JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2026
- Easy
- MCQ
A laser beam has intensity of \( 4.0\times10^{14} W/m^{2} \). The amplitude of magnetic field associated with beam is__________ T. (Take \( \epsilon_{0}=8.85\times10^{-12} C^{2}/Nm^{2} \) and \( c=3\times10^{8} m/s \))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) 1.83
Step-by-step Solution
Detailed explanation
\(I =\frac{1}{2} \epsilon_0 E _0^2 \cdot C\)
\(\therefore E _0=\sqrt{\frac{2 I }{\epsilon_0 C }}\)
\(\& \frac{ E _0}{B_0}= C\)
\(\therefore B_0=\frac{E_0}{C}=\frac{1}{C} \sqrt{\frac{2 I}{\epsilon_0 C}}\)
\(\therefore B _0=\frac{1}{3 \times 10^8} \sqrt{\frac{2 \times 4 \times 10^{14}}{8.85 \times 10^{-12} \times 3 \times 10^8}}\)
\(B _0=\frac{10}{3} \sqrt{\frac{8}{8.85 \times 3}}\)
\(B _0=1.83 T\)
\(\therefore E _0=\sqrt{\frac{2 I }{\epsilon_0 C }}\)
\(\& \frac{ E _0}{B_0}= C\)
\(\therefore B_0=\frac{E_0}{C}=\frac{1}{C} \sqrt{\frac{2 I}{\epsilon_0 C}}\)
\(\therefore B _0=\frac{1}{3 \times 10^8} \sqrt{\frac{2 \times 4 \times 10^{14}}{8.85 \times 10^{-12} \times 3 \times 10^8}}\)
\(B _0=\frac{10}{3} \sqrt{\frac{8}{8.85 \times 3}}\)
\(B _0=1.83 T\)
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