A proton accelerated through a potential of 16.0 kV enters a device which has bo

Question

A proton accelerated through a potential of 16.0 kV enters a device which has both an electric and a magnetic field, that are perpendicular to each other as shown in the figure. This device is known as a "velocity filter", because only protons with a given velocity are not deflected and continue their trajectory along the y-axis through the aperture shown in the figure.

Indicate the directions of both the electric force and the magnetic force, +/- x, +/- y, +/- z.

The E-field between the parallel plates has a magnitude of 5.000×105 N/C. Calculate the magnitude of the electric force on the proton.

Calculate the magnitude of the B-field so that the net force on the proton due to both the electric and magnetic fields is zero.

Explanation / Answer

After accelerating through 16 kV ,

Kinetic energy = qV = 1.6 x 10^-19 x 16 x 10^3 = 2.56 x 10^-15 J

1.673 x 10^-27 x v^2 /2 = 2.56 x 10^-15

v = 1.75 x 10^6 m/s

now electric force = magnetic force

Electric force = qE = 1.6 x 10^-19 x 5 x 10^5 = 8 x 10^-14 N ............Ans

8 x 10^-14 = qv B

B = 8 x 10^-14 / ( 1.6 x 10^-19 x 1.75 x 10^6 )

B = 0.286 T ....................Ans

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