Chapter 22 - Magnetism - Problems and Conceptual Exercises - Page 792: 14

$\vec{B}=(-3.0T)\hat{x}$

We know that $B=\frac{Fsin\theta}{ev}$ We plug in the known values to obtain: $B=\frac{(2.0\times 10^{-13})sin90^{\circ}}{1.6\times 10^{-19}(4.2\times 10^5)}$ $B=3.0T$ The direction along with magnitude is $\vec{B}=(-3.0T)\hat{x}$