Light wave travelling in air along $$x$$-direction is given by $$E_y = 540 \sin \pi \times 10^4(x - ct) \text{ V m}^{-1}$$. Then, the peak value of magnetic field of wave will be (Given $$c = 3 \times 10^8 \text{ m s}^{-1}$$)

The electric field of a light wave is given by $$E_y = 540 \sin \pi \times 10^4(x - ct) \text{ V m}^{-1}$$.

Identify the peak value of the electric field.

From the equation, the amplitude (peak value) of the electric field is:

$$E_0 = 540 \text{ V m}^{-1}$$

Find the peak value of the magnetic field.

For an electromagnetic wave, the relationship between the electric and magnetic field amplitudes is:

$$B_0 = \frac{E_0}{c}$$

$$B_0 = \frac{540}{3 \times 10^8} = 180 \times 10^{-8} = 18 \times 10^{-7} \text{ T}$$

The correct answer is Option A: $$18 \times 10^{-7} \text{ T}$$.