At t = 0, electron 1 is shot out of an accelerator at a speed of 2.9 × 10^3 m/s. At t = 1.0 μs, electron 2 is shot out of the accelerator and travels on a path that is parallel to and 10 mm below the path of electron 1. The speed of electron 2 is 6.8 × 10^3 m/s. At t = 3.0 μs. what is the magnitude of the magnetic field at a point that is 3.0 mm ahead of electron 1 and 5.0 mm below it? Assume that the magnetic forces the electrons exert on each other for the 2.0 μs before you make your calculations do not alter the electrons' paths significantly. Express your answer with the appropriate units. B = T

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aristocles aristocles · Answer 1 · 2019-11-17T11:17:16+01:00

Answer:

[tex]B = 2.5 \times 10^{-18} T[/tex]

Explanation:

Magnetic field due to moving electron is given by the formula

[tex]B = \frac{\mu_0 qv}{4\pi r^2}[/tex]

here we know that both moving charges are of same sing in same directions

so magnetic field at mid point between them is given as

[tex]B = \frac{\mu_o q}{4\pi r^2}(v_1 - v_2)[/tex]

here we know that

[tex]\frac{\mu_0}{4\pi} = 10^{-7}[/tex]

[tex]r = 5 mm[/tex]

[tex]v_1 = 6.8 \times 10^3 m/s[/tex]

[tex]v_2 = 2.9 \times 10^3 m/s[/tex]

now we have

[tex]B = \frac{10^{-7} (1.6 \times 10^{-19})}{(5 \times 10^{-3})^2}(6.8 \times 10^3 - 2.9 \times 10^3)[/tex]

[tex]B = 2.5 \times 10^{-18} T[/tex]