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NEET-XII-Physics

04: Moving Charges And Magnetism

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  • #19
    An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
    (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
    (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
    Ans : Magnetic field strength, B = 0.15 T

    Charge on the electron, e = 1.6 × 10-19 C

    Mass of the electron, m = 9.1 × 10-31 kg

    Potential difference, V = 2.0 kV = 2 × 103 V

    Thus, kinetic energy of the electron = eV



    Where,

    v = velocity of the electron
    (a) Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.

    Magnetic force on the electron is given by the relation,

    B ev

    Centripetal force



    From equations (1) and (2), we get



    Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.
    (b) When the field makes an angle θ of 30° with initial velocity, the initial velocity will be,



    From equation (2), we can write the expression for new radius as:



    Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.
    (a) Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.

    Magnetic force on the electron is given by the relation,

    B ev

    Centripetal force



    From equations (1) and (2), we get



    Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.
    (b) When the field makes an angle θ of 30° with initial velocity, the initial velocity will be,



    From equation (2), we can write the expression for new radius as:



    Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.
    (a) Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.

    Magnetic force on the electron is given by the relation,

    B ev

    Centripetal force



    From equations (1) and (2), we get



    Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.
    (b) When the field makes an angle θ of 30° with initial velocity, the initial velocity will be,



    From equation (2), we can write the expression for new radius as:



    Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.
  • #19-a
    is transverse to its initial velocity,
    Ans : Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.

    Magnetic force on the electron is given by the relation,

    B ev

    Centripetal force



    From equations (1) and (2), we get



    Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.
  • #19-b
    makes an angle of 30° with the initial velocity.
    Ans : When the field makes an angle θ of 30° with initial velocity, the initial velocity will be,



    From equation (2), we can write the expression for new radius as:



    Hence, the electron has a helical trajectory of radius 0.5 mm along the magnetic field direction.