Physics: Magnetic Fields?
A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha is:
The answer key says C. Can someone please explain why. I got A. Thanks.
- ejwaxxLv 61 decade agoBest Answer
Since the velocity is perpendicular to the field, we may write just F = qvB. Let the subscripts _a and _p represent the alpha particle and the proton.
(q_a)(v_a)B = (q_p)(v_p)b
(2q_p)(v_a)B = (q_p)(v_p)B
2 = (q_p)(v_p)B/[(q_p)(v_a)B]
2 = v_p/v_a
- swihartLv 43 years ago
whilst applying facebook=qv go B, undergo in suggestions q is damaging for an electron. For sensible purposes in this question, you could enable for this by treating v as despite if this is interior the -i path. There are distinctive regulations - working example you're you able to Fleming's Left hand rule for this or a correct hand go-product rule yet whilst A = B go C and you realize A and (say) B, you could artwork out C in case you're careful. (and you could consistently verify your answer by testing that A and B then supply the generic path of A.)