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I am preparing for an exam and one of the questions I have come across asks:

Define the electric field $\mathbf{E}$ and the magnetic flux density $\mathbf{B}$, in terms of the force on charges and currents.

By the Lorentz force law we have:

$$\mathbf{F}=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$$

Where $\mathbf{v}$ is the velocity of the charge carrying particle. If we then set $\mathbf{B}=\vec{0}$ we get:

$$\mathbf{E}=\lim_{q\to 0}\left(\frac{\mathbf{F}}{q}\right)$$

However, setting $\mathbf{E}=\vec{0}$ we get: $\mathbf{F}=q\mathbf{v}\times\mathbf{B}=\mathbf{I}\times\mathbf{B}$, where $\mathbf{I}$ is the current vector. However, there is no unique inversion for the cross product and therefore I am not sure how I am supposed to define $\mathbf{B}$ in terms of $\mathbf{F}$ and $\mathbf{I}$? Is there a standard definition like for the electric field?

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The question is probably simply asking you to write down the Lorentz force law, rather than rearrange it for $\mathbf{E}$ and $\mathbf{B}$ respectively. You could say: the magnetic flux density is the vector field $\mathbf{B}$ such that the force on a current $\mathbf{I}$ due to it is given by $\mathbf{F} = \mathbf{I} \times \mathbf{B}$.

The cross product, as you say, cannot be inverted. To see this, we note that the direction of $\mathbf{F}$ only tells us that $\mathbf{B}$ must lie in the plane perpendicular to $\mathbf{F}$. Then by the formula:

$$ |\mathbf{F}| = |\mathbf{I}||\mathbf{B}| \sin \theta \,, $$

we see that that the magnitude of $\mathbf{F}$ only pins down the value of $|\mathbf{B}|\sin \theta$, which involves two undetermined quantities. Hence we cannot determine $\mathbf{B}$. If we could invert the cross product, we would do something like this: consider that the cross product is linear, that is:

$$\mathbf{I} \times ( \alpha \mathbf{B}_1 + \beta \mathbf{B}_2) = \alpha \mathbf{I} \times \mathbf{B}_1 + \beta \mathbf{I} \times \mathbf{B}_2\,.$$

This means that we can write our cross product as a matrix equation:

$$ \mathbf{F} = \mathbf{I} \times \mathbf{B} \equiv \mathsf{M} \mathbf{B}\,. $$

Now what is the form of this matrix? To work this out, let's use suffix notation:

$$ F_i = \epsilon_{ijk} I_j B_k \,.$$

So we just have that

$$M_{ik} = \epsilon_{ijk} I_j\,. $$

At this point, you can check that the matrix $\mathsf{M}$ is not invertible, and so we cannot invert to give:

$$ \mathbf{B} = \mathsf{M}^{-1} \mathbf{F} \,,$$

as we would like. Consequently there's simply no way of writing $\mathbf{B}$ in terms of $\mathbf{I}$ and $\mathbf{F}$

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The Biot-Savart law expresses the magnetic field in terms of the current as, $$ \mathbf{B}=\frac{\mu_0}{4\pi}\int_C\frac{I\,d\mathbf{l}\times\mathbf{r}}{|\mathbf{r}|^3} $$ which does not account for any forces, just the current, $I$. You might be able to solve $\mathbf B$ in terms of $\mathbf F$ and $\mathbf I$ using some vector calculus identities, but I'm not convinced that it's possible due to the non-uniqueness of the inverse cross product.

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  • $\begingroup$ Thank you Kyle, I'm aware of the Biot-Savart law; I was just wondering if I'd missed something blindingly obvious in the question (I'm starting to panic about the upcoming exams), but now I'm just extremely confused as to what they want! $\endgroup$ Commented Apr 4, 2014 at 12:56
  • $\begingroup$ My guess is that they want the electric field as you've written it and the Biot-Savart law as I wrote it, but I'm not your professor! :D You may want to ask your instructor for clarification. $\endgroup$
    – Kyle Kanos
    Commented Apr 4, 2014 at 13:01
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    $\begingroup$ I disagree here: the Biot-Savart law is not defining the B-field in terms of forces on currents. $\endgroup$
    – gj255
    Commented Apr 4, 2014 at 13:11
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    $\begingroup$ I mean you could choose to read it like that, but I'm pretty convinced that's not what the examiners mean. I strongly suspect that writing the Lorentz force law would get you the marks here, and the Biot-Savart law not. $\endgroup$
    – gj255
    Commented Apr 4, 2014 at 13:15
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    $\begingroup$ I mean obviously I cannot be sure that this is what is meant by the question, but I can offer my opinion in this, the comments section. To ask, in a single sentence, the student to write $\vec{E}$ in terms of the effect it has on a charge, and $\vec{B}$ in terms of how it is generated by currents, is clearly mixing two different ideas in a way that an exam simply wouldn't. There isn't even a comma separating 'the force on charges' from 'currents', and so it strikes me as highly unlikely that this means anything other than 'the force on 'charges and currents'' $\endgroup$
    – gj255
    Commented Apr 4, 2014 at 13:21
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You might be interested in this. Lets look in case where $\vec E=0$. If you know $q \vec v$, $\vec F$ and $\alpha=\angle(\vec B;q \vec v)$ you can actually find $\vec B$. $$\vec F=q \vec v \times \vec B$$ $$\vec B=\frac{|F|}{ q|v| sin \alpha} \left (\frac{q \vec v cos \alpha}{q |v|}+\frac{q\vec v \times \vec F sin \alpha}{|q\vec v \times \vec F|} \right)= \frac{|F|}{ q|v| sin \alpha} \left (\frac{ \vec v cos \alpha}{ |v|}+\frac{\vec v \times \vec F sin \alpha}{|\vec v \times \vec F|} \right)$$

Lets look at case where $\vec E \neq0$ $$\vec F-q \vec E=q \vec v \times B$$ $$\vec B=\frac{|\vec F- q \vec E|}{ q|v| sin \alpha} \left (\frac{q \vec v cos \alpha}{q |v|}+\frac{q\vec v \times (\vec F- q \vec E) sin \alpha}{|q\vec v \times (\vec F- q \vec E)|} \right)= \frac{|\vec F- q \vec E|}{ q|v| sin \alpha} \left (\frac{ \vec v cos \alpha}{ |v|}+\frac{\vec v \times (\vec F- q \vec E) sin \alpha}{|\vec v \times (\vec F- q \vec E)|} \right)= $$

If you interested how this equations were derived look here: https://math.stackexchange.com/questions/4277293/finding-vec-b-from-vec-a-times-vec-b-vec-a-and-alpha-angle-vec.

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