An electron of kinetic energy 1.39 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 20.8 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

a)  v = 2.21 10⁷ m / s, b)  B = 6.04 10⁻⁴ T , c)  f = 1.69 10⁷ Hz , d)     T = 5.9 10⁻⁸ s

Explanation:

a) For this exercise, let's start by using the concept of kinetic energy

           K = ½ m v²

           v = √ 2K / m

Let's reduce the magnitudes to SI units

          E = K = 1.39 kev (1000 eV / 1 keV) (1.6 10⁻¹⁹ J / 1 eV) = 2,224 10⁻¹⁶ J

          r = 20.8 cm (1 m / 100 cm) = 0.208 m

let's calculate

          v = √ (2 2,224 10⁻¹⁶ / 9.1 10⁻³¹)

          v = √ (4.8879 10¹⁴)

          v = 2.21 10⁷ m / s

b) let's use Newton's second law where the force is magnetic

              F = m a

where the acceleration is centripetal

              a = v² / r

the magnetic force is

              Fm = q v x B = q v B sin θ

since the circle is perpendicular to the magnetic field, the angle is 90º and the sine is equal to one, let's substitute

          qv B = m v² / r

          B = m v / rq

let's calculate

          B = 9.1 10⁻³¹ 2.21 10⁷ / (0.208  1.6 10⁻¹⁹)

          B = 6.04 10⁻⁴ T

c) Linear and angular variables are related

          v = w r

          w = v / r

          w = 2.21 10⁷ / 0.208

          w = 1.0625 10⁸ ras / s

    angular velocity and frequency are related

          w = 2π f

          f = w / 2π

          f = 1.0625 10⁸ / (2π)

          f = 1.69 10⁷ Hz

d) the frequency is the inverse of the period

          f = 1 / T

          T = 1 / 1.69 10⁷

           T = 5.9 10⁻⁸ s

answer; ///i do believe that the correct answer is ; ///(physical change);

good looking response