A proton moving with a speed of 3.20 × 10⁵ m/s perpendicular to a uniform magnetic field of 0.0786 T will follow which of the paths described below? (q_p = 1.60 × 10−¹⁹ C and m_p = 1.67 × 10−²⁷ kg) A)a straight-line path B)a circular path of 4.25 cm radius C)a circular path of 10.9 cm radius D)a circular path of 1.52 cm radius E)a circular path of 54.1 cm radius

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Quizplus · Accepted Answer

The path of a charged particle moving perpendicular to a uniform magnetic field is circular. The radius of the path can be calculated using the formula $r = \frac{mv}{qB}$, where $m$ is the mass of the particle, $v$ is its velocity, $q$ is its charge, and $B$ is the magnetic field strength. Substituting the given values ($m = 1.67 \times 10^{-27}$ kg, $v = 3.20 \times 10^{7}$ m/s, $q = 1.60 \times 10^{-19}$ C, $B = 0.0786$ T) into the formula gives a radius of approximately 4.25 cm, which corresponds to choice B.

A Proton Moving with a Speed of 3