An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light perpendicular

Question

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light perpendicularly toward the wire. At that moment a switch is flipped, causing a current of 17.7 A 17.7 A to flow in the wire. Find the magnitude of the electron’s acceleration a a at that moment.

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Nho 5 years 2021-07-27T10:16:52+00:00 1 Answers 18 views 0

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  1. King
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    2021-07-27T10:18:49+00:00
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    Answer:

    The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

    Explanation:

    Given:

    Distance from the wire to the field point r = 2.83 \times 10^{-2} m

    Speed of electron v = 35.5 \%c

    Current I = 17.7 A

    For finding the acceleration,

    First find the magnetic field due to wire,

      B = \frac{\mu _{o}I }{2\pi r }

    Where \mu_{o} = 4\pi \times 10^{-7}

      B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

      B = 12.50 \times 10^{-5} T

    The magnetic force exerted on the electron passing through straight wire,

      F = qvB  

      F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

      F = 21.3 \times 10^{-16} N

    From the newton’s second law

      F = ma

    Where m = mass of electron = 9.1 \times 10^{-31} kg

    So acceleration is given by,

       a = \frac{F}{m}

       a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

       a = 2.34 \times 10^{15} \frac{m}{s^{2} }

    Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

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