an alpha particle moving at a speed of 5.20x10^4 m/s enters a perpendiclar magnetic field with a strength of 3.0 mT. The magnitude of the acceleration of the particle as it enters the magnetic field will be a.bc x 10^d m/s^2 ?

Question

anonymous · Answer

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F magnetic=q(v(vector)cross B(vector))=2*1.6*10^-19(5.2*10^4*3*10^-3)
                                                        =49.8*10^-18N
hence acceleration=49.8*10^-18/4*1.66*10^-27=7.5*10^9 m/sec^2
i.e a=7
b=5
c=0
d=9