# Determine Magnetic Field from Lorentz Force

1. Aug 13, 2013

### physicsphreak2

1. The problem statement, all variables and given/known data
From Zangwill, Chapter 2.

Suppose that the instantaneous forces on a charge, $q$, when it is moving with velocities $\vec{v}_1$ or $\vec{v}_2$, are $\vec{F}_1$ and $\vec{F}_2$, respectively. Without choosing a coordinate system, show that $\vec{B}(\vec{r})$ can be determined by working with the observable quantities $\vec{v}_1 \times \vec{F}_1$ and $\vec{v}_2 \times \vec{F}_2$, with $\vec{v}_1$ and $\vec{v}_2$ being suitably chosen.

2. Relevant equations
Lorentz force law: $\vec{F} = q(\vec{v} \times \vec{B}(\vec{r}))$

3. The attempt at a solution
I recognize that $\vec{v}_i \times \vec{F}_i = \vec{v}_i (\vec{v}_i \cdot \vec{B}(\vec{r})) - \vec{B}(\vec{r})(|\vec{v}_i|^2)$ by the BAC-CAB rule, with i=1 or 2 in this case.

So, I am tempted to say that if you judiciously choose the direction such that $\vec{v}_i \perp \vec{B}(\vec{r})$ we can immediately get the magnetic field (because the first term in the BAC-CAB expansion is zero, if the vectors are perpendicular). I might argue that we can do this by considering all possible velocities until we have maximized the force (since that max comes when the vectors being crossed are perpendicular)... would that be a satisfactory way to find those special directions for the velocities?

But would I also need the test velocities $\vec{v}_i$ to be perpendicular to each other? Otherwise, they could be anywhere in the plane perpendicular to $\vec{B}$. And, once I have required this, I have the vector components of $\vec{B}$ in 2 directions (the directions determined by $\vec{v}_i \times \vec{F}_i$). Actually, maybe I only have it in one direction since the velocity vectors form a plane, so their cross products with each force will be parallel??

How do I determine the magnitude of $\vec{B}$ in the 3rd direction, the one perpendicular to the plane spanned by the two velocity vectors I have chosen? I feel like I only have 2 of the three components of the the magnetic field using my method.

Am I at least close?!

Last edited: Aug 13, 2013
2. Aug 13, 2013

### TSny

Suppose you pick $\vec{v}_1$ arbitrarily and measure $\vec{F}_1$ for this velocity. Think about what you can determine about $\vec{B}$ from this measurement. You might need to consider two cases here, the case where $\vec{F}_1$ happens to be zero and the case where it is nonzero.

Last edited: Aug 13, 2013
3. Aug 14, 2013

### physicsphreak2

Having chosen the first velocity arbitrarily, let's use your suggestion about the two cases.

1) If the force happens to be zero then we have found the direction of the magnetic field (it is the direction of $\vec{v}_1$). We can then choose $\vec{v}_2$ to be any vector in the plane perpendicular to the first velocity to calculate the magnitude of the magnetic field... then we're done with this case?

2) If the force is not zero, would we still want to choose a perpendicular velocity as the second test case? Visually, I can't figure out how to extract both magnitude and direction with just one more test charge velocity.

4. Aug 14, 2013

### TSny

Could $\vec{B}$ be in the opposite direction of $\vec{v}_1$?

Yes, that seems right to me.

You can deduce the magnitude and direction of a vector if you know the plane in which it lies and also the components of the vector along two mutually perpendicular directions in that plane.

5. Aug 16, 2013

### physicsphreak2

Ok, so for the first case I should also specify that we can determine the direction of B by looking at the direction of the force exerted and using the right hand rule to deduce whether B is paralell or anti-parallel, correct?

If we choose $\vec{v}_1 \perp \vec{v}_2$ then we should at least be guaranteed to have the magnetic field lying in the plane spanned by the velocities, yes? In which case, we have two components in orthogonal directions in the plane and the plane so we've also determined B in that case?

6. Aug 17, 2013

### TSny

Yes.

I think you have it. Assuming $\vec{F}_1 \neq 0$, then you know that $\vec{B}$ as well as $\vec{v}_1$ lie in a plane $\bot$ $\vec{F}_1$. From the measurement of $\vec{F}_1$ you can deduce the component of $\vec{B}$ that is $\bot$ $\vec{v}_1$.

Then choose $\vec{v}_2$ to lie in that same plane and oriented perpendicular to $\vec{v}_1$. From the measurement of $\vec{F}_2$ you can deduce the component of $\vec{B}$ in the plane that is $\bot$ $\vec{v}_2$.

You therefore have determined a plane in which $\vec{B}$ lies as well as two perpendicular components of $\vec{B}$ in that plane. So, you can construct $\vec{B}$.