358878
The magnetic field in a plane electromagnetic wave is \({B_y} = \left( {3.5 \times {{10}^{ - 7}}} \right)\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)T.\) The corresponding electric field will be
The magnetic field of an \(EM\) wave propagating along \(x\)-axis is given by, \({B_y} = {B_0}\sin \,(kx + \omega t)\,T.\) On comparing the above equation with the given electromagnetic wave equation we get, \({B_0} = 3.5 \times {10^{ - 7}}\;\,J.\) The electric field equation is given by, \(E=E_{0} \sin (k x+\omega t) V / m\) Also, Relation between \(E\) and \(B\) is given by \(\dfrac{E}{B}=c\) \(\therefore {E_0} = {B_0}c = 3 \times 3.5 \times {10^8} \times {10^{ - 7}} = 105\;V/{m^{ - 1}}\) \(E = {E_0}\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)V/{m^{ - 1}}\) \({E_z} = 105\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)V/{m^{ - 1}}\)
JEE - 2024
PHXI15:WAVES
358879
Assertion : The changing electric field produces a magnetic field. Reason : A changing magnetic field produces an electric field.
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 changing electric field produces a magnetic field. It is accounted by AmpereMaxwell's law. The charging magnetic field also produces electric field. So option (2) is correct.
PHXI15:WAVES
358880
If a \(EM\) wave traveling in vacuum in \(y\)-direction has magnetic field \(\vec B = 8 \times {10^{ - 8}}(\hat K)\). Then value of electric field \(\vec{E}\) is:
1 \(24(\widehat i)\)
2 \(24( - \widehat i)\)
3 \(2.6 \times {10^{ - 16}}( - \widehat i)\)
4 \(3.6 \times {10^{ - 16}}( - \hat i)\)
Explanation:
\(\mathrm{E}_{0}=\mathrm{cB}_{0}\) \(\Rightarrow \mathrm{E}_{0}=8 \times 10^{-8} \times\left(3 \times 10^{8}\right)=24\) The direction of propagation for \(EM\) wave is same as \(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\). \(\because\) The wave is travelling in \( + y\) direction and \(\vec{B}\) is in \(+z\) direction, and as \((-\hat{i}) \times \hat{k}=+\hat{j}, \vec{E}\) is in \(x\) direction.
JEE - 2021
PHXI15:WAVES
358881
An electric bulb is rated as \(200\;W\). What will be the peak magnetic field at \(4\;m\) distance produced by the radiations coming from this bulb? Consider this bulb as a point source with \(3.5 \%\) efficiency.
1 \(1.19 \times {10^{ - 8}}\;T\)
2 \(1.71 \times {10^{ - 8}}\;T\)
3 \(0.84 \times {10^{ - 8}}\;T\)
4 \(3.36 \times {10^{ - 8}}\;T\)
Explanation:
We know that, power \(P\) energy density and efficiency \(\eta\) can be related by: \(\begin{aligned}& \dfrac{\eta P}{4 \pi r^{2}}=\dfrac{c B_{\max }^{2}}{2 \mu_{0}} \\& \Rightarrow B_{\max }=\dfrac{1}{r} \sqrt{\dfrac{2 \mu_{0}}{4 \pi} \dfrac{\eta P}{c}}\end{aligned}\) Here \(\eta=0.35\) \({\mu _0} = 4\pi \times {10^{ - 7}}H/m\quad {\rm{ }}({\rm{or}}){\rm{ }}N/A{{\rm{ }}^2}.\) \( = \frac{1}{4}\sqrt {\frac{{{{10}^{ - 7}} \times 2 \times 0.35 \times 200}}{{3 \times {{10}^8}}}} \) \( = 1.71 \times {10^{ - 8}}\;T\)
358878
The magnetic field in a plane electromagnetic wave is \({B_y} = \left( {3.5 \times {{10}^{ - 7}}} \right)\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)T.\) The corresponding electric field will be
The magnetic field of an \(EM\) wave propagating along \(x\)-axis is given by, \({B_y} = {B_0}\sin \,(kx + \omega t)\,T.\) On comparing the above equation with the given electromagnetic wave equation we get, \({B_0} = 3.5 \times {10^{ - 7}}\;\,J.\) The electric field equation is given by, \(E=E_{0} \sin (k x+\omega t) V / m\) Also, Relation between \(E\) and \(B\) is given by \(\dfrac{E}{B}=c\) \(\therefore {E_0} = {B_0}c = 3 \times 3.5 \times {10^8} \times {10^{ - 7}} = 105\;V/{m^{ - 1}}\) \(E = {E_0}\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)V/{m^{ - 1}}\) \({E_z} = 105\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)V/{m^{ - 1}}\)
JEE - 2024
PHXI15:WAVES
358879
Assertion : The changing electric field produces a magnetic field. Reason : A changing magnetic field produces an electric field.
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 changing electric field produces a magnetic field. It is accounted by AmpereMaxwell's law. The charging magnetic field also produces electric field. So option (2) is correct.
