Question

A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV. The beam is then entered into a region between two parallel metal plates with potential difference 120 V and a separation 8 mm, perpendicular to the direction of the field. What magnitude of magnetic field is needed so that the alpha particles emerge undeflected from between the plates?

Answer 1

Answer:

The magnetic field required required for the beam not to be deflected  is $B = 0.0036T$

Explanation:

From the question we are told that

    The charge on the particle is $q = +2e$

    The mass of the particle is  $m = 6.64 *10^{-27} kg$

    The potential difference is $V_a = 1.8 kV = 1.8 *10^{3} V$

    The potential difference between the two parallel plate is  $V_b = 120 V$

    The separation between the plate is  $d = 8 mm = \frac{8}{1000} = 8*10^{-3}m$

   

The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam  after the region having a potential difference of 1.8kV

               $KE_b = PE_b$

Generelly

              $KE_b = \frac{1}{2} m v^2$

And      $PE_b = q V_a$

 Equating this two formulas

              $\frac{1}{2} mv^2 = q V_a$

making v the subject

           $v = \sqrt{\frac{q V_a}{2 m} }$

Substituting value  

           $v = \sqrt{\frac{ 2* 1.602 *10^{-19} * 1.8 *10^{3}}{2 * 6.64 *10^{-27}} }$

           $v = 41.65*10^4 m/s$

Generally the electric field between the plates is mathematically represented as

                 $E = \frac{V_b}{d}$  

Substituting value  

                 $E = \frac{120}{8*10^{-3}}$              

                $E = 15 *10^3 NC^{-1}$

the magnetic field  is mathematically evaluate    

                     $B = \frac{E}{v}$

                   $B = \frac{15 *10^{3}}{41.65 *10^4}$

                    $B = 0.0036T$