Crossed electric and magnetic fields can be used to select one speed for which charged particles in a beam moving perpendicular to the fields will pass through the fields without being deflected. what value of the electric field e should be used in a velocity selector to pick out protons whose kinetic energy is 1,000 ev when the magnetic field b = 8.0 t? the mass of a proton is 1.67 x 10-27 kg. recall that 1.0 ev = 1.6 x 10-19 joules.
a) 3.5 x 1o v/m
b) 21,900 v/m
c) 43,800 v/m
d) 0.22 v/m
e) 3.1 x 10 v/m

true A) 3.5 10⁴  V/m

Explanation:

For the speed selector to work the electric force must be challenged to the magnetic force, so

    =

     q E = q v B

     v = E / B

This expression we need the speed of the proton, which we can find by its energy

       = K = ½ m v²

     v²2 =√ 2 / m

Let's replace

     E = v B

     E = B √ 2  / m

Let's reduce the magnitudes to the SI system

       = 1000 eV (1.6 10⁻¹⁹J / 1 eV) = 1.6 10⁻¹⁶ J

     E = 8.0 √ (2 1.6 10⁻¹⁶ / 1.67 10⁻²⁷)

     E = 8.0 √ (19.16 10¹⁰)

     E = 3.5 104 N/C

When reviewing the answer the correct one is A

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