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svet-max [94.6K]

2 years ago

3

An electron with kinetic energy 2.5 keV moves horizontally into a region of space in which there is a downward-directed uniform

electric field of magnitude 10 kV/m.
(a) What are the magnitude and direction of the (smallest) uniform magnetic field that will cause the electron to continue to move horizontally? Ignore the gravitational force, which is small.

(b) Is it possible for a proton to pass through the combination of fields undeflected? If so, under what circumstances?

1 answer:

MArishka [77]2 years ago

5 0

Answer:

(a) 2.4 x 10^-4 T

(b)

Explanation:

Kinetic energy of electron, K = 2.5 kev = 2.5 x 1.6 x 10^-16 J = 4 x 10^-16 J

Electric field , E = 10 kV/m = 10000 V/m

mass of electron, m = 9.1 x 10^-31 kg

(a)

K = 1/2 mv^2

4 x 10^-16 = 0.5 x 9.1 x 10^-31 x v^2

v = 4.2 x 10^7 m/s

v = E / B

where, B be the strength of magnetic field

B = E / v = 10000 / (4.2 x 10^7) = 2.4 x 10^-4 T

(b) No, because the mass of proton is different, so it is not undeflected in the same combination of plates.

the velocity is changed and hence the kinetic energy is changed for the same filed configuration.

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Answer:

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The given data :-

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