Question

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

Answer 1

Answer:

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

    $F_{e}$ = $F_{m}$

     q E = q v B

     v = E / B

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

     $E_{m}$  = K = ½ m v²

     v²2 =√ 2 $E_{m}$ / m

Let's replace

     E = v B

     E = B √ 2 $E_{m}$  / m

Let's reduce the magnitudes to the SI system

      $E_{m}$  = 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