A proton moving with a speed of 1.0 x 10\(^6\)ms\(^{-1}\) through a magnetic field of 1.0T experiences a magnetic force of magnitude 8.0 x 10\(^{-14}\) N. The angle between the proton's velocity and the field is
[1.6 x 10\(^{-19}\)C]
A
60°
B
90°
C
30°
D
45°
Verified Correct Answer: C.
Physical Principles & Numerical Solution
The force acting on a moving charge in the magnetic field is given by: F= qVBSin \theta
sin \theta =F=(8.0x10)-14/1.6x10-19x1.0x106x1.0 = (8.0x10)14/16 x 1013
(8.0x 10)-14/16x10-14
=0.5000
∴ \theta= sin-1 0.5000
=30^\circ
Source: Myschool Past Questions Archive (Record #13705)