Scientists study a radioactive source that produces a stream of positively-charged particles moving with different speeds along the x-axis. The stream enters a box with a uniform magnetic field B = 0.1 T directed along the z-axis and a uniform electric field E of unknown strength and direction. Scientists observe that the particles that move with the speed of 2 m/s are not deflected in the box at all. A. (6 points) Find the direction and the strength of the electric field in the box. Scientists allow only undeflected particles to leave the box through a small opening and enter a big chamber that contains only magnetic field B = 0.1 T along the z-axis. B. (2 points) Describe the trajectory of the particles in the chamber. C. (6 points) Scientists notice that the particles completely reverse their direction of motion in 0.1 ms after entering the chamber. Assuming that their particle’s charge is 1.6 x 10-19 C, determine the mass of the particles

A)

Explanation:

A)  the particles do not reflect the electric and magnetic forces, they must have the same magnitude and opposite direction.

the magnetic force is perpendicular to the speed that is on the x axis and to the magnetic field that stakes the z axis, so the force goes in the direction of the positive y axis.

With this force it must be opposite to the electric force, this implies that the electric field must go in the direction of the negative y axis.

Let's use Newton's equation of equilibrium

                = 0

                 FM = Fe

                 .q v B = q E

                  E = v B

                  E = 2 0.1

                   E = 0.2 N / C

.b) when the particular one reaches the camera with a magnetic field, the magnetic force perpendicular to the speed causes them to start a uniform circular movement since the magnitude of the speed does not change.

With the magnetic force it is on the axis and the circle described by the particles is on this axis

.c) The particles are in a circular motion the point where their motion is reversed they move in the negative direction when they are at the lowest or highest part of the path or I know that they have traveled

Let's use kinematics

               .v = va - a t

               .v = -vow

               -vow = vow - at

               .a = 2 vow / t

               .a = 2 2 / 0.1 10-3

               .a = 40 103 m / s2

Now let's use Newton's second law

                 FM = m a

                  Q v B = m a

            .m = q v B / a

           .m = 1.6 10-19 2.0 0.1 / 40 103

            .m = 8 10-25 kg