Answer
$2.60\times10^{6}m/s$
Work Step by Step
Speed $v= \frac{qBr}{m}$
Plugging in the values, we have
$v=\frac{2\times1.602\times10^{-19}C\times1.20 T\times4.50\times10^{-2}m}{4.00\times1.661\times10^{-27}kg}=2.60\times10^{6}m/s$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 28 - Magnetic Fields - Problems - Page 830: 17a
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