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An electron moves in a circular path perpendicular to a constant magnetic field with a magnitude of 2.50 mT. The angular momentum of the electron about the center of the circle is 6.00 ✕ 10-25 J·s.
(a) Determine the radius of the circular path.
(b) Determine the speed of the electron.

Respuesta :

Answer:

Explanation:

The angular momentum of electron mvR = 6 x 10⁻²⁵  Js

Magnetic field B = 2.5 x 10⁻³ T

radius of circular path R = mv / Bq

where m is mass , v is velocity and q is charge on electron

R² = mvR / Bq

R² = 6 x 10⁻²⁵ / 2.5 x 10⁻³ x 1.6 x 10⁻¹⁹

= 1.5 x 10⁻³

R = 3.87 x 10⁻² m

mvR = 6 x 10⁻²⁵

v = 6 x 10⁻²⁵  / mR

= 6 x 10⁻²⁵  / 9.1 x 10⁻³¹ x 3.87 x 10⁻²

= .17 x 10⁸

= 17 x 10⁶ m/s

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