A proton,which moves perpendicular to a magnetic field of 0.8 T in a circular path of radius 0.062 m,has what speed? (q_p = 1.6 × 10−¹⁹ C and m_p = 1.67 × 10−²⁷ kg)

Question

Quizplus · Accepted Answer

The formula to calculate the speed of a charged particle moving in a circular path in a magnetic field is:

v = B*r*m^-1

where v is the speed of the particle, B is the magnetic field, r is the radius of the circular path, and m is the mass of the particle.

Substituting the values given in the question, we get:

v = 0.8*0.062*(1.67*10^-27)^-1

v = 4.8*10^6 m/s

Therefore, the correct answer is D) 4.8E+6 m/s.

A Proton,which Moves Perpendicular to a Magnetic Field of 0