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PHXI15:WAVES

358873 A plane electromagnetic wave of frequency 25 \(MHz\) travels in free space along the \(x\)-direction. At a perpendicular point in space and time, \(E = 6.3j\,V/m\). What is \(B\) at that point?

1 \(5.1 \times {10^{ - 8}}\,\hat kT\)

2 \(4.1 \times {10^{ - 8}}\,\hat kT\)

3 \(3.1 \times {10^{ - 8}}\,\hat kT\)

4 \(2.1 \times {10^{ - 8}}\,\hat kT\)

PHXI15:WAVES

358874 A plane electromagnetic wave of frequency 20 \(MHz\) travels through a space along \(x\) direction. If the electric field vector at a certain point in space is \(6\;\,V\;{m^{ - 1}}\), what is the magnetic field vector at that point?

1 \(2\,\;T\)

2 \(2 \times {10^{ - 8}}\;\,T\)

3 \(\frac{1}{2}\;\,T\)

4 \(\frac{1}{2} \times {10^{ - 8}}\;\,T\)

KCET - 2014

PHXI15:WAVES

358875 An electromagnetic wave in vacuum has the electric and magnetic field \(\overrightarrow E \) and \(\overrightarrow B \) which are always perpendicular to each other. The direction of polarisation is given by \(\vec{X}\) and that of wave propagation by \(\overrightarrow k \). Then

1 \({\rm{\vec X||\vec B}}\,\,{\rm{and}}\,\,{\rm{\vec k||\vec B \times \vec E}}\)

2 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

3 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

4 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}}\)

PHXI15:WAVES

358876 A parallel-plate capacitor made of circular plates, each of radius \({R=6.0 {~cm}}\), has a capacitance \(C = 100\,pF\). The capacitor is connected to a \(230\,V\,a.c.\) supply with a (angular) frequency of \(300\,rad/s.\) Determine the amplitude of \({B}\) (in \({10^{-11} {~T}}\) ) at a point \(3.0\,cm\) from the axis between the plates.
supporting img

1 1.63

2 5.98

3 3.65

4 7.24

Explanation:

We are given that, \(C = 100\,pF = 100 \times {10^{ - 12}}\;F = {10^{ - 10}}\;F,\)
\({V_{rms}} = 230\,\;V,\omega = 300\,rad/s.\)
Capacitive reactance, \({X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{300 \times 10^{-10}}=\dfrac{10^{8}}{3} \Omega}\)
If \({I_{{rms}}}\) is the \(rms\) value of the conduction current,
\(I_{{rms}}=\dfrac{V_{{rms}}}{X_{C}}=\dfrac{230}{10^{8} / 3}=6.9 \times 10^{-6} {~A}=6.9 \mu {A}\)
The conduction current is equal to the displacement even in case of \(ac\).
Let \({I_{0}}\) be the peak value (amplitude) of the conduction current \((I)\) Clearly, \({I_{{rms}}=\dfrac{I_{0}}{\sqrt{2}}}\)
Or \({\quad I_{0}=I_{\text {rms }} \times \sqrt{2}=(6.9 \mu A) \times 1.414=9.76 \mu {A}}\)
Obviously, peak value (amplitude) of \({I_{d}}\) is also the same, \(i.e.,\) \({9.76 \mu {A}}\).
Consider a circular path of radius \({3.0 {~cm}(=0.03 {~m})}\) parallel to the plates of the capacitor and with its centre \({(O)}\) on the axis of the plates.
According to the Ampere's law,
\(\oint {\vec B} \cdot \overrightarrow {d\ell } = {\mu _0}\left( {I + {I_d}} \right){\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
Now \({\oint \vec{B} \cdot \overrightarrow{d \ell}=B \times}\) circumference of the circle of radius \(0.03\;m = B(2\pi \times 0.03)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
As there is no conduction current in the plates,
\(I = 0{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
Further, \({I_{d}=\dfrac{9.76(\mu {A})}{\pi(0.06)^{2}} \times \pi(0.03)^{2}}\)
\( = 2.44\,\mu A = 2.44 \times {10^{ - 6}}\;A\quad \quad (4)\)
From Eqs. (1), (2), (3) and (4),
\(B(2\pi \times 0.03) = {\mu _0}\left( {2.44 \times {{10}^{ - 6}}} \right)\)
Or \(B = \left( {\frac{{{\mu _0}}}{{2\pi }}} \right)\frac{{2.44 \times {{10}^{ - 6}}}}{{0.03}}\)
\( = \left( {2 \times {{10}^{ - 7}}} \right)\left( {81.3 \times {{10}^{ - 6}}} \right)\)
\( = 1.63 \times {10^{ - 11}}\;T\)

PHXI15:WAVES

358877 The magnetic field in travelling EM wave has a peak value of \(20nT\). The peak value of electric field strength is

1 \(6\;V/m\)

2 \(3\;V/m\)

3 \(9\;V/m\)

4 \(12\;V/m\)

PHXI15:WAVES

358873 A plane electromagnetic wave of frequency 25 \(MHz\) travels in free space along the \(x\)-direction. At a perpendicular point in space and time, \(E = 6.3j\,V/m\). What is \(B\) at that point?

