The electric field in an electromagnetic wave is given by E = 56.5 sin ω(t − x/c) NC−1. Find the intensity of the wave if it is propagating along x-axis in the free space.

(Given ϵ0 = 8.85 × 10−12 C2N−1m−2)

  1. 5.65 Wm−2
  2. 4.24 Wm−2
  3. 1.9 × 10−7 Wm−2
  4. 56.5 Wm−2

Answer (Detailed Solution Below)

Option 2 : 4.24 Wm−2
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JEE Main 04 April 2024 Shift 1
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90 Questions 300 Marks 180 Mins

Detailed Solution

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CONCEPT:

The intensity of the electromagnetic wave is written as;

I = \(\frac{1}{2} \epsilon_o E^2_oc\)      -----(1)

Here we have "I" as intensity, ϵ0 as permittivity of the free space, E is the electric field and c is the velocity of light.

CALCULATION:

Given:

E = 56.5 sin ω(t − x/c) NC-1

The velocity of light, c = 3 × 108 m/s

and permittivity of the free space,ϵ0 = 8.85 \(×\) 10-12 

Now for the calculation of intensity, we are using equation (1) we have;

I = \(\frac{1}{2} \epsilon_o E^2_oc\) 

⇒ I =  \(\frac{1}{2}× 8.85 × 10 ^{-12} × 56.5^2× 3 × 10 ^8\) 

⇒ I = \( 4.425 × 10 ^{-4} × 3192.25× 3 \)

⇒ I = 42377.11875 \(×\) 10-4

⇒ I = 4.24 Wm-2

Hence, option 2) is the correct answer.

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