PHXI15:WAVES
358880
If a \(EM\) wave traveling in vacuum in \(y\)-direction has magnetic field \(\vec B = 8 \times {10^{ - 8}}(\hat K)\). Then value of electric field \(\vec{E}\) is:
1 \(24(\widehat i)\)
2 \(24( - \widehat i)\)
3 \(2.6 \times {10^{ - 16}}( - \widehat i)\)
4 \(3.6 \times {10^{ - 16}}( - \hat i)\)
Explanation:
\(\mathrm{E}_{0}=\mathrm{cB}_{0}\) \(\Rightarrow \mathrm{E}_{0}=8 \times 10^{-8} \times\left(3 \times 10^{8}\right)=24\) The direction of propagation for \(EM\) wave is same as \(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\). \(\because\) The wave is travelling in \( + y\) direction and \(\vec{B}\) is in \(+z\) direction, and as \((-\hat{i}) \times \hat{k}=+\hat{j}, \vec{E}\) is in \(x\) direction.
JEE - 2021
PHXI15:WAVES
358881
An electric bulb is rated as \(200\;W\). What will be the peak magnetic field at \(4\;m\) distance produced by the radiations coming from this bulb? Consider this bulb as a point source with \(3.5 \%\) efficiency.
1 \(1.19 \times {10^{ - 8}}\;T\)
2 \(1.71 \times {10^{ - 8}}\;T\)
3 \(0.84 \times {10^{ - 8}}\;T\)
4 \(3.36 \times {10^{ - 8}}\;T\)
Explanation:
We know that, power \(P\) energy density and efficiency \(\eta\) can be related by: \(\begin{aligned}& \dfrac{\eta P}{4 \pi r^{2}}=\dfrac{c B_{\max }^{2}}{2 \mu_{0}} \\& \Rightarrow B_{\max }=\dfrac{1}{r} \sqrt{\dfrac{2 \mu_{0}}{4 \pi} \dfrac{\eta P}{c}}\end{aligned}\) Here \(\eta=0.35\) \({\mu _0} = 4\pi \times {10^{ - 7}}H/m\quad {\rm{ }}({\rm{or}}){\rm{ }}N/A{{\rm{ }}^2}.\) \( = \frac{1}{4}\sqrt {\frac{{{{10}^{ - 7}} \times 2 \times 0.35 \times 200}}{{3 \times {{10}^8}}}} \) \( = 1.71 \times {10^{ - 8}}\;T\)
358878
The magnetic field in a plane electromagnetic wave is \({B_y} = \left( {3.5 \times {{10}^{ - 7}}} \right)\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)T.\) The corresponding electric field will be
The magnetic field of an \(EM\) wave propagating along \(x\)-axis is given by, \({B_y} = {B_0}\sin \,(kx + \omega t)\,T.\) On comparing the above equation with the given electromagnetic wave equation we get, \({B_0} = 3.5 \times {10^{ - 7}}\;\,J.\) The electric field equation is given by, \(E=E_{0} \sin (k x+\omega t) V / m\) Also, Relation between \(E\) and \(B\) is given by \(\dfrac{E}{B}=c\) \(\therefore {E_0} = {B_0}c = 3 \times 3.5 \times {10^8} \times {10^{ - 7}} = 105\;V/{m^{ - 1}}\) \(E = {E_0}\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)V/{m^{ - 1}}\) \({E_z} = 105\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)V/{m^{ - 1}}\)
JEE - 2024
PHXI15:WAVES
358879
Assertion : The changing electric field produces a magnetic field. Reason : A changing magnetic field produces an electric field.
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 changing electric field produces a magnetic field. It is accounted by AmpereMaxwell's law. The charging magnetic field also produces electric field. So option (2) is correct.
PHXI15:WAVES
358880
If a \(EM\) wave traveling in vacuum in \(y\)-direction has magnetic field \(\vec B = 8 \times {10^{ - 8}}(\hat K)\). Then value of electric field \(\vec{E}\) is:
1 \(24(\widehat i)\)
2 \(24( - \widehat i)\)
3 \(2.6 \times {10^{ - 16}}( - \widehat i)\)
4 \(3.6 \times {10^{ - 16}}( - \hat i)\)
Explanation:
\(\mathrm{E}_{0}=\mathrm{cB}_{0}\) \(\Rightarrow \mathrm{E}_{0}=8 \times 10^{-8} \times\left(3 \times 10^{8}\right)=24\) The direction of propagation for \(EM\) wave is same as \(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\). \(\because\) The wave is travelling in \( + y\) direction and \(\vec{B}\) is in \(+z\) direction, and as \((-\hat{i}) \times \hat{k}=+\hat{j}, \vec{E}\) is in \(x\) direction.
JEE - 2021
PHXI15:WAVES
358881
An electric bulb is rated as \(200\;W\). What will be the peak magnetic field at \(4\;m\) distance produced by the radiations coming from this bulb? Consider this bulb as a point source with \(3.5 \%\) efficiency.
