358929
If \(\vec{E}\) and \(\vec{K}\) represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by (\(\omega\) - angular frequency)
1 \(\dfrac{1}{\omega}(\vec{K} \times \vec{E})\)
2 \(\vec{K} \times \vec{E}\)
3 \(\omega(\vec{E} \times \vec{K})\)
4 \(\omega(\vec{K} \times \vec{E})\)
Explanation:
Here, the magnetic field vector will be along \(\hat{K} \times \vec{E}\), Where \(\hat{K}\) is the unit vector along the direction of propagation. Also, \(|\vec{B}|=\dfrac{|\vec{E}|}{c} \Rightarrow \vec{B}=\left(\dfrac{E}{c}\right) \dfrac{\hat{K} \times \vec{E}}{E}\) \(=\dfrac{\hat{K} \times \vec{E}}{c}=\dfrac{\hat{K} \times \vec{E}}{\omega / K} \quad\left[\because c=\dfrac{\omega}{K}\right]\) \(=\dfrac{K \hat{K} \times \vec{E}}{\omega}=\dfrac{\vec{K} \times \vec{E}}{\omega} \quad[\because K \hat{K}=\vec{K}]\) So option (1) is correct.
JEE - 2023
PHXI15:WAVES
358930
Assertion : When a charged particle moves in a circular path, it produces electromagnetic wave. Reason : Charged particle moving in circular path has acceleration.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
) In circular path, direction of velocity of charge is charging as a function of time and so it has acceleration. Accelerated charges emit electromagnetic waves. So correct option is (1).
PHXI15:WAVES
358931
The electric field part of an electromagnetic wave in vacuum is \(E=3.1 N C^{-1} \cos [(1.8 \mathrm{rad} /\) \(\left.\left.m^{-1}\right) y+\left(5.4 \times 10^{8} \mathrm{rad} / \mathrm{s}^{-1}\right) t\right] i\). The wavelength of this part of electromagnetic wave is:
1 \(1.5\;m\)
2 \(2\;m\)
3 \(2.5\;m\)
4 \(3.5\;m\)
Explanation:
\(E=3.1 \cos \left[\left(1.8 y+5.4 \times 10^{8}\right] i N / C\right.\) Standard equation: \(\begin{aligned}& E=E_{0} \cos [k x-\omega t] N / C \\& \therefore k=\dfrac{2 \pi}{\lambda}=1.8 \\& \lambda=\dfrac{2 \pi}{1.8}=3.49 \simeq 3.5 .\end{aligned}\)
PHXI15:WAVES
358932
If \({v_\gamma },{v_X}\) and \({v_M}\) are the speeds of \(\gamma \)-rays, \(X\)-rays and microwaves respectively in vacuum, then
1 \({v_\gamma } > {v_M} > {v_X}\)
2 \({v_\gamma } > {v_X} > {v_M}\)
3 \({v_\gamma } = {v_X} = {v_M}\)
4 None of these
Explanation:
\(\gamma \)-rays, \(X\)-rays and microwaves respectively are electromagnetic waves. They travel with the speed of light in vacuum. Hence, \(\,\,\,\,\,{v_\gamma } = {v_X} = {v_M}\)
358929
If \(\vec{E}\) and \(\vec{K}\) represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by (\(\omega\) - angular frequency)
1 \(\dfrac{1}{\omega}(\vec{K} \times \vec{E})\)
2 \(\vec{K} \times \vec{E}\)
3 \(\omega(\vec{E} \times \vec{K})\)
4 \(\omega(\vec{K} \times \vec{E})\)
Explanation:
Here, the magnetic field vector will be along \(\hat{K} \times \vec{E}\), Where \(\hat{K}\) is the unit vector along the direction of propagation. Also, \(|\vec{B}|=\dfrac{|\vec{E}|}{c} \Rightarrow \vec{B}=\left(\dfrac{E}{c}\right) \dfrac{\hat{K} \times \vec{E}}{E}\) \(=\dfrac{\hat{K} \times \vec{E}}{c}=\dfrac{\hat{K} \times \vec{E}}{\omega / K} \quad\left[\because c=\dfrac{\omega}{K}\right]\) \(=\dfrac{K \hat{K} \times \vec{E}}{\omega}=\dfrac{\vec{K} \times \vec{E}}{\omega} \quad[\because K \hat{K}=\vec{K}]\) So option (1) is correct.
