Question

An electron is moving vertically upwards with a speed of 2.0*10^8 ms^-1,magnetic field is 0.50 N A^-1 m^-1.Calculate the acceleration of the electron

Answer 1

Magnetic force F exerted by a magnetic field E on a moving charge q with a velocity v is given by

     F = q  v Χ B      ,  cross product of velocity with  magnetic field.

Magnitude of the force is given by 
 
     F = q v B Sin Φ,   where  Φ is the angle between vectors v and B. 

You have not specified the angle between v and B.   Let us say vertical direction towards the sky is the Z-axis.

1.   Let B be horizontal, and along positive X-axis.  Then 
       F = 1.6 * 10⁻¹⁹ * 2 * 10⁸ * 0.50 Sin 90° =  =  1.6 * 10⁻¹¹ N
       Force will be along positive Y-axis,  as given by the right hand thumb rule.
       Acceleration = F/mass = 1.6 * 10⁻¹¹/9.11*10⁻³¹ = 1.756 * 10¹⁹ m/sec²

2.  Let B be horizontal , and along positive Y-axis , then
       F = 1.6 * 10⁻¹⁹ * 2 * 10⁸ * 0.50 Sin 90° =  1.6 * 10⁻¹¹ N
     Force will be along negative X-axis,  as given by the right hand thumb rule.          
       Acceleration = F/mass = 1.6 * 10⁻¹¹/9.11*10⁻³¹ = 1.756 * 10¹⁹ m/sec²

3.  Let B be along vertical direction,  then  Ф = 0°.  Hence
 
            F = 1.6 * 10⁻¹⁹ * 2 * 10⁸ * 0.50 Sin 0° = 0

Answer 2

Here's your answer. Check it out, my dear.