Question

A proton (charge +e = 1.61x10^-19 C and mass m = 1.67x10^-27 kg), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge +2e and mass 4m) which is also traveling perpendicular to the same field. The ratio of their speeds, v_alpha/v_proton, is:

A. 0.5

B. 1

C. 2

D. 4

E. 8

Answer 1

The force on the moving charge in the magnetic field is

F = qvBSin θ, where θ is the angle between thedirection of velocity and the magnetic field.

In both cases, θ = 90 degrees, so the force is

F = qvB.

For proton, F_p = ev_pB

For alpha particle, F_α =2ev_aB

The two forces are equal, then

ev_pB = 2ev_aB

v_p = 2v_a, so the ratio is 2.

C is the correct answer.

Answer 2

This problem is from topic force on moving charge in a magnetic field.

Force = È = q (W x ²) PL = q VB sino a = 90° seno=1 1ĚL= q.VB Iq K B = 9 Y ₂ B B. = B2 - q vi = 9 =) ev = 2e V₂ => V₂ - į = Valpha Vi Vproton = 0.5 option Alis correct