A radiation is emitted by 1000 W bulb and it generates an electric field and magnetic field at P, pl

Q: Why do we divide the power by $4\pi d^2$ to find the intensity?

A point source like a small bulb emits radiation uniformly in all directions in 3D space. The energy spreads out over the surface of an expanding sphere. The surface area of a sphere is $4\pi d^2$, so the intensity (power per unit area) is the total power divided by this area.

Q: What is the difference between the peak electric field ($E_0$) and the rms electric field ($E_{rms}$)?

The peak electric field $E_0$ is the maximum amplitude of the oscillating wave. The rms (root mean square) electric field is an effective value used for power calculations, given by $E_{rms} = \frac{E_0}{\sqrt{2}}$. The intensity formula $I = \frac{1}{2}\varepsilon_0 E_0^2 c$ uses the peak field. If you use the rms field, the formula becomes $I = \varepsilon_0 E_{rms}^2 c$.

Q: Why do we multiply the power by the efficiency?

A $1000\text{ W}$ bulb consumes $1000\text{ Joules}$ of electrical energy per second. However, it is not $100\%$ efficient at converting this into light; much of it is lost as heat. The efficiency tells us the fraction of power that actually becomes the electromagnetic radiation we are measuring.

Q: Does the magnetic field also contribute to the intensity?

Yes, absolutely! In an electromagnetic wave, the electric and magnetic fields carry equal amounts of energy. The formula $I = \frac{1}{2}\varepsilon_0 E_0^2 c$ actually represents the *total* intensity (the sum of the electric and magnetic contributions). We just express it entirely in terms of the electric field for convenience.

Q: How would I find the peak magnetic field ($B_0$) if the question asked for it?

Once you have the peak electric field $E_0$, finding the magnetic field is very easy. In a vacuum or air, the amplitudes of the electric and magnetic fields are related by the speed of light: $E_0 = c B_0$. So, you would simply calculate $B_0 = \frac{E_0}{c}$.

Question

A radiation is emitted by $1000	ext{ W}$ bulb and it generates an electric field and magnetic field at $P$, placed at a distance of $2	ext{ m}$. The efficiency of the bulb is $1.25\%$. The value of peak electric field at $P$ is $x 	imes 10^{-1}	ext{ V/m}$. Value of $x$ is ......... .
(Rounded-off to the nearest integer)
[Take, $\varepsilon_0 = 8.85 	imes 10^{-12}	ext{ C}^2	ext{N}^{-1}	ext{m}^{-2}$, $c = 3 	imes 10^8	ext{ ms}^{-1}$]

Sigma Percentile · Accepted Answer

### Illuminating the Physics of Electromagnetic Radiation

Imagine a light bulb glowing in a dark room. It seems simple, but there is a profound physical process happening. The bulb is emitting electromagnetic waves—ripples of electric and magnetic fields traveling at the speed of light. In this problem, we are tasked with finding the strength of that electric field at a specific distance. Let's break down the journey from electrical power to the peak electric field.

### The Concept of Effective Power

We are given a bulb with a power rating of $1000	ext{ W}$. However, not all of this power is converted into light (electromagnetic radiation). A significant portion is lost as heat. The problem states that the efficiency $\eta$ is only $1.25\%$.

Therefore, the **effective power** that actually radiates outwards as electromagnetic waves is:

$$P_{	ext{eff}} = \eta 	imes P = \frac{1.25}{100} 	imes 1000 = 12.5	ext{ W}$$

### Spherical Wavefronts and Intensity

Since the bulb is a point source, it emits radiation uniformly in all directions. Imagine a sphere expanding outwards from the bulb. As the sphere grows, the same amount of energy is spread over a larger surface area.

The **Intensity ($I$)** of the radiation at a distance $d$ is the effective power divided by the surface area of a sphere of radius $d$:

$$I = \frac{P_{	ext{eff}}}{A} = \frac{P_{	ext{eff}}}{4\pi d^2}$$

Substituting $d = 2	ext{ m}$, we get the intensity at point P.

### The Master Equation: Intensity and Electric Field

Now, how do we connect this macroscopic intensity to the microscopic electric field? The intensity of an electromagnetic wave is directly related to the square of its peak electric field ($E_0$). The formula is:

$$I = \frac{1}{2} \varepsilon_0 E_0^2 c$$

Here, $\varepsilon_0$ is the permittivity of free space, and $c$ is the speed of light. This equation beautifully bridges the gap between the energy carried by the wave and the amplitude of its oscillating fields.

### The Final Calculation

By equating our two expressions for intensity, we can solve for $E_0$:

$$\frac{12.5}{4\pi (2)^2} = \frac{1}{2} \varepsilon_0 E_0^2 c$$

Rearranging to isolate $E_0^2$:

$$E_0^2 = \frac{12.5 	imes 2}{16\pi 	imes 8.854 	imes 10^{-12} 	imes 3 	imes 10^8}$$

After carefully crunching the numbers, we find $E_0^2 \approx 187.3$. Taking the square root gives us the peak electric field:

$$E_0 \approx 13.68	ext{ V/m}$$

The question asks for the answer in the format $x 	imes 10^{-1}	ext{ V/m}$. We can rewrite our result as:

$$13.68 = 136.8 	imes 10^{-1}$$

Rounding off to the nearest integer, we get **$x = 137$**.

This problem is a fantastic exercise in connecting power, geometry, and the fundamental properties of electromagnetic waves. Always remember to account for efficiency and the spherical spreading of energy from a point source!

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Effective Power and Intensity

Intensity and Electric Field Relation

Equating and Substituting

Calculating Peak Electric Field

Final Answer Formatting

The Way Forward

The Sigma Insight: Characteristics of Electromagnetic Waves

Illuminating the Physics of Electromagnetic Radiation

The Concept of Effective Power

Spherical Wavefronts and Intensity

The Master Equation

The Final Calculation

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