An electron has an initial velocity of (17.1 + 12.7) km/s, and a constant acceleration of (1.60 * 1012 m/s2) in the positive x direction in a region in which uniform electric and magnetic fields are present. If = (529 µT) find the electric field.

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 = (17.1j + 12.7k) km/s = square root (17.1² + 12.7²) = 2.13 x 10⁴ m/s

the force acting on electron is

F = qvBsinΦ

F = (1.6 x 10⁻¹⁹C) (2.13. x 10⁴ m/s) (526 x 10⁻⁶ T) (sin90º)

F = 1.793x 10⁻¹⁸ N

The net force acting on electron is

F = e (E + (vXB)

= ( - 1.6 * 10⁻¹⁹) (E + (17.1 * 10³j + 12.7 * 10³ k) X (529 * 10⁻⁶) (i)

= (-1.6 * 10⁻¹⁹) (E - 6.7k + 9.0j)

a = F/m

1.60 * 10¹² i = (-1.6 * 10⁻¹⁹) (E - 6.9 k + 7.56 j) / 9.11 * 10⁻³¹

9.11 i = - (E - 6.7 k + 9.0 j)

E = - 9.11i + 6.7k - 9.0j