A proton of charge 1.6 x 10\(^{19}\)C is projected into a uniform magnetic field of flux density 5.0 x 10\(^5\)T. If the proton moves parallel to the field with a constant speed of 1.6 x 10\(^6\)ms\(^{-1}\), calculate the magnitude of the force exerted on it by the field
- 0.0N
- 2.0 x 10-21N
-
1.3 x 10-17N
- 5.1 x 10-14N
- 2.3 x 1013N
The correct answer is: C
Explanation
F = Bqv = 5.0 x 10\(^{-5}\) x 1.6 x 10\(^{-19}\) x 10\(^6\)
= 1.3 x 10\(^{-17}\)N