A proton experiences an eastward force of 30 × N while traveling in a magnetic field of 5 × T north.
F=qvBsinα=qvB,F=qvB\sin \alpha=qvB,F=qvBsinα=qvB,
v=FqB,v=\frac{F}{qB},v=qBF,
v=301.6⋅10−19⋅5=37.5⋅1018 ms.v=\frac{30}{1.6\cdot 10^{-19}\cdot 5}=37.5\cdot 10^{18}~\frac ms.v=1.6⋅10−19⋅530=37.5⋅1018 sm.
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