Glencoe Physics: Principles and Problems, Student Edition
Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN: 9780078807213
Author: Paul W. Zitzewitz
Publisher: Glencoe/McGraw-Hill
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Question
Chapter 26.1, Problem 1PP
To determine

The radius of the circular path if the speed of a proton is 7.5×104m/s .

Expert Solution & Answer
Check Mark

Answer to Problem 1PP

The radius of the circular path is r=9.79×105m .

Explanation of Solution

Given:

Speed of the proton is v=7.5×104m/s .

Strength of magnetic field is B=0.080T

Formula used:

The expression for radius of the circular path is

  r=mvBq  ...... (1)

Here, m,v,B and q are mass of a particle, velocity, magnetic field strength and charge of a particle respectively.

Calculation:

Charge of a proton is q=1.6×1016C ,

Mass of a proton is m=1.67×1027kg ,

Substitute 7.5×104m/s for v , 0.080T for B , 1.6×1016 for q and 1.67×1027kg for m in equation (1),

  r=(1.67×1027)×(7.5×104)(0.08×(1.6×1019))r=12.53×10230.128×1019r=97.89×104mr=9.79×105m

Conclusion:

Hence, the radius of the circular path is r=9.79×105m .

Chapter 26 Solutions

Glencoe Physics: Principles and Problems, Student Edition

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