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Manitude of the electric and magnetic fi...

Manitude of the electric and magnetic field in an electromagnetic wave radiated by a `200 W` bulb at a distance 2m from it is assuming efficiency of bulb is `5%` and it behaves like a point source.

A

`12.27 N//C, 4.09 xx 10^(-8) T`

B

`10 N//C, 10^(-5) T`

C

`11.7 N//C, 5 xx 10^(-6)T`

D

`5N//C, 10^(-4)T`

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The correct Answer is:
To find the magnitude of the electric and magnetic fields in an electromagnetic wave radiated by a 200 W bulb at a distance of 2 m, we can follow these steps: ### Step 1: Calculate the Effective Power The bulb has an efficiency of 5%. Therefore, the effective power (P) radiated by the bulb can be calculated as: \[ P = \text{Efficiency} \times \text{Total Power} = \frac{5}{100} \times 200 \, \text{W} = 10 \, \text{W} \] ### Step 2: Calculate the Intensity of the Wave The intensity (I) of the electromagnetic wave at a distance (r) from a point source is given by: \[ I = \frac{P}{A} = \frac{P}{4\pi r^2} \] Substituting the values: \[ I = \frac{10}{4\pi(2^2)} = \frac{10}{4\pi(4)} = \frac{10}{16\pi} \] Using \(\pi \approx 3.142\): \[ I \approx \frac{10}{16 \times 3.142} \approx \frac{10}{50.272} \approx 0.199 \, \text{W/m}^2 \] ### Step 3: Calculate the Maximum Electric Field (E₀) The intensity is also related to the electric field by the formula: \[ I = \frac{1}{2} \epsilon_0 c E_0^2 \] Rearranging for \(E_0\): \[ E_0^2 = \frac{2I}{\epsilon_0 c} \] Taking the square root: \[ E_0 = \sqrt{\frac{2I}{\epsilon_0 c}} \] Substituting the known values (where \(\epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m}\) and \(c \approx 3 \times 10^8 \, \text{m/s}\)): \[ E_0 = \sqrt{\frac{2 \times 0.199}{(8.85 \times 10^{-12})(3 \times 10^8)}} \] Calculating: \[ E_0 = \sqrt{\frac{0.398}{2.655 \times 10^{-3}}} \approx \sqrt{150.0} \approx 12.27 \, \text{N/C} \] ### Step 4: Calculate the Maximum Magnetic Field (B₀) The relationship between the electric field and magnetic field in an electromagnetic wave is given by: \[ B_0 = \frac{E_0}{c} \] Substituting the values: \[ B_0 = \frac{12.27}{3 \times 10^8} \approx 4.09 \times 10^{-8} \, \text{T} \] ### Final Results - Maximum Electric Field (E₀) = 12.27 N/C - Maximum Magnetic Field (B₀) = \(4.09 \times 10^{-8}\) T

To find the magnitude of the electric and magnetic fields in an electromagnetic wave radiated by a 200 W bulb at a distance of 2 m, we can follow these steps: ### Step 1: Calculate the Effective Power The bulb has an efficiency of 5%. Therefore, the effective power (P) radiated by the bulb can be calculated as: \[ P = \text{Efficiency} \times \text{Total Power} = \frac{5}{100} \times 200 \, \text{W} = 10 \, \text{W} \] ...
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