JEE - 2023
PHXI15:WAVES
358930
Assertion : When a charged particle moves in a circular path, it produces electromagnetic wave. Reason : Charged particle moving in circular path has acceleration.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
) In circular path, direction of velocity of charge is charging as a function of time and so it has acceleration. Accelerated charges emit electromagnetic waves. So correct option is (1).
PHXI15:WAVES
358931
The electric field part of an electromagnetic wave in vacuum is \(E=3.1 N C^{-1} \cos [(1.8 \mathrm{rad} /\) \(\left.\left.m^{-1}\right) y+\left(5.4 \times 10^{8} \mathrm{rad} / \mathrm{s}^{-1}\right) t\right] i\). The wavelength of this part of electromagnetic wave is:
1 \(1.5\;m\)
2 \(2\;m\)
3 \(2.5\;m\)
4 \(3.5\;m\)
Explanation:
\(E=3.1 \cos \left[\left(1.8 y+5.4 \times 10^{8}\right] i N / C\right.\) Standard equation: \(\begin{aligned}& E=E_{0} \cos [k x-\omega t] N / C \\& \therefore k=\dfrac{2 \pi}{\lambda}=1.8 \\& \lambda=\dfrac{2 \pi}{1.8}=3.49 \simeq 3.5 .\end{aligned}\)
PHXI15:WAVES
358932
If \({v_\gamma },{v_X}\) and \({v_M}\) are the speeds of \(\gamma \)-rays, \(X\)-rays and microwaves respectively in vacuum, then
1 \({v_\gamma } > {v_M} > {v_X}\)
2 \({v_\gamma } > {v_X} > {v_M}\)
3 \({v_\gamma } = {v_X} = {v_M}\)
4 None of these
Explanation:
\(\gamma \)-rays, \(X\)-rays and microwaves respectively are electromagnetic waves. They travel with the speed of light in vacuum. Hence, \(\,\,\,\,\,{v_\gamma } = {v_X} = {v_M}\)
358929
If \(\vec{E}\) and \(\vec{K}\) represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by (\(\omega\) - angular frequency)
1 \(\dfrac{1}{\omega}(\vec{K} \times \vec{E})\)
2 \(\vec{K} \times \vec{E}\)
3 \(\omega(\vec{E} \times \vec{K})\)
4 \(\omega(\vec{K} \times \vec{E})\)
Explanation:
Here, the magnetic field vector will be along \(\hat{K} \times \vec{E}\), Where \(\hat{K}\) is the unit vector along the direction of propagation. Also, \(|\vec{B}|=\dfrac{|\vec{E}|}{c} \Rightarrow \vec{B}=\left(\dfrac{E}{c}\right) \dfrac{\hat{K} \times \vec{E}}{E}\) \(=\dfrac{\hat{K} \times \vec{E}}{c}=\dfrac{\hat{K} \times \vec{E}}{\omega / K} \quad\left[\because c=\dfrac{\omega}{K}\right]\) \(=\dfrac{K \hat{K} \times \vec{E}}{\omega}=\dfrac{\vec{K} \times \vec{E}}{\omega} \quad[\because K \hat{K}=\vec{K}]\) So option (1) is correct.
JEE - 2023
PHXI15:WAVES
358930
Assertion : When a charged particle moves in a circular path, it produces electromagnetic wave. Reason : Charged particle moving in circular path has acceleration.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
) In circular path, direction of velocity of charge is charging as a function of time and so it has acceleration. Accelerated charges emit electromagnetic waves. So correct option is (1).
