Question

A 1000 $W$ electric bulb emits radiation uniformly in all directions with the efficiency of 1.25%. A point $P$ is situated $2m$ away from it. Take $\epsilon_0 = 8.85 \times 10^{-12} C^2N^{-1}m^{-2}$, $c = 3 \times 10^8 m/s$

The value of peak electric field at $P$ due to the radiation is $x \times 10^{-1} V/m$. Value of $x$ is: (Rounded off to nearest integer)

• A. 137
• B. 136
• C. 138
• D. 135

Answer 1

Answer: (A)

137

Answer 2

The electric bulb consumes a power of $P_{bulb} = 1000 \, W$. The efficiency of the bulb in emitting radiation is $\eta = 1.25\% = 0.0125$. The power radiated as electromagnetic waves is $P_{rad} = P_{bulb} \times \eta = 1000 \, W \times 0.0125 = 12.5 \, W$.

Since the radiation is emitted uniformly in all directions, it is an isotropic source. The intensity ($I$) of the radiation at a distance $r$ from the source is given by the power per unit area of a sphere of radius $r$: $I = \frac{P_{rad}}{4\pi r^2}$

At point $P$, the distance from the bulb is $r = 2 \, m$. So, the intensity at point $P$ is: $I = \frac{12.5 \, W}{4\pi (2 \, m)^2} = \frac{12.5}{16\pi} \, W/m^2$

The intensity of an electromagnetic wave is related to the peak electric field ($E_0$) by the formula: $I = \frac{1}{2} c \epsilon_0 E_0^2$ where $c$ is the speed of light and $\epsilon_0$ is the permittivity of free space.

We can rearrange this formula to solve for $E_0$: $E_0^2 = \frac{2I}{c \epsilon_0}$ $E_0 = \sqrt{\frac{2I}{c \epsilon_0}}$

Substitute the values: $I = \frac{12.5}{16\pi} \, W/m^2$ $c = 3 \times 10^8 \, m/s$ $\epsilon_0 = 8.85 \times 10^{-12} \, C^2N^{-1}m^{-2}$

$E_0 = \sqrt{\frac{2 \times \frac{12.5}{16\pi}}{(3 \times 10^8) \times (8.85 \times 10^{-12})}}$ $E_0 = \sqrt{\frac{\frac{25}{16\pi}}{26.55 \times 10^{-4}}}$ $E_0 = \sqrt{\frac{25}{16\pi \times 26.55 \times 10^{-4}}}$

Using $\pi \approx 3.14159$: $16\pi \approx 50.265$ $E_0 = \sqrt{\frac{25}{50.265 \times 26.55 \times 10^{-4}}}$ $E_0 = \sqrt{\frac{25}{1334.55 \times 10^{-4}}}$ $E_0 = \sqrt{\frac{25 \times 10^4}{1334.55}}$ $E_0 = \sqrt{\frac{250000}{1334.55}} \approx \sqrt{187.325}$ $E_0 \approx 13.6866 \, V/m$

The problem states that the peak electric field is $x \times 10^{-1} \, V/m$. So, $E_0 = x \times 10^{-1} \, V/m$. $13.6866 = x \times 10^{-1}$ $x = \frac{13.6866}{10^{-1}} = 13.6866 \times 10 = 136.866$

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