2. Refer to Sample Problem No. 2. Prove that r= 2.6 x 10-2 m using the equation mv r = Iq|B Show your solution and answer.

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter30: Magnetic Fields And Forces
Section: Chapter Questions
Problem 43PQ
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Sample #2 is attached

A proton moves through a magnetic field of magnitude 2.0 T at a speed of
5.00 x 106 m/s perpendicular to the field. Find the (a) centripetal
acceleration and (b) radius of the circular path of the proton.
2.
Given:
B = 2.0 T; v = 5.00 x 106 m/s;
Mass and charge of a proton: m = 1.67 x 10-27 kg' q = 1.60 x 10-19 C
Solution and Answer:
(a) The magnetic force is computed first.
F = lq|vB sin 6 = |1.60 x 10-19 CI(5.00 x 106 m/s)(2 T) sin 90°
F = 1.6 x 10-12 N
Using Newton's second law of motion,
1.6 x 10-12 N
m 1.67 x 10-27 kg
= 9.58 × 1014
(b) Using the formula for centripetal acceleration a =
v2 (5.00 x 106 m)²
r =
a
9.58 x 1014 m= 0.026 = 2.6 × 10-2 m
Transcribed Image Text:A proton moves through a magnetic field of magnitude 2.0 T at a speed of 5.00 x 106 m/s perpendicular to the field. Find the (a) centripetal acceleration and (b) radius of the circular path of the proton. 2. Given: B = 2.0 T; v = 5.00 x 106 m/s; Mass and charge of a proton: m = 1.67 x 10-27 kg' q = 1.60 x 10-19 C Solution and Answer: (a) The magnetic force is computed first. F = lq|vB sin 6 = |1.60 x 10-19 CI(5.00 x 106 m/s)(2 T) sin 90° F = 1.6 x 10-12 N Using Newton's second law of motion, 1.6 x 10-12 N m 1.67 x 10-27 kg = 9.58 × 1014 (b) Using the formula for centripetal acceleration a = v2 (5.00 x 106 m)² r = a 9.58 x 1014 m= 0.026 = 2.6 × 10-2 m
2. Refer to Sample Problem No. 2. Prove that r= 2.6 x 10-2 m using the equation
%3D
mv
r=-
Iq|B
Show your solution and answer.
Transcribed Image Text:2. Refer to Sample Problem No. 2. Prove that r= 2.6 x 10-2 m using the equation %3D mv r=- Iq|B Show your solution and answer.
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