What is the radius of the path of an electron (mass 9 × 10-31 kg and

1.	What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6 × 10–19 C) moving at a speed of 3 ×107 m/s in a magnetic field of 6 × 10–4 T perpendicular to it? What is its frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10–19 J).
Answer» We find r = m v / (qB) = 9 ×10^–31 kg × 3 × 10⁷ m s^–1 / (1.6 × 10^–19C × 6 × 10^–4 T) = 26 × 10^–2m = 26 cm n = v / (2 pr) = 2×10⁶ s^–1 = 2×10⁶ Hz = 2 MHz. E = (½ )mv² = (½ ) 9 × 10^–31 kg × 9 × 10¹⁴ m²/s² = 40.5 ×10^–17 J » 4×10^–16 J = 2.5 keV.

What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6 × 10–19 C) moving at a speed of 3 ×107 m/s in a magnetic field of 6 × 10–4 T perpendicular to it? What is its frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10–19 J).

Answer»

We find r = m v / (qB)

= 9 ×10^–31 kg × 3 × 10⁷ m s^–1 / (1.6 × 10^–19C × 6 × 10^–4 T)

= 26 × 10^–2m = 26 cm

n = v / (2 pr) = 2×10⁶ s^–1 = 2×10⁶ Hz = 2 MHz.

E = (½ )mv² = (½ ) 9 × 10^–31 kg × 9 × 10¹⁴ m²/s²

= 40.5 ×10^–17 J » 4×10^–16 J = 2.5 keV.