An electron has a velocity of 1.50 km/s (in the positive x direction) and an acceleration of 2.00 ✕ 1012 m/s2 (in the positive z direction)

Question

An electron has a velocity of 1.50 km/s (in the positive x direction) and an acceleration of 2.00 ✕ 1012 m/s2 (in the positive z direction) in uniform electric and magnetic fields. If the electric field has a magnitude of strength of 18.0 N/C (in the positive z direction), determine the following components of the magnetic field. If a component cannot be determined, enter ‘undetermined’.

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Thanh Thu 1 month 2021-07-31T16:44:23+00:00 1 Answers 2 views 0

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    2021-07-31T16:45:36+00:00

    Answer:

    see explanation

    Explanation:

    Given that,

    velocity of 1.50 km/s = 1.50 × 10³m/s

    acceleration of 2.00 ✕ 1012 m/s2

    electric field has a magnitude of strength of 18.0 N/C

    \bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]

    9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)][tex]42.2 \times 10^-^1^9 \hat k = -2.4 \times 10^1^6B_y \hat k + 2.4 \times 10 ^1^6 \hat j B_z\\[/tex]

    B_x = undetermined

    B_y = \frac{42.2 \times 10^-^1^9}{-2.4 \times 10^-^1^6} \\\\= - 0.0176 T

    B_z = 0T

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