A proton moves through a magnetic field of magnitude 4.5 T at a speed of 7.08×10^6 m/s
Perpendicular to the field. Find the
a. Centripetal acceleration
b. Radius of the circular path of the proton
qvB=ma, ⟹ a=qvBm=5.6⋅1018 ms2,qvB=ma,\implies a=\frac{qvB}m=5.6\cdot 10^{18}~\frac{m}{s^2},qvB=ma,⟹a=mqvB=5.6⋅1018 s2m,
r=mvqB=9⋅10−6 m.r=\frac{mv}{qB}=9\cdot 10^{-6}~m.r=qBmv=9⋅10−6 m.
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