A 2MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T.the force on a proton is

Question

A 2MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T.the force on a proton is

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Hải Đăng 4 years 2021-08-04T23:26:24+00:00 2 Answers 26 views 0

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  1. nguyetanh
    0
    2021-08-04T23:27:27+00:00
    Reply

    Answer:

    8*10^-12

    Explanation:

    Given that

    Energy of proton, K = 2 MeV = 2 * 1.6*10^-19 *10^6 = 3.2*10^-13

    magnetic field strength, B = 2.5 T

    mass of proton, m = 1.67*10^-27 kg

    K = ½mv², making v² the subject of formula by rearranging, we have

    v² = 2k/m

    v² = (2 * 3.2*10^-13) / 1.67*10^-27

    v² = 6.4*10^-13 / 1.6*10^-27

    v² = 4*10^14

    v = √4*10^14

    v = 2*10^7 m/s

    f = qvbsinθ, where

    θ = 90

    v = 2*10^7 m/s

    b = 2.5 T

    q = 1.6*10^-19

    f = 1.6*10^-19 * 2*10^7 * 2.5 sin 90

    f = 8*10^-12 N

    thus, the force on the proton is 8*10^-12

  2. trungdung22
    0
    2021-08-04T23:27:54+00:00
    Reply

    Answer:

    7.8×10-12N

    Explanation:

    We know that

    Magnetic force = F = qVB

    And

    Also Kinetic energy K.E is

    E = (1/2)mV²

    So making v subject

    V = √(2E / m)

    And

    E = KE = 2MeV

    = 2 × 106 eV

    = 2 × 106 × 1.6 × 10–19 J

    = 3.2 × 10–13 J

    And then

    V= √2×3.2E-13/1.6E-27

    1.9E7m/s

    Given that

    mass of proton = 1.6 × 10–27 kg,

    Magnetic field strength B = 2.5 T.

    So F= qBv sinစ

    =

    So F = 1.6 × 10–19 × 2.5 × 1.9 x10^7 x sin 90°

    = 7.8 x 10^-12N

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