A proton moving with a speed of 2.9 × 10⁵ m/s perpendicular to a uniform magnetic field of 0.35 T will follow which of the paths described below? (q_p = 1.6 × 10−¹⁹ C and m_p = 1.67 × 10−²⁷ kg)

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle moving in a magnetic field is given by F = qvB, where q is the charge of the particle, v is its velocity, and B is the magnetic field strength. The direction of the force is perpendicular to both v and B. In this case, the proton is moving perpendicular to the magnetic field, so the force on it will be perpendicular to both its velocity and the magnetic field. This will cause the proton to move in a circular path. The radius of the circular path can be calculated using the formula r = mv/qB, where m is the mass of the particle. Plugging in the given values, we get r = (1.67 × 10−27 kg)(2.9 × 105 m/s)/(1.6 × 10−19 C)(0.35 T) = 0.9 cm. Therefore, the correct answer is B.

A Proton Moving with a Speed of 2