Question

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 .

Answer 1

Step-by-step explanation:

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