358835
A long straight wire of resistance \(R\), radius ' \(a\) ' and length ' \(l\) ' carries a constant current ' \(I\) '. The poynting vector for the wire will be-
1 \(\dfrac{I R^{2}}{a l}\)
2 \(\dfrac{I^{2} R}{a l}\)
3 \(\dfrac{I R}{2 \pi a l}\)
4 \(\dfrac{I^{2} R}{2 \pi a l}\)
Explanation:
If \(V\) is the potential difference across the wire, then \(\mathrm{E}=\dfrac{\mathrm{V}}{\ell}=\dfrac{\mathrm{IR}}{\ell}\) and magnetic field at the surface of wire \(B=\dfrac{\mu_{0} I}{2 \pi a}\). Hence pointing vector, directed radially inwards, is given by \(S=\dfrac{E B}{\mu_{0}}=\dfrac{I R}{\mu_{0} \ell} \times \dfrac{\mu_{0} I}{2 \pi a}=\dfrac{I^{2} R}{2 \pi a l}\) Hence the correct answer will be (4).
PHXI15:WAVES
358836
Assertion : In an electromagnetic wave, magnitude of magnetic field vector is equal to the magnitude of electric field vector. Reason : Energy of electromagnetic waves is shared equally by the electric and magnetic fields.
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:
The equation \(\frac{E}{B} = c = 3 \times {10^8}\;m{\rm{/}}s\) signifies that the electric field's magnitude \((E)\) is much greater than the magnetic field's magnitude \((B)\) in electromagnetic waves. Despite this, both fields share energy equally during wave propagation, emphasizing their collaborative role in transporting electromagnetic energy. So correct option is (4).
PHXI15:WAVES
358837
An electromagnetic wave passes through space and its equation is given by \(E = {E_0}\sin (\omega t - kx)\) where \(E\) is electric field. Energy density of electromagnetic wave in space is
1 \(\frac{1}{2}{\varepsilon _0}E_0^2\)
2 \(\frac{1}{4}{\varepsilon _0}E_0^2\)
3 \({\varepsilon _0}E_0^2\)
4 \(2{\varepsilon _0}E_0^2\)
Explanation:
Energy density for the wave, \({u_E} = {\varepsilon _0}E_{rms}^2 = {\varepsilon _0}{\left( {\frac{{{E_0}}}{{\sqrt 2 }}} \right)^2} = \frac{1}{2}{\varepsilon _0}E_0^2\)
MHTCET - 2021
PHXI15:WAVES
358838
In the given electromagnetic wave \(E_{y}=600\) sin \((\omega t-k x) V m^{-1}\), intensity of the associated light beam is \(({\text{in}}\,\,W/{m^2})\) (Given \({\varepsilon _0} = 9 \times {10^{ - 12}}{C^2}{N^{ - 1}}\;{m^{ - 2}})\)
1 729
2 243
3 486
4 972
Explanation:
Given, \(E_{y}=600 \sin (\omega t-k x)\) and \({\varepsilon _0} = 9 \times {10^{ - 12}}{C^2}/N{m^2}\) Intensity of \(EM\) wave is given by, \(I=\dfrac{1}{2} \varepsilon_{0} E_{0}^{2} c=\dfrac{1}{2} \dfrac{B_{0}^{2}}{\mu_{0}} c\) Intensity, \(I=\dfrac{1}{2} \varepsilon_{0} E_{0}^{2} c\) \( = \frac{1}{2} \times 9 \times {10^{ - 12}} \times {600^2} \times 3 \times {10^8} = 486\;W/{m^2}\) So, correct option is (3).
358835
A long straight wire of resistance \(R\), radius ' \(a\) ' and length ' \(l\) ' carries a constant current ' \(I\) '. The poynting vector for the wire will be-
1 \(\dfrac{I R^{2}}{a l}\)
2 \(\dfrac{I^{2} R}{a l}\)
3 \(\dfrac{I R}{2 \pi a l}\)
4 \(\dfrac{I^{2} R}{2 \pi a l}\)
Explanation:
If \(V\) is the potential difference across the wire, then \(\mathrm{E}=\dfrac{\mathrm{V}}{\ell}=\dfrac{\mathrm{IR}}{\ell}\) and magnetic field at the surface of wire \(B=\dfrac{\mu_{0} I}{2 \pi a}\). Hence pointing vector, directed radially inwards, is given by \(S=\dfrac{E B}{\mu_{0}}=\dfrac{I R}{\mu_{0} \ell} \times \dfrac{\mu_{0} I}{2 \pi a}=\dfrac{I^{2} R}{2 \pi a l}\) Hence the correct answer will be (4).
PHXI15:WAVES
358836
Assertion : In an electromagnetic wave, magnitude of magnetic field vector is equal to the magnitude of electric field vector. Reason : Energy of electromagnetic waves is shared equally by the electric and magnetic fields.
