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.

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

a=d/t

a="triangle"v/t

explanation:

a = acceleration

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2