1 \begin{problem*}{27.64}
2 A particle of charge $q>0$ is moving at speed $v$ in the
3 $+z$-direction through a region of uniform magnetic field \vect{B}.
4 The magnetic force on the particle is $\vect{F}=F_0(3\ihat+4\jhat)$,
5 where $F_0$ is a positive constant. \Part{a} Determine the components
6 $B_x$, $B_y$, and $B_z$, or at least as many of the three components
7 as is possible from the information given. \Part{b} If it is given in
8 addition that the magnetic field has magnitude $6F_0/qv$, determine as
9 much as you can about the remaining components of \vect{B}.
14 Writing $\vect{F}_B=q\vect{v}\cdot\vect{B}$ in terms of components
17 &= q[\ihat(v_yB_z-v_zB_y)-\jhat(v_xB_z-v_zB_x)+\khat(v_xB_y-v_yB_x)]
18 = q(-v_zB_y\ihat+v_zB_x\jhat) = F_0(3\ihat+4\jhat) \;.
20 Matching components and writing $v_z=v$ we have
22 -qvB_y &= 3F_0 & qvB_x &= 4F_0 \\
23 B_y &= \ans{\frac{-3F_0}{qv}} & B_x &= \ans{\frac{4F_0}{qv}} \;.
25 We can't find $B_z$ because it does not contribute to the force felt
26 by the charge, which is currently our only handle on \vect{B}.
29 With $|\vect{B}|$, we can solve for $|B_z|$.
31 |\vect{B}|^2 &= B_x^2 + B_y^2 + B_z^2 \\
32 B_z^2 &= |\vect{B}^2| - B_x^2 - B_y^2
33 = 36\frac{F_0^2}{q^2v^2} - 9\frac{F_0^2}{q^2v^2} - 16\frac{F_0^2}{q^2v^2}
34 = (36-9-16)\frac{F_0^2}{q^2v^2}
35 = 11\frac{F_0^2}{q^2v^2} \\
36 B_z &= \ans{\sqrt{11}\frac{F_0}{qv}} \;.
38 However, we still cannot find the direction of $B_z$.