Chapter 22 - Magnetism - Problems and Conceptual Exercises - Page 796: 72

$3.26\times 10^{-6}T$

We know that $v=\frac{2K}{m}$ $v=\sqrt{\frac{2(4.5\times 10^4\times 1.6\times 10^{-19})}{9.11\times 10^{-31}}}$ $v=1.26\times 10^8\frac{m}{s}$ Now we can find the magnetic field as $B=\frac{mv}{er}$ We plug in the known values to obtain: $B=\frac{(9.11\times 10^{-31})(1.26\times 10^8)}{1.6\times 10^{-19}(220)}$ $B=3.26\times 10^{-6}T$