Question

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 x107 m/s in a magnetic field of 6 x 104 T perpendicular to it? What is its frequency? Calculate its energy in keV. (1 eV = 1.6 × 10-19 J).​

Answer 1

Answer:

We find r = m v / (qB) = 9 ×10–31 kg × 3 × 107 m s–1 / (1.6 × 10–19 C × 6 × 10–4 T) = 26 × 10–2 m = 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. Read more on Sarthaks.com - https://www.sarthaks.com/426631/what-the-radius-of-the-path-of-electron-mass-and-charge-moving-speed-of-10-in-magnetic-field

Explanation:

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