A laser beam has intensity of $$4.0 \times 10^{14} W/m^{2}.$$ The amplitude of magnetic field associated with beam is____________T. (Take $$\epsilon_{0}= 8.85 \times 10^{-12} C^{2}/Nm^{2}$$ and $$c= 3 \times 10^{8} m/s$$ )

We are given a laser intensity $$I = 4.0 \times 10^{14}$$ W/m$$^2$$, permittivity of free space $$\epsilon_0 = 8.85 \times 10^{-12}$$ C$$^2$$/(N m$$^2$$), and the speed of light $$c = 3 \times 10^8$$ m/s. The intensity relates to the electric field amplitude by $$I = \frac{1}{2}\epsilon_0 c E_0^2$$ and the electric and magnetic field amplitudes satisfy $$E_0 = cB_0$$.

Solving the intensity equation for $$E_0$$ gives $$E_0^2 = \frac{2I}{\epsilon_0 c} = \frac{2 \times 4.0 \times 10^{14}}{8.85 \times 10^{-12} \times 3 \times 10^8}$$ which simplifies to $$= \frac{8.0 \times 10^{14}}{26.55 \times 10^{-4}} = \frac{8.0 \times 10^{14}}{2.655 \times 10^{-3}} = 3.013 \times 10^{17}$$ and hence $$E_0 = \sqrt{3.013 \times 10^{17}} = 5.49 \times 10^{8} \text{ V/m}$$.

Using the relation $$B_0 = \frac{E_0}{c}$$ we find $$B_0 = \frac{5.49 \times 10^8}{3 \times 10^8} = 1.83 \text{ T}$$. The correct answer is Option B: 1.83.