1 \(1.19 \times {10^{ - 8}}\;T\)
2 \(1.71 \times {10^{ - 8}}\;T\)
3 \(0.84 \times {10^{ - 8}}\;T\)
4 \(3.36 \times {10^{ - 8}}\;T\)
Explanation:
We know that, power \(P\) energy density and efficiency \(\eta\) can be related by: \(\begin{aligned}& \dfrac{\eta P}{4 \pi r^{2}}=\dfrac{c B_{\max }^{2}}{2 \mu_{0}} \\& \Rightarrow B_{\max }=\dfrac{1}{r} \sqrt{\dfrac{2 \mu_{0}}{4 \pi} \dfrac{\eta P}{c}}\end{aligned}\) Here \(\eta=0.35\) \({\mu _0} = 4\pi \times {10^{ - 7}}H/m\quad {\rm{ }}({\rm{or}}){\rm{ }}N/A{{\rm{ }}^2}.\) \( = \frac{1}{4}\sqrt {\frac{{{{10}^{ - 7}} \times 2 \times 0.35 \times 200}}{{3 \times {{10}^8}}}} \) \( = 1.71 \times {10^{ - 8}}\;T\)
358878
The magnetic field in a plane electromagnetic wave is \({B_y} = \left( {3.5 \times {{10}^{ - 7}}} \right)\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)T.\) The corresponding electric field will be
The magnetic field of an \(EM\) wave propagating along \(x\)-axis is given by, \({B_y} = {B_0}\sin \,(kx + \omega t)\,T.\) On comparing the above equation with the given electromagnetic wave equation we get, \({B_0} = 3.5 \times {10^{ - 7}}\;\,J.\) The electric field equation is given by, \(E=E_{0} \sin (k x+\omega t) V / m\) Also, Relation between \(E\) and \(B\) is given by \(\dfrac{E}{B}=c\) \(\therefore {E_0} = {B_0}c = 3 \times 3.5 \times {10^8} \times {10^{ - 7}} = 105\;V/{m^{ - 1}}\) \(E = {E_0}\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)V/{m^{ - 1}}\) \({E_z} = 105\sin \left( {1.5 \times {{10}^3}x + 0.5 \times {{10}^{11}}t} \right)V/{m^{ - 1}}\)
JEE - 2024
PHXI15:WAVES
358879
Assertion : The changing electric field produces a magnetic field. Reason : A changing magnetic field produces an electric field.
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 changing electric field produces a magnetic field. It is accounted by AmpereMaxwell's law. The charging magnetic field also produces electric field. So option (2) is correct.
PHXI15:WAVES
358880
If a \(EM\) wave traveling in vacuum in \(y\)-direction has magnetic field \(\vec B = 8 \times {10^{ - 8}}(\hat K)\). Then value of electric field \(\vec{E}\) is:
1 \(24(\widehat i)\)
2 \(24( - \widehat i)\)
3 \(2.6 \times {10^{ - 16}}( - \widehat i)\)
4 \(3.6 \times {10^{ - 16}}( - \hat i)\)
Explanation:
\(\mathrm{E}_{0}=\mathrm{cB}_{0}\) \(\Rightarrow \mathrm{E}_{0}=8 \times 10^{-8} \times\left(3 \times 10^{8}\right)=24\) The direction of propagation for \(EM\) wave is same as \(\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\). \(\because\) The wave is travelling in \( + y\) direction and \(\vec{B}\) is in \(+z\) direction, and as \((-\hat{i}) \times \hat{k}=+\hat{j}, \vec{E}\) is in \(x\) direction.
JEE - 2021
PHXI15:WAVES
358881
An electric bulb is rated as \(200\;W\). What will be the peak magnetic field at \(4\;m\) distance produced by the radiations coming from this bulb? Consider this bulb as a point source with \(3.5 \%\) efficiency.
1 \(1.19 \times {10^{ - 8}}\;T\)
2 \(1.71 \times {10^{ - 8}}\;T\)
3 \(0.84 \times {10^{ - 8}}\;T\)
4 \(3.36 \times {10^{ - 8}}\;T\)
Explanation:
We know that, power \(P\) energy density and efficiency \(\eta\) can be related by: \(\begin{aligned}& \dfrac{\eta P}{4 \pi r^{2}}=\dfrac{c B_{\max }^{2}}{2 \mu_{0}} \\& \Rightarrow B_{\max }=\dfrac{1}{r} \sqrt{\dfrac{2 \mu_{0}}{4 \pi} \dfrac{\eta P}{c}}\end{aligned}\) Here \(\eta=0.35\) \({\mu _0} = 4\pi \times {10^{ - 7}}H/m\quad {\rm{ }}({\rm{or}}){\rm{ }}N/A{{\rm{ }}^2}.\) \( = \frac{1}{4}\sqrt {\frac{{{{10}^{ - 7}} \times 2 \times 0.35 \times 200}}{{3 \times {{10}^8}}}} \) \( = 1.71 \times {10^{ - 8}}\;T\)
