A proton travels at a speed of 5.0 Times 10^7 m/s through a 1.0-T magnetic field

Question

A proton travels at a speed of 5.0 Times 10^7 m/s through a 1.0-T magnetic field. What is the magnitude of the magnetic force on the proton if the angle between the proton's velocity and the magnetic field vector is 30 degrees? (e = 1.60 Times 10^-19 C) 2.0 Times 10^-14 N 4.0 Times 10^-14 N 2.0 Times 10^-12 N 4.0 Times 10^-12 N An electron moves with a speed of 8.0 Times 10^6 m/s along the +x-axis. It enters a region where there is a magnetic field of 2.5 T, directed at an angle of 60 degrees to the +x-axis and lying in the xy-plane. What is the magnitude of the acceleration of the electron? (e = 1.60 Times 10^-19 C, m_e1 = 9.11 Times 10^-31 kg) 1.3 Times 10^18m/s^2 3.0 Times 10^18m/s^2 1.3 Times 10-^18 m/s^2 3.0 Times 10^-18 m/s^2 0 m/s^2

Explanation / Answer

We know that the force in a magnetic field is given by

F = q v X B

F = q vB sin (30)

F = 1.6 * 10-19 * 5 *107 * 1 sin (30) = 4 10-14 N

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