(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?

(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

answer: what are you talking about?

explanation:

f=ma newton's second law 7x8=56n