Force density from an electromagnetic field

[dykuma]

Homework Statement

I am given the electric and magnetic field, $\vec E= \alpha x \hat x$ and $\vec B= \beta x \hat y$, where $\alpha$ and $\beta$ are constants. It is assumed that neither field is changing in time.

I am then asked to find the electromagnetic 'force per volume' (which I believe to be the 'force density'), as well as the 'power per volume' that's converted from electromagnetic energy into mechanical energy. Lastly, I'm asked to find a form of $\alpha$ and $\beta$ that makes the force density equal zero, besides both of them equaling zero.

Relevant Equations

$$\vec F_V = \rho \vec E + \vec J \times \vec B$$
possibly (not sure if this is correct) that work per volume is
$$ W_V = \frac {\varepsilon_0 \vec E^2} { \ 2 } + \frac {\vec B^2} { \ \mu_0 2 } $$

My guess is that the force per volume is:
$$ \vec F_V = \rho \alpha x \hat x + \vec J \times \beta x \hat y$$
but I'm not sure where to go after that. I'm not given a value for either the charge density or the current density, so I can't simplify the relation much. Further, I'm not sure if my equation for the "work per volume" is correct. however, it's my best guess for what's being asked for.

Edit:
I've made some progress, which I've attached below. The issue is that I'm suppose to find conditions for that make the force vanish, without having the terms themselves be zero. I can't do this, because the only way for the force volume to be zero is for 'x' to be zero, or for alpha to be zero. (my understanding is that $\alpha$ and $\beta$ need to be related to each other so that the constant's themselves sum the force volume relation to zero).

 

[dykuma]

Nevermind, I figured it out. The problem was that I took the cross product incorrectly. it should have been:
$$(\nabla \times \vec B) \times \vec B$$
not
$$\nabla \times (\vec B \times \vec B)$$

 

[Delta2]

Yes, well the operation of cross product $\times$ is not associative, though the equality $$(\nabla\times\vec{B})\times\vec{B}=\nabla\times(\vec{B}\times\vec{B})$$ might hold in some special cases, like for example when $\vec{B}=\vec{c}$ is a constant vector, or when $\nabla\times\vec{B}=\vec{0}$.

 

[dykuma]

That's what I thought. However, when I tried to simplify it, I got:

$$\nabla \times (\vec B \times \vec B) = \vec B(\nabla \cdot \vec B) - \vec B(\nabla \cdot \vec B) + (\nabla \cdot \vec B)\vec B - (\nabla \cdot \vec B)\vec B = 0$$

But when I actually carried out the work, I got an answer. I guess I have that identity wrong?

 

[Delta2]

No the identity seems correct. Another way to view the result directly is to notice that $\vec{B}\times\vec{B}=\vec{0}$

 

[dykuma]

So then, what did I do wrong here?

 

[Delta2]

you do nothing wrong here (except that you use the symbol $d$ for derivatives while I believe it is more formally correct to use the symbol $\partial$ for partial derivatives), you calculate $\frac{(\nabla\times \vec{B})\times\vec{B}}{\mu_0}$ instead which you found correctly to be $-\frac{\beta^2x\hat x}{\mu_0}$

The equality presented at post #3, holds only in some special cases, it doesn't generally hold because the cross product doesn't have the associative property.

 

[TSny]

possibly (not sure if this is correct) that work per volume is
$$ W_V = \frac {\varepsilon_0 \vec E^2} { \ 2 } + \frac {\vec B^2} { \ \mu_0 2 } $$

The right-hand side of this equation represents the field energy per unit volume. But you need to find the rate, per unit volume, that the fields are doing work on the material that has charge density $\rho$ and current density $\vec J$. Recall from mechanics that the rate at which a force $\vec F$ does work on a particle moving with velocity $\vec v$ is $\vec F \cdot \vec v$. Try to extend this to your situation. Hint: How are $\vec v$, $\rho$ and $\vec J$ related?

 

[dykuma]

Right, so because:
$$ \vec J = \rho \vec v$$

So I want:

$$ P_V = \vec F \cdot \frac {\vec J} {\rho}$$
?

 

[TSny]

$$ P_V = \vec F \cdot \frac {\vec J} {\rho}$$

Yes. This will be the rate of doing work per unit volume if $\vec F$ is the force per unit volume $\vec F_V$. Using the expression $\vec F_V = \rho \vec E + \vec J \times \vec B$, this will reduce to a simple expression for $P_V$.

 

[dykuma]

To go through all of this then, does this look correct to you?

 

[dykuma]

Actually, I see that I made an error there.

I'm not sure this makes sense though. I would expect there to be some mechanical work done (because the energy per volume is not zero), but because of the dot product (I'm dotting everything that's in the $\hat x$ direction with the current density which is in the $\hat z$ direction), it vanishes.

My next question is how is current density is related to the power density? Are they related at all, or is this zero because they are unrelated?

 

[Delta2]

It is a well known result, that the power (per unit volume) delivered from the EM field to matter is $$P=\vec{E}\cdot\vec{J}$$ so your work and result seems consistent with this.

 

[TSny]

Substitute $\bf f_V = \rho \bf E + \bf J \times \bf B$ into $P_V = \bf f_V \cdot \frac{\bf J}{\rho}$. It should simplify in just a couple of steps to the simple formula given by @Delta2 in post #13. Then it will be easy to obtain $P_V = 0$ for this problem.