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olga2289 [7]
3 years ago
5

A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5

× 1013 m/s 2 in the positive x direction when its velocity is in the positive z direction. the mass of a proton is 1.673 × 10−27 kg. find the magnitude of the field. answer in units of t.
Physics
1 answer:
svlad2 [7]3 years ago
4 0
The magnetic force experienced by the proton is given by
F=qvB \sin \theta
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and \theta the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so \sin \theta=1 and we can ignore it in the formula.

For Netwon's second law, the force is also equal to the proton mass times its acceleration:
F=ma

So we have
ma=qvB
from which we can find the magnitude of the field:
B= \frac{ma}{qv}= \frac{(1.67 \cdot 10^{-27}kg)(5\cdot 10^{13}m/s^2)}{(1-6 \cdot 10^{-19}C)(3.7 \cdot 10^7 m/s)}=0.014 T
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