How to express magnetic field vector in terms of force on current

Question

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?

Answer 1

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 $\mathbf{F}$

Answer 2

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.

Answer 3

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.