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:
The equation \(\frac{E}{B} = c = 3 \times {10^8}\;m{\rm{/}}s\) signifies that the electric field's magnitude \((E)\) is much greater than the magnetic field's magnitude \((B)\) in electromagnetic waves. Despite this, both fields share energy equally during wave propagation, emphasizing their collaborative role in transporting electromagnetic energy. So correct option is (4).
PHXI15:WAVES
358837
An electromagnetic wave passes through space and its equation is given by \(E = {E_0}\sin (\omega t - kx)\) where \(E\) is electric field. Energy density of electromagnetic wave in space is
1 \(\frac{1}{2}{\varepsilon _0}E_0^2\)
2 \(\frac{1}{4}{\varepsilon _0}E_0^2\)
3 \({\varepsilon _0}E_0^2\)
4 \(2{\varepsilon _0}E_0^2\)
Explanation:
Energy density for the wave, \({u_E} = {\varepsilon _0}E_{rms}^2 = {\varepsilon _0}{\left( {\frac{{{E_0}}}{{\sqrt 2 }}} \right)^2} = \frac{1}{2}{\varepsilon _0}E_0^2\)
MHTCET - 2021
PHXI15:WAVES
358838
In the given electromagnetic wave \(E_{y}=600\) sin \((\omega t-k x) V m^{-1}\), intensity of the associated light beam is \(({\text{in}}\,\,W/{m^2})\) (Given \({\varepsilon _0} = 9 \times {10^{ - 12}}{C^2}{N^{ - 1}}\;{m^{ - 2}})\)
1 729
2 243
3 486
4 972
Explanation:
Given, \(E_{y}=600 \sin (\omega t-k x)\) and \({\varepsilon _0} = 9 \times {10^{ - 12}}{C^2}/N{m^2}\) Intensity of \(EM\) wave is given by, \(I=\dfrac{1}{2} \varepsilon_{0} E_{0}^{2} c=\dfrac{1}{2} \dfrac{B_{0}^{2}}{\mu_{0}} c\) Intensity, \(I=\dfrac{1}{2} \varepsilon_{0} E_{0}^{2} c\) \( = \frac{1}{2} \times 9 \times {10^{ - 12}} \times {600^2} \times 3 \times {10^8} = 486\;W/{m^2}\) So, correct option is (3).
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PHXI15:WAVES
358835
A long straight wire of resistance \(R\), radius ' \(a\) ' and length ' \(l\) ' carries a constant current ' \(I\) '. The poynting vector for the wire will be-
1 \(\dfrac{I R^{2}}{a l}\)
2 \(\dfrac{I^{2} R}{a l}\)
3 \(\dfrac{I R}{2 \pi a l}\)
4 \(\dfrac{I^{2} R}{2 \pi a l}\)
Explanation:
If \(V\) is the potential difference across the wire, then \(\mathrm{E}=\dfrac{\mathrm{V}}{\ell}=\dfrac{\mathrm{IR}}{\ell}\) and magnetic field at the surface of wire \(B=\dfrac{\mu_{0} I}{2 \pi a}\). Hence pointing vector, directed radially inwards, is given by \(S=\dfrac{E B}{\mu_{0}}=\dfrac{I R}{\mu_{0} \ell} \times \dfrac{\mu_{0} I}{2 \pi a}=\dfrac{I^{2} R}{2 \pi a l}\) Hence the correct answer will be (4).
PHXI15:WAVES
358836
Assertion : In an electromagnetic wave, magnitude of magnetic field vector is equal to the magnitude of electric field vector. Reason : Energy of electromagnetic waves is shared equally by the electric and magnetic fields.
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:
The equation \(\frac{E}{B} = c = 3 \times {10^8}\;m{\rm{/}}s\) signifies that the electric field's magnitude \((E)\) is much greater than the magnetic field's magnitude \((B)\) in electromagnetic waves. Despite this, both fields share energy equally during wave propagation, emphasizing their collaborative role in transporting electromagnetic energy. So correct option is (4).
