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The electric field part of an electromag...

The electric field part of an electromagnetic wave in vacuum is given by E = 6 cos `[1.2 ("rad")/(m) y + 3.6 xx 10^(8) ("red")/(s) t ] hat(i) (N)/(C ) ,` then
find
i) frequency of propagation (f)
(ii) Ampitude of magnetic field in electromagnetic wave

A

`5.7 xx 10^(7) Hz, 2 xx 10^(-8)` T

B

`57 xx 10^(7) Hz, 0.2 xx 10^(-8)` T

C

`5.7 xx 10^(4) Hz, 2 xx 10^(-6)` T

D

`5.7 xx 10^(10) Hz, 2 xx 10^(-10)` T

Text Solution

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The correct Answer is:
To solve the problem, we need to find the frequency of propagation (f) and the amplitude of the magnetic field (B0) for the given electromagnetic wave. The electric field is given as: \[ E = 6 \cos(1.2y + 3.6 \times 10^8 t) \hat{i} \, \text{(N/C)} \] ### Step 1: Identify the angular frequency (ω) From the equation of the electric field, we can identify the angular frequency (ω) from the term associated with time (t): \[ \omega = 3.6 \times 10^8 \, \text{rad/s} \] ### Step 2: Calculate the frequency (f) The relationship between angular frequency (ω) and frequency (f) is given by: \[ \omega = 2\pi f \] We can rearrange this to solve for f: \[ f = \frac{\omega}{2\pi} \] Substituting the value of ω: \[ f = \frac{3.6 \times 10^8}{2\pi} \] Calculating this gives: \[ f \approx \frac{3.6 \times 10^8}{6.2832} \approx 5.73 \times 10^7 \, \text{Hz} \] ### Step 3: Find the amplitude of the magnetic field (B0) The relationship between the amplitude of the electric field (E0) and the amplitude of the magnetic field (B0) in an electromagnetic wave is given by: \[ E_0 = B_0 \cdot c \] Where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \). From the electric field equation, we know: \[ E_0 = 6 \, \text{N/C} \] Now, we can rearrange the equation to solve for B0: \[ B_0 = \frac{E_0}{c} \] Substituting the values: \[ B_0 = \frac{6}{3 \times 10^8} \] Calculating this gives: \[ B_0 = 2 \times 10^{-8} \, \text{T} \] ### Final Answers: i) Frequency of propagation (f) = \( 5.73 \times 10^7 \, \text{Hz} \) ii) Amplitude of magnetic field (B0) = \( 2 \times 10^{-8} \, \text{T} \) ---
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