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
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
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