Q69 / 75 · JEE Main 2026 (22 January Shift 2) · Physics
A laser beam has intensity of $4.0 \times 10^{14} \mathrm{~W} / \mathrm{m}^{2}$. The amplitude of magnetic field associated with beam is $\_\_\_\_$ T.
(Take $\epsilon_{\mathrm{o}}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$)
(Take $\epsilon_{\mathrm{o}}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$)
- A18.3
- Correct answerB1.83
- C5.5
- D2.0
Solution
The intensity of an electromagnetic wave is
$I = \frac{1}{2} \epsilon_0 c E_0^2$.
From the relation $E_0 = c B_0$, we get $I = \frac{1}{2} \epsilon_0 c^3 B_0^2$.
Solving for magnetic field amplitude:
$B_0 = \sqrt{\frac{2I}{\epsilon_0 c^3}} = \sqrt{\frac{2 \times 4.0 \times 10^{14}}{8.85 \times 10^{-12} \times (3 \times 10^8)^3}} = \sqrt{\frac{8.0 \times 10^{14}}{2.39 \times 10^{14}}}$
$= \sqrt{3.347} = 1.83$ T.
$I = \frac{1}{2} \epsilon_0 c E_0^2$.
From the relation $E_0 = c B_0$, we get $I = \frac{1}{2} \epsilon_0 c^3 B_0^2$.
Solving for magnetic field amplitude:
$B_0 = \sqrt{\frac{2I}{\epsilon_0 c^3}} = \sqrt{\frac{2 \times 4.0 \times 10^{14}}{8.85 \times 10^{-12} \times (3 \times 10^8)^3}} = \sqrt{\frac{8.0 \times 10^{14}}{2.39 \times 10^{14}}}$
$= \sqrt{3.347} = 1.83$ T.
Hence, the correct answer is option: 'B'.