A proton moving with a speed of 1.0 x 10\(^6\)ms\(^{-1}\) through a magnetic field of 1.0T...

A proton moving with a speed of 1.0 x 10\(^6\)ms\(^{-1}\) through a magnetic field of 1.0T...

JAMB 2005

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°

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Correct Answer: Option C

Explanation

The force acting on a moving charge in the magnetic field is given by: F= qVBSin 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°

There is an explanation video available below.

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