Need help to calculate this physics about kinetic energy

timer Asked: Nov 20th, 2015
account_balance_wallet $5

Question Description

A proton with a kinetic energy of 4.9*10^-16J movesw perpendicularto a magnetic field of .26T. What is the radius of its circularpath?

this answer is on the physics homework help as #20 physics jameswalker chap 22, but i just don't get it. Could somebody leave me avery detailed explanation but make it easy for me to understand,thank you.
will rate high

Tutor Answer

School: Rice University

Thank you for the opportunity to help you with your question!

kinetic energy K = mv2/2
mass of a proton m = 1.67*10-27 kg
so 4.9*10-16 J = 1.67*10-27 kg *v2/2
v = √(2*4.9*10-16/1.67*10-27) =7.66*105 m/s
magnetic force F = qvB
Newton 2nd law: F = ma = mv2/r
so qvB = mv2/r
qB = mv/r
qBr = mv
∴r = mv/(qB) =1.67*10-27*7.66*105/(1.6*10-19*0.26)= 3.08*10-2 m = 3.08 cm

Please let me know if you need any clarification. I'm always happy to answer your questions.

flag Report DMCA

Thank you! Reasonably priced given the quality not just of the tutors but the moderators too. They were helpful and accommodating given my needs.

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors