Answer
$B_1$ is stronger.
Work Step by Step
We can write an expression for the radius of curvature of the electron:
$r = \frac{mv}{\vert q\vert~B}$
When $B$ is greater, the radius of curvature is smaller.
Therefore, $B_1$ is stronger.
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 - Questions - Page 828: 7a
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