Answer:
The magnitude of the acceleration  is [tex]a = 0.33 m/s^2[/tex]
The direction is [tex]- \r k[/tex] i.e the negative direction of the z-axis
Explanation:
 From  the question we are that
    The mass of the particle [tex]m = 1.8*10^{-3} kg[/tex]
     The charge on the particle is [tex]q = 1.22*10^{-8}C[/tex]
     The velocity is [tex]\= v = (3.0*10^4 m/s ) j[/tex]
    The the magnetic field is  [tex]\= B = (1.63T)\r i + (0.980T) \r j[/tex]
The charge experienced  a force which is mathematically represented as
    Â
          [tex]F = q (\= v * \= B)[/tex]
  Substituting value
     [tex]F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)[/tex]
      [tex]= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))[/tex]
      [tex]= (-5.966*10^4 N) \r k[/tex]
Note :
      [tex]i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\[/tex]
Now force is also mathematically represented as
    [tex]F = ma[/tex]
Making a the subject
   [tex]a = \frac{F}{m}[/tex]
  Substituting values
   [tex]a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}[/tex]
    [tex]= (-0.33m/s^2)\r k[/tex]
    [tex]= 0.33m/s^2 * (- \r k)[/tex]
