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

60°

90°

30°

45°

Correct answer is C

The force acting on moving charge in magnetic field is given by: F=q VBSin θ
sin θ =F=(8.0x10)^-14/1.6x10^-19x1.0x10⁶x1.0 = (8.0x10)¹⁴/16 x 10¹³
(8.0x 10)^-14/16x10^-14
=0.5000
∴ θ= sin^-1 0.5000
=30°

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