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 10^{5}) i + (1.54 x 10^{5}) 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 10^{5} = - 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

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