An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.14 mT. If the speed of the electron is

Question

An electron moves in a circular path perpendicular to a uniform magnetic field with a magnitude of 2.14 mT. If the speed of the electron is 1.48 107 m/s, determine the following.
(a) the radius of the circular path ………… cm
(b) the time interval required to complete one revolution ………… s

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Ngọc Hoa 5 months 2021-08-15T15:42:49+00:00 1 Answers 1 views 0

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    2021-08-15T15:44:14+00:00

    Answer:

    (a) 3.9cm

    (b) 1.66 x 10⁻⁸s

    Explanation:

    Since the electron is moving in a circular path, the centripetal acceleration needed to keep it from slipping off is provided by the magnetic force. This force (F), according to Newton’s second law of motion is given by,

    F = m x a          ————–(i)

    Where;

    m = mass of the particle

    a = acceleration of the mass

    The centripetal acceleration is given by;

    a = v² / r          [v = linear velocity of particle, r = radius of circular path]

    Therefore, equation (i) becomes;

    F = m v²/ r             ——————–(ii)

    The magnitude of the magnetic force on a moving charge in a magnetic field as stated by Lorentz’s law is given by;

    F = qvBsinθ          ————-(iii)

    Where;

    q = charge of the particle

    v = velocity of the particle

    B = magnetic field

    θ = angle between the velocity and the magnetic field

    Combine equations (ii) and (iii) as follows;

    m (v² / r) = qvBsinθ         [divide both side by v]

    m v / r = qBsinθ              [make r subject of the formula]

    r = (m v) / (qBsinθ)              ———(iv)

    (a) From the question;

    v = 1.48 x 10⁷m/s

    B = 2.14mT = 2.14 x 10⁻³T

    θ = 90°          [since the direction of velocity is perpendicular to magnetic field]

    m = mass of electron = 9.11 x 10⁻³¹kg

    q = charge of electron = 1.6 x 10⁻¹⁹C

    Substitute these values into equation (iv) as follows;

    r = (9.11 x 10⁻³¹ x 1.48 x 10⁷) / (1.6 x 10⁻¹⁹ x 2.14 x 10⁻³ sin 90°)

    r = 3.9 x 10⁻²m

    r = 3.9cm

    Therefore, the radius of the circular path is 3.9cm

    (b) The time interval required to complete one revolution is the period (T) of the motion of the electron and it is given by

    T = d / v          ————–(*)

    Where;

    d = distance traveled in the circular path in one complete turn = 2πr

    v = velocity of the motion = 1.48 x 10⁷m/s

    d = 2 π (3.9 x 10⁻²)            [Take π = 22/7 = 3.142]

    d = 2(3.142)(3.9 x 10⁻²) = 0.245m

    Substitute the values of d and v into equation (*) as follows;

    T = 0.245 / 1.48 x 10⁷

    T = 0.166 x 10⁻⁷s

    T = 1.66 x 10⁻⁸s

    Therefore, the time interval is 1.66 x 10⁻⁸s

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