A proton, which moves perpendicular to a magnetic field of 0.60 T in a circular path of radius 0.080 m, has what speed? (q_p = 1.6 *10^-¹⁹ C and m_p = 1.67 *10^-²⁷ kg) A) 9.2 * 10⁶ m/s B) 4.6 * 10⁶ m/s C) 9.6 * 10⁶ m/s D) 3.4 * 10⁶ m/s

Question

Quizplus · Accepted Answer

The speed of the proton can be calculated using the formula for the radius of the circular path in a magnetic field: r = (m*v)/(q*B). Solving for v gives v = (r*q*B)/m. Substituting the given values: r = 0.080 m, q = 1.6 x 10^-19 C, B = 0.60 T, and m = 1.67 x 10^-27 kg, we find v ≈ 4.6 x 10^6 m/s.

A Proton, Which Moves Perpendicular to a Magnetic Field of 0.60