Asked by viper
The magnetic field of the Earth at a certain location is directed vertically downward and has a magnitude of 54.0 µT. A proton is moving horizontally towards the west in this field with a speed of 6.00 ✕ 106 m/s. What is the radius of the circular arc followed by this proton?
km
km
Answers
Answered by
Elena
F=qvBsinα
sinα=1
q=e=1.6•10⁻¹⁹ C
F=evB
B↓v←F⊙
mv²/R=evB
R=mv/eB
sinα=1
q=e=1.6•10⁻¹⁹ C
F=evB
B↓v←F⊙
mv²/R=evB
R=mv/eB
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