A proton having a speed of 3.0 × 106 m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within

Question

A proton having a speed of 3.0 × 106 m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within the field. What is the magnitude of the magnetic field?

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Latifah 4 years 2021-08-05T03:15:35+00:00 2 Answers 7 views 0

Answers ( )

  1. kimchi2
    0
    2021-08-05T03:17:31+00:00
    Reply

    Answer:

    0.16T

    Explanation:

  2. Ben
    0
    2021-08-05T03:17:34+00:00
    Reply

    Answer:

    B=0.15T

    Explanation:

    In a magnetic field we have that the force generated by the magnetic field equals the centripetal force

    Fm=qvB=mv^2/r

    Hence, this problem can be solved using the expression that relates the radius of a curved path described by a particle and a magnetic field

    B=\frac{m_pv}{q_pr}

    mp: mass of a proton = 1.67*10^{-27}kg

    qp: charge of the proton = 1.6*10^{-19}C

    r: radius = 0.20m

    v: speed of the proton = 3*10^{6}m/s

    – By replacing in the expression we have

    B = \frac{(1.67*10^{-27}kg)(3*10^{6}\frac{m}{s})}{(1.6*10^{-19}C)(0.2m)}\\\\B=0.15T

    Hope this helps!!

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