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 5.15ModifyingAbove i With caret minus 17.8ModifyingAbove j With caret plus 25.3ModifyingAbove 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 1.98 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 5.47times 10 Superscript negative 17 Baseline Upper Nright-parenthesis i Overscript caret EndScripts plus left-parenthesis 1.58times 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

B = 5.15 i -17.8 j + 25.3 k

v = vx i + vy j +(1.98km/s) k

   = vx i + vy j + 1980 k

F = (5.47x10 -17 ) i +(1.58x10 -17) j

We know F = q ( v X B )                Where q = charge of proton = 1.6 x10 -19 C

v X B =        i                            j                                k

                 vx                         v y 1980

5.15 -17.8 25.3

        = i[25.3vy +(1980x17.8)] -j [25.3vx -(1980x5.15)] + k[-17.8 vx -5.15 vy ]

       = i [25.3 vy +35244] - j[25.3 vx - 10197] -k[ 17.8vx +5.15 vy ]

q(vXB) = (1.6x10 -19 ) { i [25.3 vy +35244] - j[25.3 vx - 10197] -k[ 17.8vx +5.15 vy ]}

       F = (1.6x10 -19 ) { i [25.3 vy +35244] - j[25.3 vx - 10197] -k[ 17.8vx +5.15 vy ]}

(5.47x10 -17 ) i +(1.58x10 -17) j = (1.6x10 -19 ) { i [25.3 vy +35244] - j[25.3 vx - 10197] -k[ 17.8vx +5.15 vy ]}

compare we get , (1.6x10 -19 ) [25.3 vy +35244]= (5.47x10 -17 )

[25.3 vy +35244] = 341.875

25.3vy = -34902.125

            vy = -1379.53 m/s

                = -1.37953 km/s

(1.6x10 -19 )[25.3 vx - 10197] = (1.58x10 -17)

[25.3 vx - 10197] = 98.75

25.3vx = 10295.75

          vx = 406.94 m/s

              = 0.40694 km/s

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