PHXI15:WAVES
358931
The electric field part of an electromagnetic wave in vacuum is \(E=3.1 N C^{-1} \cos [(1.8 \mathrm{rad} /\) \(\left.\left.m^{-1}\right) y+\left(5.4 \times 10^{8} \mathrm{rad} / \mathrm{s}^{-1}\right) t\right] i\). The wavelength of this part of electromagnetic wave is:
1 \(1.5\;m\)
2 \(2\;m\)
3 \(2.5\;m\)
4 \(3.5\;m\)
Explanation:
\(E=3.1 \cos \left[\left(1.8 y+5.4 \times 10^{8}\right] i N / C\right.\) Standard equation: \(\begin{aligned}& E=E_{0} \cos [k x-\omega t] N / C \\& \therefore k=\dfrac{2 \pi}{\lambda}=1.8 \\& \lambda=\dfrac{2 \pi}{1.8}=3.49 \simeq 3.5 .\end{aligned}\)
PHXI15:WAVES
358932
If \({v_\gamma },{v_X}\) and \({v_M}\) are the speeds of \(\gamma \)-rays, \(X\)-rays and microwaves respectively in vacuum, then
1 \({v_\gamma } > {v_M} > {v_X}\)
2 \({v_\gamma } > {v_X} > {v_M}\)
3 \({v_\gamma } = {v_X} = {v_M}\)
4 None of these
Explanation:
\(\gamma \)-rays, \(X\)-rays and microwaves respectively are electromagnetic waves. They travel with the speed of light in vacuum. Hence, \(\,\,\,\,\,{v_\gamma } = {v_X} = {v_M}\)
358929
If \(\vec{E}\) and \(\vec{K}\) represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by (\(\omega\) - angular frequency)
1 \(\dfrac{1}{\omega}(\vec{K} \times \vec{E})\)
2 \(\vec{K} \times \vec{E}\)
3 \(\omega(\vec{E} \times \vec{K})\)
4 \(\omega(\vec{K} \times \vec{E})\)
Explanation:
Here, the magnetic field vector will be along \(\hat{K} \times \vec{E}\), Where \(\hat{K}\) is the unit vector along the direction of propagation. Also, \(|\vec{B}|=\dfrac{|\vec{E}|}{c} \Rightarrow \vec{B}=\left(\dfrac{E}{c}\right) \dfrac{\hat{K} \times \vec{E}}{E}\) \(=\dfrac{\hat{K} \times \vec{E}}{c}=\dfrac{\hat{K} \times \vec{E}}{\omega / K} \quad\left[\because c=\dfrac{\omega}{K}\right]\) \(=\dfrac{K \hat{K} \times \vec{E}}{\omega}=\dfrac{\vec{K} \times \vec{E}}{\omega} \quad[\because K \hat{K}=\vec{K}]\) So option (1) is correct.
JEE - 2023
PHXI15:WAVES
358930
Assertion : When a charged particle moves in a circular path, it produces electromagnetic wave. Reason : Charged particle moving in circular path has acceleration.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
) In circular path, direction of velocity of charge is charging as a function of time and so it has acceleration. Accelerated charges emit electromagnetic waves. So correct option is (1).
PHXI15:WAVES
358931
The electric field part of an electromagnetic wave in vacuum is \(E=3.1 N C^{-1} \cos [(1.8 \mathrm{rad} /\) \(\left.\left.m^{-1}\right) y+\left(5.4 \times 10^{8} \mathrm{rad} / \mathrm{s}^{-1}\right) t\right] i\). The wavelength of this part of electromagnetic wave is:
1 \(1.5\;m\)
2 \(2\;m\)
3 \(2.5\;m\)
4 \(3.5\;m\)
Explanation:
\(E=3.1 \cos \left[\left(1.8 y+5.4 \times 10^{8}\right] i N / C\right.\) Standard equation: \(\begin{aligned}& E=E_{0} \cos [k x-\omega t] N / C \\& \therefore k=\dfrac{2 \pi}{\lambda}=1.8 \\& \lambda=\dfrac{2 \pi}{1.8}=3.49 \simeq 3.5 .\end{aligned}\)
PHXI15:WAVES
358932
If \({v_\gamma },{v_X}\) and \({v_M}\) are the speeds of \(\gamma \)-rays, \(X\)-rays and microwaves respectively in vacuum, then
1 \({v_\gamma } > {v_M} > {v_X}\)
2 \({v_\gamma } > {v_X} > {v_M}\)
3 \({v_\gamma } = {v_X} = {v_M}\)
4 None of these
Explanation:
\(\gamma \)-rays, \(X\)-rays and microwaves respectively are electromagnetic waves. They travel with the speed of light in vacuum. Hence, \(\,\,\,\,\,{v_\gamma } = {v_X} = {v_M}\)
