Question

An electron beam enters a crossed-field velocity selector with magnetic and electric fields of 2.0 mT and 6.0×10^3 N/C, respectively. (a) What must the velocity of the electron beam be to traverse the crossed fields undeflected? If the electric field is turned off? (b) What is the acceleration of the electron beam?(c) What is the radius of the circular motion that results?

Answer 1

Answer:

a)   v = 3 10⁶ m / s , b)   a = 1.055 10¹² m / s² , c)    r = 8.53 m

Explanation:

a) Let's use Newton's Second Law of Balance, so that electrons do not deviate

      $F_{e}$ - $F_{m}$ = 0

      $F_{e}$ =  $F_{m}$

     q E = q v B

     v = E / B

Let's calculate

    v = 6.0 10³ / 2.0 10⁻³

    v = 3 10⁶ m / s

b) If the electric field is disconnected, the only force left is the magnetic one

    $F_{m}$ = m a

    q v B = m a

    a = q / m v B

    a = q / m (E / B) B

    a = q / m E

    a = 1.6 10⁻¹⁹ /9.1 10⁻³¹ 6.0 10³

    a = 1.055 10¹² m / s²

c) Acceleration is centripetal

    a = v² / r

    r = v² / a

    r = (3 10⁶)² /1.055 10¹²

    r = 8.53 m