Question

(a) A velocity selector consists of electric and magnetic fields described by the expressions E = E and B = B ĵ, with B = 10.0 mT. Find the value of E (in kV/m) such that a 740 eV electron moving in the negative x-direction is undeflected. kV/m (b) What If? For the value of E found in part (a), what would the kinetic energy of a proton have to be (in MeV) for it to move undeflected in the negative x-direction?

Answer 1

Answer:

(a) 160000 kV/m

(b) 1336 keV

Explanation:

(a) magnetic filed, B = 10 T

energy of electron, E = 740 eV

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

Let v be the velocity of electron.

E = 1/2 mv^2

740 x 1.6 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 1.6 x 10^7 m/s

v = E / B

E = v x B = 1.6 x 10^7 x 10 = 16 x 10^7 V/m

E = 160000 kV/m

(b) E = 16 x 10^7 V/m

B = 10 T

Let v be the velocity of protons.

v = E / B = 16 x 10^7 / 10 = 1.6 x 10^7 m/s

Kinetic energy of proton, E = 1/2 mv^2

                                          = 0.5 x 1.67 x 10^-27 x 1.6 x 1.6 x 10^14

                                          = 2.14 x 10^-13 J = 1336000 eV = 1336 keV