The following text is a method for finding the magnetic field as described in Purcell's Electricity and Magnetism (page $151$, the top part).

Measure the force on the particle when its velocity is $\bf{v}$; repeat with $\bf{v}$ in some other direction. Now find a $\bf{B}$ that will make $\textbf{f}=q\textbf{v}\times\textbf{B}$ fit all these results.

Why do we need two velocity vectors?

For a given velocity $\textbf{v}_{1}=[v_{1} v_{2} v_{3}]^{T}$, there will be a corresponding force $\textbf{f}=[f_{1} f_{2} f_{3}]^{T}$. Therefore, the equation $\textbf{f}=q\textbf{v}\times\textbf{B}$ will be equivalent to




Three equations, three unknowns. Am I missing something?

  • $\begingroup$ I don’t have the text. Maybe he is thinking of measuring the E field too? That would give 6 unknowns. $\endgroup$ – Dale Nov 14 '19 at 11:14
  • 1
    $\begingroup$ @Dale Under the section Definitions, units, and measurements of this wiki page, there is a quoted section that is exactly taken from Purcell's book and it is the one I am reffering to above. The $E-$field was taken care of by measuring the force at rest (so three velocities in total). $\endgroup$ – Hilbert Nov 14 '19 at 11:24
  • $\begingroup$ Try to solve this system of equations. Make sure the determinant is zero. $\endgroup$ – Alex Trounev Nov 14 '19 at 11:31

The linear system you've written, $$ \vec f= \begin{pmatrix} f_1 \\ f_2 \\ f_3 \end{pmatrix} =q \begin{pmatrix} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \end{pmatrix} \begin{pmatrix} B_1 \\ B_2 \\ B_3 \end{pmatrix} =q(\vec v\times) \vec B $$ is indeterminate, and it does not have a unique solution. You can easily verify this by noticing that the determinant is zero, since the matrix $M=(\vec v\times)$ is antisymmetric, but the transpose can't change the determinant, so $$ \det(M) = \det(M^T) = \det(-M) = (-1)^3\det(M) = -\det(M), $$ which requires $\det(M)=0$.

Alternatively, you can just notice that it is unable to distinguish a magnetic field parallel to the velocity from a vanishing field (since both give zero force). This is seen most simply by choosing your coordinate axes so that $\vec v$ lies along the $x$ axis, so that the system of equations reads $$ \begin{pmatrix} f_1 \\ f_2 \\ f_3 \end{pmatrix} = qv_1 \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} B_1 \\ B_2 \\ B_3 \end{pmatrix}, $$ which is clearly sufficient to figure out $B_2$ and $B_3$ from $f_2$ and $f_3$, but cannot say anything about $B_1$.


The system of equations

$$f_{1}=q(v_{2}B_{3}-v_{3}B_{2})$$ $$f_{2}=q(v_{1}B_{3}-v_{3}B_{1})$$ $$f_{3}=q(v_{1}B_{2}-v_{2}B_{1})$$

Can be rewritten as $$\mathbf{f}=q\mathbf{M} \mathbf{B}$$ where $$\mathbf{M}=\left( \begin{array}{ccc} 0 & -v_3 & v_2 \\ v_3 & 0 & -v_1 \\ -v_2 & v_1 & 0 \\ \end{array} \right)$$

Note that $\det(\mathbf{M})=0$ so the system is not linearly independent. So even though you do have three equations in three unknowns there is not a unique solution. You need at least two non-zero $\mathbf{v}$ to solve it.


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