WAEC - Physics (1998 - No. 44)
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
Penjelasan
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