JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2026
- Hard
- Numerical
An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, \(\sigma=10 mho / m\). The ratio of maximum conducting current density to maximum displacement current density is _________ .
\(\left[\right.\) Take \(\left.\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 Nm ^2 / C ^2\right]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 1800
Step-by-step Solution
Detailed explanation
\(j _{ c }=\sigma E\)
\(E \Rightarrow E _0 \sin (\omega t - kx )\)
\(j _{ c }=\sigma E _0 \sin (\omega t - kx )\)
\(\Rightarrow\left( j _{ c }\right)_{\max }=\sigma E _0\)....(i)
\(J_d \Rightarrow \frac{i_d}{A}=\frac{1}{A} \times \varepsilon_0 \frac{A d E}{d t}\)
\(\Rightarrow \varepsilon_0 \times E _0 \omega \cos (\omega t- kx )\)
\(\left( j _{ d }\right)_{\max } \Rightarrow \varepsilon_0 E _0 \omega\)....(ii)
(i)/(ii)
\(\frac{\left( j _{ c }\right)_{\max }}{\left( j _{ d }\right)_{\max }}=\frac{\sigma E _0}{\varepsilon_0 \omega E _0} \Rightarrow \frac{\sigma}{\varepsilon_0 \omega}\)
\(\Rightarrow \frac{10 \times 4 \pi \times 9 \times 10^9}{2 \pi \times 100 \times 10^6}\)
\(\Rightarrow 1800\)
\(E \Rightarrow E _0 \sin (\omega t - kx )\)
\(j _{ c }=\sigma E _0 \sin (\omega t - kx )\)
\(\Rightarrow\left( j _{ c }\right)_{\max }=\sigma E _0\)....(i)
\(J_d \Rightarrow \frac{i_d}{A}=\frac{1}{A} \times \varepsilon_0 \frac{A d E}{d t}\)
\(\Rightarrow \varepsilon_0 \times E _0 \omega \cos (\omega t- kx )\)
\(\left( j _{ d }\right)_{\max } \Rightarrow \varepsilon_0 E _0 \omega\)....(ii)
(i)/(ii)
\(\frac{\left( j _{ c }\right)_{\max }}{\left( j _{ d }\right)_{\max }}=\frac{\sigma E _0}{\varepsilon_0 \omega E _0} \Rightarrow \frac{\sigma}{\varepsilon_0 \omega}\)
\(\Rightarrow \frac{10 \times 4 \pi \times 9 \times 10^9}{2 \pi \times 100 \times 10^6}\)
\(\Rightarrow 1800\)
More questions from Physics
- The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea \((7.8\, mm)\). This surface separates two media of refractive indices \(1\) and \(1.34\). Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.......\(cm\)JEE Mains 2019 Medium
- If \(\in\), \( E \) and \( t \) represent the free space permittivity, electric field and time respectively, then the unit of \( \frac{eE}{t} \) will be :JEE Mains 2026 Medium
- An element \(\Delta l=\Delta \mathrm{xi}\) is placed at the origin and carries a large current \(\mathrm{I}=10 \mathrm{~A}\). The magnetic field on the \(y\)-axis at a distance of \(0.5 \mathrm{~m}\) from the elements \(\Delta \mathrm{x}\) of \(1 \mathrm{~cm}\) length is _______.
JEE Mains 2024 Hard - A particle starts from origin at \(t=0\) with a velocity \(5 \hat{i} \mathrm{~m} / \mathrm{s}\) and moves in \(x-y\) plane under action of a force which produces a constant acceleration of \((3 \hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}^2\). If the \(x\)-coordinate of the particle at that instant is \(84 \mathrm{~m}\), then the speed of the particle at this time is \(\sqrt{\alpha} \mathrm{m} / \mathrm{s}\). The value of \(\alpha\) is___________.JEE Mains 2024 Hard
- A series \(LCR\) circuit is designed to resonate at an angular frequency \(\omega_{0}=10^{5} \,rad / s\). The circuit draws \(16\, W\) power from \(120\, V\) source at resonance. The value of resistance \('R'\) in the circuit is ...... \(\Omega\).JEE Mains 2021 Medium
- A capacitor is connected to a \(20\, {V}\) battery through a resistance of \(10\, \Omega .\) It is found that the potential difference across the capacitor rises to \(2\, {V}\) in \(1\, \mu {s}\). The capacitance of the capacitor is \(....\,\mu {F}\) Given : \(\ln \left(\frac{10}{9}\right)=0.105\)JEE Mains 2021 Hard