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A proton moves north at $10^6$ m/s in a magnetic field of $0.1$ T pointing east. The magnetic force on it is:
AZero, since proton velocity is along the field direction here
B$1.6\times 10^{-14}$ N along east, in the field direction itself
C$1.6\times 10^{-14}$ N upward, by right-hand rule for the cross
D$1.6\times 10^{-12}$ N downward, larger by a factor of hundred
Answer & Solution
Correct answer: C. $1.6\times 10^{-14}$ N upward, by right-hand rule for the cross
$F = qvB\sin 90 = 1.6\times 10^{-19}\cdot 10^6\cdot 0.1 = 1.6\times 10^{-14}$ N; direction upward by right-hand rule.
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