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.

Answer: 1.057*10^-12 N, towards the negative x axis

Explanation: An electron moving in a uniform magnetic field experiences a force.

This force is defined by the formulae below.

F = qvB sinx°

Where F = force on electronic charge

q = magnitude of electronic charge = 1.602*10^-19c

v = velocity of electron = 1.1*10^6 m/s

B = strength of magnetic field = 6.0T

x° = angle between the velocity of the electron and strength of magnetic field = 90° (this is because the axis of the Cartesian plane are perpendicular)

By substituting the parameters, we have that

F = 1.602*10^-19 * 1.1*10^6 * 6 * sin 90

F = 1.602*10^-19 * 1.1*10^6 * 6 * 1

F = 1.057*10^-12 N

The direction of this force is gotten by using cross product on the direction of the velocity and the magnetic field.

Magnetic field is directed towards the negative z axis (-k) and velocity is directed towards the positive y axis (j)

j * (-k) = - i

Hence the force is directed towards the negative axis

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