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26 November, 07:55

The magnetic field over a certain range is given by B~ = Bx ˆı + By ˆ, where Bx = 2 T and By = 4 T. An electron moves into the field with a velocity ~v = vx ˆı+vy ˆ +vz ˆk, where vx = 2 m/s, vy = 6 m/s and vz = 8 m/s. The charge on the electron is - 1.602 * 10-19 C. What is the ˆı component of the force exerted on the electron by the magnetic field? Answer in units of N.

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  1. 26 November, 08:07
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    The force exerted in a magnetic field is given as

    F = q (v * B)

    Where

    F is the force entered

    q is the charge

    v is the velocity

    B is the magnetic field

    Given that,

    The magnetic field is

    B = 2•i + 4•j. T

    The velocity of the electron is

    v = 2•i + 6•j + 8•k. m/s

    Also, the charge of an electron is

    q = - 1.602 * 10^-19 C.

    Then note that,

    V*B is the cross product of the speed and the magnetic field

    Then,

    F = q (V*B)

    F = - 1.602 * 10^-19 (2•i + 4•j + 8•k * 2•i + 4•j)

    Note

    i*i=j*j*k*k=0

    i*j=k. j*i=-k

    j*k=i. k*j=-i

    k*i=j. i*k=-j

    F = - 1.602 * 10^-19[ (2•i + 4•j + 8•k) * (2•i + 4•j) ]

    F = - 1.602 * 10^-19 [2*2• (i*i) + 2*4• (i*j) + 4*2• (j*i) + 4*4• (j*j) + 8*2• (k*i) + 8*4• (k*j) ]

    F = - 1.602 * 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]

    F = - 1.602 * 10^-19 (0 + 8•k - 8•k + 0 + 16•j - 32•i)

    F = - 1.602 * 10^-19 (16•j - 32•i)

    F = - 1.602 * 10^-19 * (-32•i + 16•j)

    F = 5.126 * 10^-18 •i - 2.563 * 10^-18 •j

    Then, the x component of the force is

    Fx = 5.126 * 10^-18 N

    Also, the y component of the force is

    Fy = - 2.563 * 10^-18 N
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