Question

The ionosphere contains free electrons. What is the amplitude of oscillation of the electrons when

subject to a 200 kHz electromagnetic wave in which the oscillations of electric filed have

amplitude 5 x 10-3 V m-1

?​

Answer 1

72 v2 is the answer

Answer 2

0.557 × 10^-3 m is the required answer.

GIVEN:

• Frequency, f = 200 kHz = 200 × 10³ Hz = 2 × 10⁵ Hz

• Electric field, E = 5 × 10^-3 V/m.

TO FIND:

Amplitude of oscillation, y

FORMULAE:

• Acceleration, a = - ω² y

Here ω is the angular frequency.

• ω = 2πf

• F = qE

Here q is the charge of electron.

q = - 1.6 × 10^-19 C

• According to Newton's second law of motion, F = ma

Here m is the mass of electron, m = 9.11 × 10 ^-31 kg

SOLUTION:

STEP 1: TO FIND FORCE

F = qE

F = - 1.6 × 10^-19 × 5 × 10^-3

F = - 8 × 10^-22 N

STEP 2: TO FIND ACCELERATION

F = ma

- 8 × 10^-22  = 9.11 × 10 ^-31 × a

$a=\dfrac{-8\times10^{-22}}{9.11\times10^{-31}} \\\\a = \dfrac{-8}{9.11} \times10^{-22+31}\\\\a= -0.878 \times10^9\: m/s^2$

STEP 3: TO FIND ANGULAR FREQUENCY

ω = 2πf

ω = 2 × 3.14 × 2 × 10⁵

ω = 12.56 × 10⁵

STEP 4: TO FIND AMPLITUDE OF OSCILLATION

a = - ω² y

$-0.878 \times10^9 =- {(12.56\times10^5) }^2\times y\\\\0.878 \times10^9 = 12.56 \times 12.56 \times 10^{10} \times y\\\\0.878 \times10^9 =157.7536\times10^{10}\times y\\\\y = \dfrac{0.878 \times10^9}{157.7536\times10^{10}} \\\\y = \dfrac{0.878}{157.7536} \times 10^{9-10}\\\\y = 0.00557 \times10^{-1}\\\\y = 0.557 \times10^{-2}\times10^{-1}\\\\y = 0.557 \times10^{-3}\: m$

ANSWER:

Amplitude of oscillation is 0.557 × 10^-3 m