A proton moves through a uniform magnetic field given by ModifyingAbove Upper B

Question

A proton moves through a uniform magnetic field given by ModifyingAbove Upper B With right-arrow equals left-parenthesis 11.6ModifyingAbove i With caret minus 19.5ModifyingAbove j With caret plus 12.7ModifyingAbove k With caret right-parenthesis mT. At time t1, the proton has a velocity given by v Overscript right-arrow EndScripts equals v Subscript x Baseline i Overscript caret EndScripts plus v Subscript y Baseline j Overscript caret EndScripts plus left-parenthesis 1.77 km/s right-parenthesis k Overscript caret EndScripts and the magnetic force on the proton is Upper F Overscript right-arrow EndScripts Subscript Upper B Baseline equals left-parenthesis 3.95times 10 Superscript negative 17 Baseline Upper Nright-parenthesis i Overscript caret EndScripts plus left-parenthesis 2.35times 10 Superscript negative 17 Baseline Upper Nright-parenthesis j Overscript caret EndScripts . (a) At that instant, what is vx? (b) At that instant, what is vy? show step by step instructions please. including how the cross works.

Explanation / Answer

B = (11.6 i -19.5 j +12.7 k ) mT
V = (VXi + Vy j + 1.77 k ) km/s
F = 3.95*10-17 i + 2.35*10-17 j
We know that
Force = q(V x B)
So we have to calculate the cross product of the v and B
v x B = (12.7Vy + 34.515) i - (12.7Vx - 20.532)j + (19.5vx -11.6 vy) k
F = 3.95*10-17 i + 2.35*10-17 j = q[(12.7Vy + 34.515) i - (12.7Vx - 20.532)j + (19.5vx -11.6 vy) k ]
3.95*10-17 i + 2.35*10-17 j = 1.6*10-19[(12.7Vy + 34.515) i - (12.7Vx - 20.532)j + (19.5vx -11.6 vy) k ]
2.468*102 i + 1.468*102 j = [(12.7Vy + 34.515) i - (12.7Vx - 20.532)j + (19.5vx -11.6 vy) k ]
now commparing both side
2.468*102 = 12.7Vy + 34.515
Vy = 16.715 km/s
1.468*102 = - (12.7Vx - 20.532)
Vx = -9.94 km/s

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