A particle with charge q and kinetic energy K travels in a uniform magnetic field of magnitude B. If the particle moves in a circular path o

Question

A particle with charge q and kinetic energy K travels in a uniform magnetic field of magnitude B. If the particle moves in a circular path of radius R, find expressions for:

a. its speed
b. its mass.

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Tryphena 1 year 2021-08-06T13:19:41+00:00 2 Answers 306 views 0

Answers ( )

    0
    2021-08-06T13:21:36+00:00

    Answer:

    Given that K.E is

    1/2mv²

    So to find speed v,

    Make it subject

    K.E= 1.2mv²

    However radial force = magnetic force

    So mv²/r= qvB

    So v subject

    V= 2K.E/ qBr that is speed

    To find mass

    K.E = 1/2mv²

    Puy value of v

    So KE= 1/2m(2K.E/qBr)

    m= (qBr)/2K.E

    That is mass

    0
    2021-08-06T13:21:38+00:00

    Answer:

    m = qbr/v

    v = 2k/qbr

    Explanation:

    When a charged particle enters a magnetic field, it experiences a force that is always perpendicular to the velocity. This force provides a centripetal force, and thus, we have

    qvb = mv²/r

    if we make m the subject of the formula, we will have

    m = qbr/v

    Recall that the kinetic energy, KE = ½mv²

    Now, let’s make v² the subject of formula, we have

    v² = 2K/m

    now, we substitute for m from the equation we got earlier

    v² = 2K / (qbr/v)

    v² = 2Kv / qbr, if we simplify further, we have

    v = 2k / qbr

    Therefore, we can say that the expression for the mass and speed is respectively,

    m = qbr/v

    v = 2k/qbr

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