An electron moves along the z-axis with vz=4.1Ã107m/s. As it passes the origin, what are the strength and direction of the magnetic field at

Question

An electron moves along the z-axis with vz=4.1Ã107m/s. As it passes the origin, what are the strength and direction of the magnetic field at the following (xx, yy, zz) positions?

a. (1 cm , 0 cm, 0 cm)
b. (0 cm, 0 cm, 2 cm)

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Helga 6 months 2021-07-29T23:48:23+00:00 1 Answers 11 views 0

Answers ( )

    0
    2021-07-29T23:49:33+00:00

    Answer:

    a

    The value of magnetic field is B_a=6.56*10^{-19} T

    In the direction of the positive x – axis

    b

    The value of magnetic field is B_a=6.56*10^{-19} T

    In the direction of the positive z – axis

    Explanation:

    From the question we are told that

             The velocity of the electron is v_z = 4.1*10^{7}m/s

    Considering the first position

        The equation for the magnetic field is

                        \= B = \frac{\mu_0}{ 4 \pi} \frac{q \=v * \= r}{r^2}

     Now r = \sqrt{i^2 + j^2 + k^2}

      substituting values

               r = \sqrt{1^2 +0^2 +0^2}

                  = 1

              \= r = \frac{\r r}{r}

            \r r = 1 i + 0j + 0k

    Therefore \= r = \frac{1i + 0j + 0k}{1}

                       = i

    So   Substituting  4 \pi *10^ {-7} for \mu_o , 1.602 *10^{-19} for q

                \= B_a = \frac{4\pi *10^{-7}}{4 \pi} \frac{1.602 *10^{-19} * 4.1*10^{7} * i}{1^2}

                     \= B_a=6.56*10^{-19} T (i)

    Considering the second  position

        Here

                   r = \sqrt{0^2 + 0^2 + 2^2}

                      =2

                  \= r = \frac{\r r}{r}

                  \r r = 0 i + 0j + 2k

                \= r = \frac{0i + 0j + 2k}{2}

                   = k

     So   Substituting  4 \pi *10^ {-7} for \mu_o , 1.602 *10^{-19} for q

                \= B_a = \frac{4\pi *10^{-7}}{4 \pi} \frac{1.602 *10^{-19} * 4.1*10^{7} * k}{1^2}

                     \= B_a=6.56*10^{-19} T (k)

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