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
A
0.0N
B
2.0 x 10-21N
C
1.3 x 10-17N
D
5.1 x 10-14N
E
2.3 x 1013N
Verified Correct Answer: A.
Physical Principles & Numerical Solution
F = BqvSin\(\theta\) = 5.0 x 10\(^{-5}\) x 1.6 x 10\(^{-19}\) x 10\(^6\)Sin(0)
= 1.3 x 10\(^{-17}\) x 0 = 0N
NOTE:(Since the proton moves parallel to the magnetic field,\(\theta\) = 0º
Source: Myschool Past Questions Archive (Record #33679)