An electron with kinetic energy 3.6 keV moves horizontally into aregion of space

Question

An electron with kinetic energy 3.6 keV moves horizontally into aregion of space in which there is a downward-directed uniformelectric field of magnitude 11 kV/m. (a) What are the magnitude and direction of the(smallest) magnetic field that will cause the electron to continueto move horizontally? Ignore the gravitational force, which israther small. (a) What are the magnitude and direction of the(smallest) magnetic field that will cause the electron to continueto move horizontally? Ignore the gravitational force, which israther small.

Explanation / Answer

Kinetic nergy of electron K = 3.6 keV

So, Potential difference V = 3.6 kV = 3600 V

Speed of electron v = [ 2Vq / m]

Whre q = charge = 1.6* 10 ^ -19 C

         M = mass = 9.11* 10 ^ -31 kg

Plug ther values we get   v = 35.56 * 10 ^ 6 m / s

Electric field E = 11 kV / m = 11000 V / m

we know E = Bv

from thisrequire magnetic field B = E / v

                                                 = 309.33 * 10 ^ -6 T

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