Download Please review my solution to the problem and explain in

Please review my solution to the problem and explain in detail what I may be doing wrong and what concepts I
may not be applying correctly. I am not sure if I am apply the crossed fields concept correct in presuming that B
is perpendicular.
The problem states:
An electron has an initial velocity of (12.0j + 15.0k) km/s and a constant acceleration of (2.00 x 1012 m/s2)i in a
region in which uniform electric and magnetic fields are present. If B = (400μT)i, find the electric field E.
We know:
This is a crossed fields problem with the two fields perpendicular to each other.
B is perpendicular to v, out of the page.
Since B is perpendicular, it does no work on the electron but instead deflects it in a circular path.
q = 1.6 x 10-19 C
v = (12.0j + 15.0k) km/s = square root(12.02 + 15.02) = 19.21 km/s = 1.921 x 104 m/s
B = 400μT = 400 x 10-6 T = 400 x 10-6 N/Cm/s
Φ = 90º
E = qvBsinΦ
E = (1.6 x 10-19 C)(1.921 x 104 m/s)(400 x 10-6 N/Cm/s)(sin90º)
E = 1.229 x 10-18 N
The force due to the magnetic field is normal to plane containing the magnetic field and velocity vectors and
that due to electric field is along the direction of the field (for positive charge and opposite to the negative
charge), The total force is the resultant of both and acceleration is due to resultant force.
This can be done easily using the vector form of the quantities and the formula.
The force on a moving charged particle in presence of E and B is given by the Lorentz formula as

F  qE  q v  B

Hence the net force on the electron will be


F  e E  v  B 


Or
F   1.6 *1019 C   E  12.0 *103 j + 15.0 *103 k    400 *106 i 
[I, j and k are unit vectors in x, y and z directions respectively]
Or
F   1.6 *1019 C   E  12.0 *103 j   400 *106 i   15.0 *103 k    400 *10 6 i 
Or
F   1.6 *1019 C   E  4.8  k   6.0 j 
Or
F   1.6 *1019 C   E  4.8k  6.0 j 
Now the acceleration of the particle is given by
a
Or
F
m
2.00 *10 i 
This gives
12
 1.6 *10
19
C   E  4.8k  6.0 j 
9.11*1031
9.11*1031 * 2.00 *1012 i
1.6 *10
19
C
   E  4.8k  6.0 j 
Or
11.4i    E  4.8k  6.0 j 
Or
E  11.4i  4.8k  6.0 j (N/C)
This is the electric field in the region.