. A proton, which moves perpendicular to a magnetic field of 1.2 T in a circular path of radius 0.080 m, has what speed? (qp = 1.6 · 10-19 C

Question

. A proton, which moves perpendicular to a magnetic field of 1.2 T in a circular path of radius 0.080 m, has what speed? (qp = 1.6 · 10-19 C and mp = 1.67 · 10-27 kg)

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RuslanHeatt 1 year 2021-09-04T13:01:40+00:00 1 Answers 141 views 0

Answers ( )

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    2021-09-04T13:03:07+00:00

    Answer:

    [tex]9.198\times 10^6 m/s[/tex]

    Explanation:

    We are given that

    Magnetic field, B=1.2 T

    Radius of circular path, r=0.080 m

    [tex]q_p=1.6\times 10^{-19} C[/tex]

    [tex]m_p=1.67\times 10^{-27} kg[/tex]

    [tex]\theta=90^{\circ}[/tex]

    We have to find the speed of proton.

    We know that

    Magnetic force, F=[tex]qvBsin\theta[/tex]

    According to question

    Magnetic force=Centripetal force

    [tex]q_pvBsin90^{\circ}=\frac{m_pv^2}{r}[/tex]

    [tex]1.6\times 10^{-19}\times 1.2=\frac{1.67\times 10^{-27}v}{0.08}[/tex]

    [tex]v=\frac{1.6\times 10^{-19}\times 1.2\times 0.08}{1.67\times 10^{-27}}[/tex]

    [tex]v=9.198\times 10^6 m/s[/tex]

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