a proton moving with a speed of 1.0 x 106ms-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
-
A.
60° -
B.
90° -
C.
30° -
D.
45°
Correct Answer: Option C
Explanation
The force acting on moving charge in magnetic field is given by: F=q VBSin θ
sin θ =F=(8.0×10)-14/1.6×10-19x1.0x106x1.0 = (8.0×10)14/16 x 1013
(8.0x 10)-14/16×10-14
=0.5000
∴ θ= sin-1 0.5000
=30°