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21 January, 14:15

An α-particle has a charge of + 2e and a mass of 6.64 * 10-27 kg. It is accelerated from rest through a potential difference that has a value of 1.22 * 106 V and then enters a uniform magnetic field whose magnitude is 1.80 T. The α-particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the α-particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path?

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  1. 21 January, 18:00
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    charge on alpha particle q = 2 x 1.6 x 10⁻¹⁹ C

    mass m = 6.64 x 10⁻²⁷

    potential diff V = 1.22 x 10⁶ v

    magnetic field B = 1.8 T

    speed of particle = v

    kinetic energy will be acquired by charged particle in electric field

    1/2 m v² = q V

    1/2 x 6.64 x 10⁻²⁷ x v² = 1.22 x 10⁶ x 2 x 1.6 x 10⁻¹⁹

    v² = 1.176 x 10¹⁴

    v = 1.08 x 10⁷ m / s

    speed of particle v = 1.08 x 10⁷ m / s

    b) magnetic force in magnetic field

    = Bqv

    = 1.8 x 2 x 1.6 x 10⁻¹⁹ x 1.08 x 10⁷

    = 6.22 x 10⁻¹² N

    c)

    radius of circular path r

    m v² / r = Bqv

    r = mv / Bq

    = 6.64 x 10⁻²⁷ x 1.08 x 10⁷ / (1.8 x 2 x 1.6 x 10⁻¹⁹)

    = 7.17 x 10⁻²⁰ / 5.76 x 10⁻¹⁹

    =.124 m

    12.4 cm
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