Question

A proton moves through a uniform magnetic field given by ModifyingAbove Upper B With right-arrow equals left-parenthesis 10.5ModifyingAbove i With caret minus 21.6ModifyingAbove j With caret plus 22.1ModifyingAbove 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 1.52? 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 5.08times 10 Superscript negative 17 Baseline ? Upper Nright-parenthesis i Overscript caret EndScripts plus left-parenthesis 2.47times 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?

Answer 1

= (10.5 i - 21.6 j + 22.1 k) (10^-3) T

= v_x i + v_y j + 1520 k

= (5.08 x 10^-17) i + (2.47 x 10^-17) j

q = charge on the proton = 1.6 x 10^-19

magnetic force is given as

= q (x )

(5.08 x 10^-17) i + (2.47 x 10^-17) j = (1.6 x 10^-19) (10^-3) ((v_x i + v_y j + 1520 k) x (10.5 i - 21.6 j + 22.1 k))

(3.175 x 10⁵) i + (1.54 x 10⁵) j =  ((-21.6 v_x - 10.5 v_y) k + (- 22.1 v_x - 15960) j + (- 10.5 v_y + 32832) i

comparing the coefficient of "j" both side

1.54 x 10⁵ = - 22.1 v_x - 15960

v_x = -7690.5 m/s

comparing the coefficient of "k" both side

-21.6 v_x - 10.5 v_y = 0

-21.6 (-7690.5) - 10.5 v_y = 0

v_y = 15820.5 m/s