A cosmic ray electron moves at 7.50×106 m/s perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.00×10−5 T .

Question

A cosmic ray electron moves at 7.50×106 m/s perpendicular to the Earth’s magnetic field at an altitude where field strength is 1.00×10−5 T . What is the radius of the circular path the electron follows?

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Thu Nguyệt 5 months 2021-08-15T00:07:05+00:00 1 Answers 105 views 0

Answers ( )

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    2021-08-15T00:08:45+00:00

    Answer:

    4.266 m.

    Explanation:

    Given,

    Speed of the electron = 7.50 x 10⁶ m/s

    Magnetic field, B = 1 x 10⁻⁵ T

    radius of circular path = ?

    Using formula of radius of curvature of the charge particle

    r = \dfrac{mv}{qB}

    m is mass of electron

    q is charge of electron

    r = \dfrac{9.11 \times 10^{-31}\times 7.50\times 10^6}{1.6\times 10^{-19}\times 1 \times 10^{-5}}

    r = 4.266 m

    Hence, radius of circular path is equal to 4.266 m.

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