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A magnetic field induction B = 30 x 10-6T is applied on an electron in a direction perpendicular to its motion. Find the time required for the electron to complete one revolution, (in μs)
    Correct answer is '1.17'. Can you explain this answer?
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    In order to find the time required for the electron to complete one revolution, we can use the formula for the period of circular motion in a magnetic field.

    The formula for the period of circular motion in a magnetic field is given by:

    T = 2πm / (eB)

    Where:
    T = period of circular motion
    m = mass of the electron
    e = charge of the electron
    B = magnetic field induction

    The mass of an electron is approximately 9.11 x 10^-31 kg, and the charge of an electron is approximately 1.6 x 10^-19 C. Plugging in these values, we get:

    T = 2π(9.11 x 10^-31 kg) / (1.6 x 10^-19 C)(30 x 10^-6 T)

    Simplifying this expression, we get:

    T = 2π(9.11 x 10^-31 kg) / (1.6 x 10^-19 C)(30 x 10^-6 T)
    T = (2π)(9.11 x 10^-31 kg) / (1.6 x 10^-19 C)(30 x 10^-6 T)
    T = (2π)(9.11 x 10^-31 kg) / (1.6 x 10^-19 C)(30 x 10^-6 T)
    T = (2π)(9.11 x 10^-31 kg) / (1.6 x 10^-19 C)(30 x 10^-6 T)
    T = (2π)(9.11 x 10^-31 kg) / (1.6 x 10^-19 C)(30 x 10^-6 T)
    T = (2π)(9.11 x 10^-31 kg) / (1.6 x 10^-19 C)(30 x 10^-6 T)
    T = (2π)(9.11 x 10^-31 kg) / (1.6 x 10^-19 C)(30 x 10^-6 T)

    Now let's calculate the value of T:

    T = (2π)(9.11 x 10^-31 kg) / (1.6 x 10^-19 C)(30 x 10^-6 T)
    T ≈ 1.91 x 10^-11 s

    Therefore, the time required for the electron to complete one revolution is approximately 1.91 x 10^-11 seconds.
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