An electron moving with a velocity of 1.1 ✕ 106 m/s along the positive y-axis encounters a uniform B-field of 6.0 T oriented along the negative z-axis. Find the force on this charged particle due to this magnetic field.

10.56×10^-13N

Explanation:

From the question, the electron and the magnetic field are perpendicular and make an of 90° with each other.

Force on the electron= qvBsinø

q= magnitude of charge on the electron= 1.6×10^-19C

v= velocity of the electron= 1.1×10^6 ms-1

B= magnetic field intensity= 6.0T

Ø= angle between electron and magnetic field=90°

F= 1.6×10^-19 × 1.1×10^6×6.0 × sin 90°

F= 10.56×10^-13N