Can Magnetic Fields Actually Do Work?

[fricke]

Why magnetic field never do work?

 

[fricke]

Magnetic force is always perpendicular to velocity so there is no work. (cos 90°=0)

if you pull a loop of wire (to the right) with velocity v and charge is moving perpendicular to your motion (upright) with velocity u, so which velocity does the magnetic force refer to? the loop of wire is immersed in an uniform magnetic field (toward the page)

 

[Noctisdark]

$dW = \vec F \cdot d\vec x,$ so $\frac{dW}{dt} = \vec F \cdot \frac{d\vec x}{dt} + \frac{d\vec F}{dt}\cdot \vec x$ in case of magnetic fields $\frac{dW}{dt} = \vec F \cdot \vec v + \vec x \cdot \frac{d\vec F}{dt} = 0 + \vec x \cdot q(\vec a \times \vec B + \vec v \times \frac{d\vec B}{dt} )$ note that $\frac{d\vec B}{dt} = 0$ since it's static so the result is $\frac{dW}{dt} = q\vec x \cdot (\vec a \times \vec B) = q\vec B \cdot (\vec a \times \vec x)$ since this is a circular motion then $\vec a$ and $\vec x$ have opposites direction so the whole thing simplifies into $\frac{dW}{dt} = 0$ the potential difference is zero, so you might expect that it doesn't do work

 

[Dale]

Why magnetic field never do work?

Are you familiar with Poyntings theorem?

 

[fricke]

Are you familiar with Poyntings theorem?

No, I haven't learned that theorem yet.

 

[Dale]

OK, so Poynting's theorem is the key theorem about energy and work in electromagnetism. I will briefly derive it here (using natural units so that I don't have to keep track of constants) for the "microscopic" Maxwell's equations. Often it is derived for the macroscopic equations, so you can easily look those up also.

Start from Ampere's law.
$\nabla \times B = \frac{\partial}{\partial t} E + J$

Then take the dot product with E and move everything over to one side.
$0 = - E \cdot \nabla \times B + E \cdot \frac{\partial}{\partial t} E + E \cdot J$

Then there is a vector identity for the divergence of a cross product https://en.wikipedia.org/wiki/Vector_calculus_identities#Divergence_2
$0 = \nabla \cdot (E \times B) - B \cdot \nabla \times E + E \cdot \frac{\partial}{\partial t} E + E \cdot J$

Then substituting in Faraday's law
$0 = \nabla \cdot (E \times B) + B \cdot \frac{\partial}{\partial t}B + E \cdot \frac{\partial}{\partial t} E + E \cdot J$

and simplifying gives
$0 = \nabla \cdot (E \times B) + \frac{\partial}{\partial t} (\frac{E^2}{2}+\frac{B^2}{2}) + E \cdot J$

The second term is the energy density of the electromagnetic field, and the third term is the power density of the work done on matter. So the first term is interpreted as the flow of electromagnetic energy from one location to another. So basically what it says is that any change in the electromagnetic energy density must be associated with a corresponding flux of electromagnetic energy or work done on matter.

Regarding your specific question in the OP, note that the term for work done on matter is $E \cdot J$ so the magnetic field is not involved.

 

[Jano L.]

OK, so Poynting's theorem is the key theorem about energy and work in electromagnetism. I will briefly derive it here (using natural units so that I don't have to keep track of constants) for the "microscopic" Maxwell's equations. Often it is derived for the macroscopic equations, so you can easily look those up also.

...

The second term is the energy density of the electromagnetic field, and the third term is the power density of the work done on matter. So the first term is interpreted as the flow of electromagnetic energy from one location to another. So basically what it says is that any change in the electromagnetic energy density must be associated with a corresponding flux of electromagnetic energy or work done on matter.

Regarding your specific question in the OP, note that the term for work done on matter is $E \cdot J$ so the magnetic field is not involved.

That is wrong answer to the original question. Poynting's theorem is a theorem about fields governed by macroscopic Maxwell equations. There is no concept of work involved and so there cannot be any way to derive the idea that magnetic force does no work.

The work and energy interpretation of the Poynting theorem is an additional assumption, based on the idea that matter is subject to force due to field and work of this force (energy supplied to matter) per unit time and unit volume is

$$
\mathbf j \cdot \mathbf E.
$$

The magnetic field does not occur because of an assumption. One cannot derive that it does no work from Poynting theorem here.

To answer the original question: the statement magnetic field does no work is a little foggy, it is sometimes true, sometimes not. It is true for charged particles, because there the magnetic force is always given by the vector expression

$$
\mathbf F_m = q \mathbf v\times \mathbf B
$$

where $q$ is charge of the particle, $\mathbf v$ its velocity and $\mathbf B$ is the external magnetic field at the position of the particle. This force is always perpendicular to velocity of the particle. Originally an experimental fact, but nowadays this can even be taken as definition of magnetic field in theory. If total force has parallel component to the velocity, its origin is sought elsewhere, in electric or mechanical forces.

The rate of work is given by the formula

$$
P = \mathbf F_m \cdot \mathbf v.
$$

Inserting the expression for magnetic force, we obtain

$$
P = q (\mathbf v \times \mathbf B) \cdot \mathbf v.
$$

Now, this expression is always 0, irrespective of vectors $\mathbf B$ or $\mathbf v$. Hence the rate of work of magnetic force on charged particle is always 0.

 

[Dale]

There is no concept of work involved

Work is a non-thermal transfer of energy. Poyntings theorem involves a non-thermal transfer of energy. Therefore it describes work.

 

[Jano L.]

Work is a non-thermal transfer of energy. Poyntings theorem involves a non-thermal transfer of energy. Therefore it describes work.

Poynting's theorem is an equation relating EM fields with charge and current densities. It does not "involve a transfer of energy" by itself.

Adopting the terms occurring in this theorem as EM energy and work done on matter is an interpretation introduced on top of this theorem; it is not part of it. This interpretation assumes power given to matter is given by $\mathbf j\cdot \mathbf E$. This assumption is valid in some cases like Ohmic dissipation in conductor but deficient in others, like when electro-chemical and other non-EM electromotive and ponderomotive phenomena are involved.

In macroscopic theory with regular sources, Poynting's theorem is always valid. Its work-energy interpretation is not always valid; there are electro-chemical, thermo-electric effects where work on matter cannot be handled properly with simple expressions from the Poynting theorem.

 

[Dale]

Adopting the terms occurring in this theorem as EM energy and work done on matter is an interpretation introduced on top of this theorem; it is not part of it.

Sure. similar statements can be made of any equation in physics.

This interpretation assumes power given to matter is given by j⋅E\mathbf j\cdot \mathbf E. This assumption is valid in some cases like Ohmic dissipation in conductor but deficient in others, like when electro-chemical and other non-EM electromotive and ponderomotive phenomena are involved.

Well, the particular one that I showed was the microscopic expression, so it works for those situations at a microscopic level.

In macroscopic theory with regular sources, Poynting's theorem is always valid. Its work-energy interpretation is not always valid; there are electro-chemical, thermo-electric effects where work on matter cannot be handled properly with simple expressions from the Poynting theorem.

I don't doubt that.

 

[Dale]

In macroscopic theory with regular sources, Poynting's theorem is always valid.

By the way, what is your favorite version of the macroscopic Poyntings?

 

[Jano L.]

By the way, what is your favorite version of the macroscopic Poyntings?

I don't think discussion of Poynting's theorem would be enlightening to the original question. If you have a specific question on Poynting's theorem, please post a new question here on the forum and I will check it out.