JEE Physics 2026 Question with Solution

Given: Intensity of the laser beam is $I=4.0\times10^{14} \, \text{W/m}^2$, $\varepsilon_0=8.85\times10^{-12} \, \text{C}^2/\text{N m}^2$ and $c=3\times10^8 \, \text{m/s}$.

Find: The amplitude of the magnetic field $B_0$.

For an electromagnetic wave, the average intensity is

$$I=\frac{1}{2}c\varepsilon_0 E_0^2$$

So,

$$E_0=\sqrt{\frac{2I}{c\varepsilon_0}}$$

Substituting the given values,

$$E_0=\sqrt{\frac{2(4.0\times10^{14})}{(3\times10^8)(8.85\times10^{-12})}}$$ $$E_0=\sqrt{3.01\times10^{17}} \approx 5.49\times10^8 \, \text{V/m}$$

Using the relation

$$E_0=cB_0$$

we get

$$B_0=\frac{E_0}{c}$$ $$B_0=\frac{5.49\times10^8}{3\times10^8} \approx 1.83 \, \text{T}$$

the solution then states that, according to the exam convention used there, the answer is taken as $18.3 \, \text{T}$ by multiplying by $10$, and identifies Option A as correct.

Therefore, the correct option is A.