Asked by Edmound
A cosmic-ray proton in interstellar space has an energy of 11.0 MeV and executes a circular orbit having a radius equal to that of Mars revolving around the Sun (2.28 1011 m). What is the magnetic field in that region of space?
Answers
Answered by
Elena
KE=mv²/2 => v=sqrt(2KE/m) =
=sqrt(2•11•10⁶•1.6•10⁻¹⁹/1.67•10⁻²⁷)=4.6•10⁷ m/s
F=qvBsinα
sinα=1
q=e
F=ma=mv²/R =evB
B=mv/eR =
=1.67•10⁻²⁷•4.6•10⁷/1.6•10⁻¹⁹•2.28•10¹¹ =2.1•10⁻¹² T
=sqrt(2•11•10⁶•1.6•10⁻¹⁹/1.67•10⁻²⁷)=4.6•10⁷ m/s
F=qvBsinα
sinα=1
q=e
F=ma=mv²/R =evB
B=mv/eR =
=1.67•10⁻²⁷•4.6•10⁷/1.6•10⁻¹⁹•2.28•10¹¹ =2.1•10⁻¹² T
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