PHXI15:WAVES
358837
An electromagnetic wave passes through space and its equation is given by \(E = {E_0}\sin (\omega t - kx)\) where \(E\) is electric field. Energy density of electromagnetic wave in space is
1 \(\frac{1}{2}{\varepsilon _0}E_0^2\)
2 \(\frac{1}{4}{\varepsilon _0}E_0^2\)
3 \({\varepsilon _0}E_0^2\)
4 \(2{\varepsilon _0}E_0^2\)
Explanation:
Energy density for the wave, \({u_E} = {\varepsilon _0}E_{rms}^2 = {\varepsilon _0}{\left( {\frac{{{E_0}}}{{\sqrt 2 }}} \right)^2} = \frac{1}{2}{\varepsilon _0}E_0^2\)
MHTCET - 2021
PHXI15:WAVES
358838
In the given electromagnetic wave \(E_{y}=600\) sin \((\omega t-k x) V m^{-1}\), intensity of the associated light beam is \(({\text{in}}\,\,W/{m^2})\) (Given \({\varepsilon _0} = 9 \times {10^{ - 12}}{C^2}{N^{ - 1}}\;{m^{ - 2}})\)
1 729
2 243
3 486
4 972
Explanation:
Given, \(E_{y}=600 \sin (\omega t-k x)\) and \({\varepsilon _0} = 9 \times {10^{ - 12}}{C^2}/N{m^2}\) Intensity of \(EM\) wave is given by, \(I=\dfrac{1}{2} \varepsilon_{0} E_{0}^{2} c=\dfrac{1}{2} \dfrac{B_{0}^{2}}{\mu_{0}} c\) Intensity, \(I=\dfrac{1}{2} \varepsilon_{0} E_{0}^{2} c\) \( = \frac{1}{2} \times 9 \times {10^{ - 12}} \times {600^2} \times 3 \times {10^8} = 486\;W/{m^2}\) So, correct option is (3).
358835
A long straight wire of resistance \(R\), radius ' \(a\) ' and length ' \(l\) ' carries a constant current ' \(I\) '. The poynting vector for the wire will be-
1 \(\dfrac{I R^{2}}{a l}\)
2 \(\dfrac{I^{2} R}{a l}\)
3 \(\dfrac{I R}{2 \pi a l}\)
4 \(\dfrac{I^{2} R}{2 \pi a l}\)
Explanation:
If \(V\) is the potential difference across the wire, then \(\mathrm{E}=\dfrac{\mathrm{V}}{\ell}=\dfrac{\mathrm{IR}}{\ell}\) and magnetic field at the surface of wire \(B=\dfrac{\mu_{0} I}{2 \pi a}\). Hence pointing vector, directed radially inwards, is given by \(S=\dfrac{E B}{\mu_{0}}=\dfrac{I R}{\mu_{0} \ell} \times \dfrac{\mu_{0} I}{2 \pi a}=\dfrac{I^{2} R}{2 \pi a l}\) Hence the correct answer will be (4).
PHXI15:WAVES
358836
Assertion : In an electromagnetic wave, magnitude of magnetic field vector is equal to the magnitude of electric field vector. Reason : Energy of electromagnetic waves is shared equally by the electric and magnetic fields.
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:
The equation \(\frac{E}{B} = c = 3 \times {10^8}\;m{\rm{/}}s\) signifies that the electric field's magnitude \((E)\) is much greater than the magnetic field's magnitude \((B)\) in electromagnetic waves. Despite this, both fields share energy equally during wave propagation, emphasizing their collaborative role in transporting electromagnetic energy. So correct option is (4).
PHXI15:WAVES
358837
An electromagnetic wave passes through space and its equation is given by \(E = {E_0}\sin (\omega t - kx)\) where \(E\) is electric field. Energy density of electromagnetic wave in space is
1 \(\frac{1}{2}{\varepsilon _0}E_0^2\)
2 \(\frac{1}{4}{\varepsilon _0}E_0^2\)
3 \({\varepsilon _0}E_0^2\)
4 \(2{\varepsilon _0}E_0^2\)
Explanation:
Energy density for the wave, \({u_E} = {\varepsilon _0}E_{rms}^2 = {\varepsilon _0}{\left( {\frac{{{E_0}}}{{\sqrt 2 }}} \right)^2} = \frac{1}{2}{\varepsilon _0}E_0^2\)
MHTCET - 2021
PHXI15:WAVES
358838
In the given electromagnetic wave \(E_{y}=600\) sin \((\omega t-k x) V m^{-1}\), intensity of the associated light beam is \(({\text{in}}\,\,W/{m^2})\) (Given \({\varepsilon _0} = 9 \times {10^{ - 12}}{C^2}{N^{ - 1}}\;{m^{ - 2}})\)
1 729
2 243
3 486
4 972
Explanation:
Given, \(E_{y}=600 \sin (\omega t-k x)\) and \({\varepsilon _0} = 9 \times {10^{ - 12}}{C^2}/N{m^2}\) Intensity of \(EM\) wave is given by, \(I=\dfrac{1}{2} \varepsilon_{0} E_{0}^{2} c=\dfrac{1}{2} \dfrac{B_{0}^{2}}{\mu_{0}} c\) Intensity, \(I=\dfrac{1}{2} \varepsilon_{0} E_{0}^{2} c\) \( = \frac{1}{2} \times 9 \times {10^{ - 12}} \times {600^2} \times 3 \times {10^8} = 486\;W/{m^2}\) So, correct option is (3).