InterviewSolution
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InterviewSolution
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Two protons move parallel to each other with an equal velcity `v = 300 km//s`. Find the ration of forces of magentic and electricla of the protons. |
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Answer» Force of magentic interaction `vec(F)_(mag) = e(vec(v) xx vec(B))` Where, `vec(B) = (mu_(0))/(4pi) (e (vec(v) xx vec(r)))/(r^(3))` So, `vec(F)_(mag) = (mu_(0))/(4pi) (e^(2))/(r^(3)) [vec(v) xx (vec(v) xx vec(r))]` `= (mu_(0))/(4pi) (e^(2))/(r^(3)) [(vec(v). vec(r)) xx vec(v) - (vec(v) . vec(v)) xx vec(r)] = (mu_(0))/(4pi) (e^(2))/(r^(3)) (-v^(2) vec(r))` And `vec(F)_("ele") = e vec(E) = e (1)/(4pi epsilon_(0)) (e vec(r))/(|vec(r)|^(3))` Hence, `(|vec(F)_(mag)|)/(|vec(F)_("electric")|) = v^(2) mu_(0) epsilon_(0) = ((v)/(c))^(2) = 1.00xx10^(-6)` |
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