ssiding

10+ Year Member
Aug 20, 2008
41
0
Status
Pre-Medical
Traveling at an initial speed of 1.5 e 6 m/s, a proton enters a region of constant magnetic field of magnitue 1.5T. It the proton's initial velocity vector makes an angle of 30 degrees with the magnetic field, compute the proton's speed 4 seconds after entering th emagnetic field.

A. 5 e 5 m/s
B. 7.5 e 5 m/s
C. 1.5 e 6 m/s
D. 3.0 e 6 m/s
 

soby10

10+ Year Member
5+ Year Member
Apr 27, 2008
56
0
Status
Pre-Medical
The answer is c 1.5 X 10^6 m/s. The same as the initial velocity b/c at a constatnt B field the velocity is constant. This is due to def. of magnetic filed B=F/q(v*sine angle). Force is perpendicular to both velocity and B field. Since force is always perpendicular to veolcity it does no work. W=Force *distance*(cos 90) so it changes direction but never changes magnitude of velocity. Hope that explains it.
 

isaacmn

10+ Year Member
May 5, 2007
29
0
International programs
Status
Medical Student
when charges move in a constant magnetic field ..it undergoes a uniformed circular motion in the form of centripetal fore with constant velocity

qvbsintheta= v2/r
 
About the Ads