The intensity of light from a source is $\left( {\frac{{500}}{\pi }} \right)W/{m^2}$ . Find the amplitude of electric field in this wave
The intensity of light from a source is $\left( {\frac{{500}}{\pi }} \right)W/{m^2}$ . Find the amplitude of electric field in this wave
- A
$\sqrt 3 \times {10^{2\,}}\,N/C$
- B
$2\sqrt 3 \times {10^{2\,}}\,N/C$
- C
$\frac{{\sqrt 3 }}{2} \times {10^{2\,}}\,N/C$
- D
$2\sqrt 3 \times {10^{1\,}}\,N/C$
Similar Questions
The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{ B }=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ j }$, then the associated electric field will be :
The magnetic field of a plane electromagnetic wave is given by $\overrightarrow{ B }=3 \times 10^{-8} \cos \left(1.6 \times 10^3 x +48 \times 10^{10} t \right) \hat{ j }$, then the associated electric field will be :
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Intensity of sunlight is observed as $0.092\, {Wm}^{-2}$ at a point in free space. What will be the peak value of magnetic field at that point? $\left(\sigma_{0}=8.85 \times 10^{-12}\, {C}^{2} \,{N}^{-1} \,{m}^{-2}\right.$ )
Intensity of sunlight is observed as $0.092\, {Wm}^{-2}$ at a point in free space. What will be the peak value of magnetic field at that point? $\left(\sigma_{0}=8.85 \times 10^{-12}\, {C}^{2} \,{N}^{-1} \,{m}^{-2}\right.$ )
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An $LC$ resonant circuit contains a $400 pF$ capacitor and a $100\mu H$ inductor. It is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is
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A particle of mass $M$ and positive charge $Q$, moving with a constant velocity $\overrightarrow{ u }_1=4 \hat{ i } ms ^{-1}$, enters a region of uniform static magnetic field normal to the $x-y$ plane. The region of the magnetic field extends from $x=0$ to $x$ $=L$ for all values of $y$. After passing through this region, the particle emerges on the other side after $10$ milliseconds with a velocity $\overline{ u }_2=2(\sqrt{3} \hat{ i }+\hat{ j }) ms ^{-1}$. The correct statement$(s)$ is (are) :
$(A)$ The direction of the magnetic field is $-z$ direction.
$(B)$ The direction of the magnetic field is $+z$ direction
$(C)$ The magnitude of the magnetic field $\frac{50 \pi M }{3 Q }$ units.
$(D)$ The magnitude of the magnetic field is $\frac{100 \pi M}{3 Q}$ units.
A particle of mass $M$ and positive charge $Q$, moving with a constant velocity $\overrightarrow{ u }_1=4 \hat{ i } ms ^{-1}$, enters a region of uniform static magnetic field normal to the $x-y$ plane. The region of the magnetic field extends from $x=0$ to $x$ $=L$ for all values of $y$. After passing through this region, the particle emerges on the other side after $10$ milliseconds with a velocity $\overline{ u }_2=2(\sqrt{3} \hat{ i }+\hat{ j }) ms ^{-1}$. The correct statement$(s)$ is (are) :
$(A)$ The direction of the magnetic field is $-z$ direction.
$(B)$ The direction of the magnetic field is $+z$ direction
$(C)$ The magnitude of the magnetic field $\frac{50 \pi M }{3 Q }$ units.
$(D)$ The magnitude of the magnetic field is $\frac{100 \pi M}{3 Q}$ units.
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Electric field in a plane electromagnetic wave is given by ${E}=50 \sin \left(500 {x}-10 \times 10^{10} {t}\right) \,{V} / {m}$ The velocity of electromagnetic wave in this medium is :
(Given ${C}=$ speed of light in vacuum)
Electric field in a plane electromagnetic wave is given by ${E}=50 \sin \left(500 {x}-10 \times 10^{10} {t}\right) \,{V} / {m}$ The velocity of electromagnetic wave in this medium is :
(Given ${C}=$ speed of light in vacuum)
- [JEE MAIN 2021]