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)

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’.

0 thoughts on “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)”

  1. 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

    [tex]\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 )][/tex]

    [tex]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][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]

    [tex]B_x = undetermined[/tex]

    [tex]B_y = \frac{42.2 \times 10^-^1^9}{-2.4 \times 10^-^1^6} \\\\= – 0.0176 T[/tex]

    [tex]B_z = 0T[/tex]

    Reply

Leave a Comment