Determine Magnetic Field from Lorentz Force

In summary, the instantaneous forces on a charge, q, when it is moving with velocities \vec{v}_1 or \vec{v}_2, can be used to determine the magnetic field, \vec{B}(\vec{r}), without choosing a coordinate system. By measuring the observable quantities \vec{v}_1 \times \vec{F}_1 and \vec{v}_2 \times \vec{F}_2, we can determine a plane in which \vec{B}(\vec{r}) lies as well as two perpendicular components of \vec{B}(\vec{r}) in that plane, allowing us to construct \vec{B}(\vec{r}). This can be done
  • #1
physicsphreak2
13
0

Homework Statement


From Zangwill, Chapter 2.

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

Homework Equations


Lorentz force law: [itex] \vec{F} = q(\vec{v} \times \vec{B}(\vec{r})) [/itex]

The Attempt at a Solution


I recognize that [itex] \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)[/itex] 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 [itex] \vec{v}_i \perp \vec{B}(\vec{r}) [/itex] 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 [itex] \vec{v}_i [/itex] to be perpendicular to each other? Otherwise, they could be anywhere in the plane perpendicular to [itex] \vec{B} [/itex]. And, once I have required this, I have the vector components of [itex] \vec{B} [/itex] in 2 directions (the directions determined by [itex] \vec{v}_i \times \vec{F}_i [/itex]). 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 [itex] \vec{B} [/itex] 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?!
 
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  • #2
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.
 
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  • #3
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 [itex] \vec{v}_1 [/itex]). We can then choose [itex] \vec{v}_2 [/itex] 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
physicsphreak2 said:
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 [itex] \vec{v}_1 [/itex]).

Could ##\vec{B}## be in the opposite direction of [itex] \vec{v}_1 [/itex]?

We can then choose [itex] \vec{v}_2 [/itex] 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?

Yes, that seems right to me.

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.

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
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 parallel or anti-parallel, correct?

If we choose [itex]\vec{v}_1 \perp \vec{v}_2 [/itex] 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
physicsphreak2 said:
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 parallel or anti-parallel, correct?

Yes.

If we choose [itex]\vec{v}_1 \perp \vec{v}_2 [/itex] 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?

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 [itex]\bot[/itex] ##\vec{F}_1##. From the measurement of ##\vec{F}_1## you can deduce the component of ##\vec{B}## that is [itex]\bot[/itex] ##\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 [itex]\bot[/itex] ##\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}##.
 

1. What is the Lorentz Force?

The Lorentz Force is a fundamental law in electromagnetism that describes the force exerted on a charged particle moving through a magnetic field. It is given by the vector cross product of the particle's velocity and the magnetic field.

2. How is magnetic field determined from the Lorentz Force?

The magnetic field can be determined from the Lorentz Force by rearranging the equation to solve for the magnetic field. This involves using the known values of the charge, velocity, and force, and using the right-hand rule to determine the direction of the magnetic field.

3. What is the equation for determining the magnetic field from the Lorentz Force?

The equation for determining the magnetic field from the Lorentz Force is B = F / (q*v), where B is the magnetic field, F is the Lorentz Force, q is the charge of the particle, and v is the velocity of the particle.

4. What are the units of measurement for the magnetic field determined from the Lorentz Force?

The units of measurement for the magnetic field determined from the Lorentz Force are Tesla (T) in the SI system and Gauss (G) in the CGS system. 1 Tesla is equal to 10,000 Gauss.

5. Can the Lorentz Force be used to determine the strength of a magnetic field at any point?

Yes, the Lorentz Force can be used to determine the strength of a magnetic field at any point, as long as the charge and velocity of the particle are known. This can be useful in various scientific and engineering applications where the magnetic field needs to be measured or controlled.

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