# On the method described by Purcell for finding the magnetic field by measuring the force on a test particle

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.