Question

A velocity selector has a magnetic field of magnitude 0.28 T perpendicular to an electric field of magnitude 0.48 MV/m.(a) What must the speed of a particle be for it to pass through undeflected? (__________ )m/s(b)What energy must protons have to pass through undeflected? ( __________)keV(c) What energy must electrons have to pass through undeflected? ( __________)eV

Answer 1

Answer:

a) v = 1,714 10⁶ m / s , b)   K = 13.37 10⁻¹⁹ J , c)   K= 8.36 eV

Explanation:

a) A speed selector uses an electric force and a magnetic force that opposes it, so we can use the equilibrium equation

       $F_{e}$ - $F_{e}$ = 0

       $F_{e}$ = $F_{m}$

      q E = q v B

      E = v B

      v = E / B

Let's calculate

     v = 0.48 10⁶ / 0.28

     v = 1,714 10⁶ m / s

b) the energy of the particles is kinetic energy

     K = ½ m v2

The mass of the proton is

    mp = 1.67 10⁻²⁷ kg

     K = ½ 1.67 10⁻²⁷ (1,714 10⁶)²

     K = 2.46 10⁻¹⁵ J

Let's reduce

    K = 2.45 10⁻¹⁵ J (1eV / 1.6 10⁻¹⁹J)

    K = 1.53 10⁴ eV

   K = 15.3 KeV

The mass of the electron

    me = 9.1 10-31 kg

     K = ½ 9.1 10⁻³¹ (1,714 10⁶)²

     K = 13.37 10⁻¹⁹ J

Let's reduce

    K = 13.37 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹ J)

    K= 8.36 eV