JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2021
- Hard
- Numerical
A radiation is emitted by \(1000\, W\) bulb and it generates an electric field and magnetic field at \(P\), placed at a distance of \(2\, m\). The efficiency of the bulb is \(1.25 \%\). The value of peak electric field at \(P\) is \(x \times 10^{-1} \,V / m\). Value of \(x\) is. (Rounded-off to the nearest integer) [Take \(\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}, c =3 \times 10^{8}\) \(ms ^{-1}\) ]
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(137\)
Step-by-step Solution
Detailed explanation
\(I _{ avg }=\frac{1}{2} \varepsilon_{0} E _{0}^{2} C\)
\(\frac{1.25}{100} \times \frac{1000}{4 \pi(2)^{2}}=\frac{1}{2} \times 8.85 \times 10^{-12} \times 3\times 10^{8} \times E _{0}^{2}\)
\(E _{0}^{2}=187.4\)
\(\therefore E _{0}=13.689 V / m\)
\(=136.89 \times 10^{-1} V / m\)
\(\therefore x =136.89\)
Rounding off to nearest integer \(x =137\)
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