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

The bulb has a power of 1000 W and an efficiency of 1.25%, so the radiated power is $$P = 1000 \times 0.0125 = 12.5$$ W.

The radiation spreads uniformly in all directions. At a distance $$r = 2$$ m, the intensity is:

$$I = \frac{P}{4\pi r^2} = \frac{12.5}{4\pi \times 4} = \frac{12.5}{16\pi}$$ W m$$^{-2}$$

The intensity of an electromagnetic wave is related to the peak electric field by $$I = \frac{1}{2}\varepsilon_0 c E_0^2$$. Solving for $$E_0$$:

$$E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} = \sqrt{\frac{2 \times \frac{12.5}{16\pi}}{8.85 \times 10^{-12} \times 3 \times 10^8}}$$

Computing the denominator: $$\varepsilon_0 c = 8.85 \times 10^{-12} \times 3 \times 10^8 = 2.655 \times 10^{-3}$$.

Computing the numerator: $$2 \times \frac{12.5}{16\pi} = \frac{25}{16\pi} = \frac{25}{50.265} = 0.4974$$.

$$E_0 = \sqrt{\frac{0.4974}{2.655 \times 10^{-3}}} = \sqrt{187.3} = 13.69$$ V m$$^{-1}$$

Expressing as $$x \times 10^{-1}$$ V m$$^{-1}$$: $$13.69 = 136.9 \times 10^{-1}$$, so $$x \approx \mathbf{137}$$.