Question

An electron with a kinetic energy of 22.5 eV moves into a region of uniform magnetic field B of magnitude 4.55 x 104 T. The angle between the directions of the magnetic field and the electron's velocity is 65.5 degrees. What is the pitch of the helical path taken by the electron?

Answer 1

Answer:

9.17 x 10^-10 m

Explanation:

q = 1.6 x 10^-19 C

B = 4.55 x 10^4 T

m = 9.1 x 10^-31 kg

K = 22.5 eV = 22.5 x 1.6 x 10^-19 J = 3.6 x 10^-18 J

θ = 65.5°

Use the formula for the kinetic energy

Let r be the radius of circular path and v be th evelocity

$K = \frac{B^{2}q^{2}r^{2}}{2m}$

By substituting the values

$3.6 \times 10^{-18} = \frac{\left ( 4.55 \times 10^{4} \right )^{2}(1.6\times 10^{-19} \right )^{2}r^{2}}{2\times 9.1\times 10^{-31}$

r = 3.52 x 10^-10 m

Use the formula for the speed

$v =\frac{Bqr}{m}$

$v =\frac{4.55 \times 10^{4}\times 1.6\times 10^{-19}\times 3.52 \times 10^{-10}}{9.1\times 10^{-31}}$

v = 2.82 x 10^6 m/s

The pitch of the helix = $v Cos\theta \times \frac {2\pi r}{v}$

 = $2\pi rCos\theta =2 \times 3.14 \times 3.52 \times 10^{-10} \ Cos65.5$

= 9.17 x 10^-10 m