistive work that basically just generates heat. Usually we want to maximize the mechanical work and we want to minimize the Ohmic work. So suppose, instead of trying to find where the $\mathbf E$ field is, we try to separate out the Ohmic work from the mechanical work.

To do so, we will transform to the rest-frame of the conductor, since in that frame there is no mechanical work. Assuming that $v<<c$ the transformation equations are: $$\mathbf E' = \mathbf E + \mathbf v \times \mathbf B$$ $$\mathbf J' = \mathbf J - \rho \mathbf v$$ where the primed quantities are quantities in the rest frame of the conductor. Substituting those into $\mathbf E \cdot \mathbf J$ and simplifying, we get: $$\mathbf E \cdot \mathbf J = \mathbf E' \cdot \mathbf J' + \mathbf v \cdot (\rho \mathbf E + \mathbf J \times \mathbf B)$$

So, that means that the $\mathbf E \cdot \mathbf J$ term itself contains within it the mechanical work due to the magnetic field that you are interested in. If you pull out the Ohmic work, then the mechanical work is exactly what you would expect including a term from the magnetic field.

This result may seem a little surprising or confusing, as it appears to contradict the above, but it does not. The thing is that all of the fields in electromagnetism are closely related to each other. You can often express the same thing in multiple ways, or dig out hidden dependencies. So although Poynting's theorem holds and although it clearly states that the total work is always $\mathbf E \cdot \mathbf J$, it is not a mere coincidence that formulas describing only the mechanical work correctly include the $\mathbf B$ field as doing mechanical work.