JEE Physics 2026 Question with Solution

Given: The electric field is

$$E = \sqrt{377} \sin(6.27 \times 10^3 t - 2.09 \times 10^{-5} x)$$

So the peak electric field is $E_0 = \sqrt{377} \text{ N/C}$. Also, the impedance of free space is $\eta = \sqrt{\mu_0/\varepsilon_0} = 377\,\Omega$.

Find: The value of $a$ if the average power per unit area is $1/a$.

For an electromagnetic wave, the average intensity is

$$I = \frac{E_0^2}{2\eta}$$

Substituting the given values,

$$I = \frac{(\sqrt{377})^2}{2 \times 377}$$ $$I = \frac{377}{754} = \frac{1}{2} \text{ W/m}^2$$

Comparing with $1/a$,

$$\frac{1}{a} = \frac{1}{2}$$ $$a = 2$$

Therefore, the value of $a$ is $2$.

Given: For a sinusoidal electromagnetic wave, the time-averaged Poynting vector gives the intensity. The hint states that the average value is half the peak value, i.e. proportional to $\frac{1}{2} E_0 H_0$.

Find: The numerical value of $a$.

From the wave equation, the amplitude is

$$E_0 = \sqrt{377}$$

Using the free-space impedance relation,

$$\eta = \frac{E_0}{H_0} = 377$$

Hence,

$$I = \frac{1}{2} E_0 H_0 = \frac{1}{2} E_0 \left(\frac{E_0}{\eta}\right)$$ $$I = \frac{E_0^2}{2\eta}$$

Now substitute:

$$I = \frac{377}{2 \times 377} = \frac{1}{2} \text{ W/m}^2$$

So,

$$\frac{1}{a} = \frac{1}{2}$$

Therefore, the value of $a$ is $2$.