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 9.40ModifyingAbove i With caret minus 21.2ModifyingAbove j With caret plus 10.2ModifyingAbove 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 1.72 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 3.92times 10 Superscript negative 17 Baseline Upper Nright-parenthesis i Overscript caret EndScripts plus left-parenthesis 1.74times 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?

Explanation / Answer

V = Vx i^ + Vy j^ + 1720 k^

F = 3.92 x 10-17 i^ + 1.74 x 10-17 j^

B = 9.40 i^ - 21.2 j^ + 10.2 k^

q = charge on proton = 1.6 x 10-19 C

magnetic force is given as

F = q (V x B)

3.92 x 10-17 i^ + 1.74 x 10-17 j^ = (1.6 x 10-19) [(Vx i^ + Vy j^ + 1720 k^ ) x (9.40 i^ - 21.2 j^ + 10.2 k^)]

3.92 x 10-17 i^ + 1.74 x 10-17 j^ = (1.6 x 10-19) [- 21.2 Vx k^ - 10.2 Vx j^ - 9.40 Vy k^ + 10.2 Vy i^ + (1720 x 9.40) j^ + (1720 x 21.2)]

245 i^ + 108.75 j^ = [- 21.2 Vx k^ - 10.2 Vx j^ - 9.40 Vy k^ + 10.2 Vy i^ + (1720 x 9.40) j^ + (1720 x 21.2) i^]

comparing the x-component

245 = 10.2 Vy + (1720 x 21.2)

Vy = - 3550.8

comparing the Y-component

108.75 = - 10.2 Vx + (1720 x 9.40)

Vx = 1574.44 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.