1 \(5.1 \times {10^{ - 8}}\,\hat kT\)

2 \(4.1 \times {10^{ - 8}}\,\hat kT\)

3 \(3.1 \times {10^{ - 8}}\,\hat kT\)

4 \(2.1 \times {10^{ - 8}}\,\hat kT\)

PHXI15:WAVES

358874 A plane electromagnetic wave of frequency 20 \(MHz\) travels through a space along \(x\) direction. If the electric field vector at a certain point in space is \(6\;\,V\;{m^{ - 1}}\), what is the magnetic field vector at that point?

1 \(2\,\;T\)

2 \(2 \times {10^{ - 8}}\;\,T\)

3 \(\frac{1}{2}\;\,T\)

4 \(\frac{1}{2} \times {10^{ - 8}}\;\,T\)

KCET - 2014

PHXI15:WAVES

358875 An electromagnetic wave in vacuum has the electric and magnetic field \(\overrightarrow E \) and \(\overrightarrow B \) which are always perpendicular to each other. The direction of polarisation is given by \(\vec{X}\) and that of wave propagation by \(\overrightarrow k \). Then

1 \({\rm{\vec X||\vec B}}\,\,{\rm{and}}\,\,{\rm{\vec k||\vec B \times \vec E}}\)

2 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

3 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

4 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}}\)

PHXI15:WAVES

358876 A parallel-plate capacitor made of circular plates, each of radius \({R=6.0 {~cm}}\), has a capacitance \(C = 100\,pF\). The capacitor is connected to a \(230\,V\,a.c.\) supply with a (angular) frequency of \(300\,rad/s.\) Determine the amplitude of \({B}\) (in \({10^{-11} {~T}}\) ) at a point \(3.0\,cm\) from the axis between the plates.
supporting img

1 1.63

2 5.98

3 3.65

4 7.24

Explanation:

We are given that, \(C = 100\,pF = 100 \times {10^{ - 12}}\;F = {10^{ - 10}}\;F,\)
\({V_{rms}} = 230\,\;V,\omega = 300\,rad/s.\)
Capacitive reactance, \({X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{300 \times 10^{-10}}=\dfrac{10^{8}}{3} \Omega}\)
If \({I_{{rms}}}\) is the \(rms\) value of the conduction current,
\(I_{{rms}}=\dfrac{V_{{rms}}}{X_{C}}=\dfrac{230}{10^{8} / 3}=6.9 \times 10^{-6} {~A}=6.9 \mu {A}\)
The conduction current is equal to the displacement even in case of \(ac\).
Let \({I_{0}}\) be the peak value (amplitude) of the conduction current \((I)\) Clearly, \({I_{{rms}}=\dfrac{I_{0}}{\sqrt{2}}}\)
Or \({\quad I_{0}=I_{\text {rms }} \times \sqrt{2}=(6.9 \mu A) \times 1.414=9.76 \mu {A}}\)
Obviously, peak value (amplitude) of \({I_{d}}\) is also the same, \(i.e.,\) \({9.76 \mu {A}}\).
Consider a circular path of radius \({3.0 {~cm}(=0.03 {~m})}\) parallel to the plates of the capacitor and with its centre \({(O)}\) on the axis of the plates.
According to the Ampere's law,
\(\oint {\vec B} \cdot \overrightarrow {d\ell } = {\mu _0}\left( {I + {I_d}} \right){\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
Now \({\oint \vec{B} \cdot \overrightarrow{d \ell}=B \times}\) circumference of the circle of radius \(0.03\;m = B(2\pi \times 0.03)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
As there is no conduction current in the plates,
\(I = 0{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
Further, \({I_{d}=\dfrac{9.76(\mu {A})}{\pi(0.06)^{2}} \times \pi(0.03)^{2}}\)
\( = 2.44\,\mu A = 2.44 \times {10^{ - 6}}\;A\quad \quad (4)\)
From Eqs. (1), (2), (3) and (4),
\(B(2\pi \times 0.03) = {\mu _0}\left( {2.44 \times {{10}^{ - 6}}} \right)\)
Or \(B = \left( {\frac{{{\mu _0}}}{{2\pi }}} \right)\frac{{2.44 \times {{10}^{ - 6}}}}{{0.03}}\)
\( = \left( {2 \times {{10}^{ - 7}}} \right)\left( {81.3 \times {{10}^{ - 6}}} \right)\)
\( = 1.63 \times {10^{ - 11}}\;T\)

PHXI15:WAVES

358877 The magnetic field in travelling EM wave has a peak value of \(20nT\). The peak value of electric field strength is

1 \(6\;V/m\)

2 \(3\;V/m\)

3 \(9\;V/m\)

4 \(12\;V/m\)

PHXI15:WAVES

358873 A plane electromagnetic wave of frequency 25 \(MHz\) travels in free space along the \(x\)-direction. At a perpendicular point in space and time, \(E = 6.3j\,V/m\). What is \(B\) at that point?

1 \(5.1 \times {10^{ - 8}}\,\hat kT\)

2 \(4.1 \times {10^{ - 8}}\,\hat kT\)

3 \(3.1 \times {10^{ - 8}}\,\hat kT\)

4 \(2.1 \times {10^{ - 8}}\,\hat kT\)

PHXI15:WAVES

358874 A plane electromagnetic wave of frequency 20 \(MHz\) travels through a space along \(x\) direction. If the electric field vector at a certain point in space is \(6\;\,V\;{m^{ - 1}}\), what is the magnetic field vector at that point?

1 \(2\,\;T\)

2 \(2 \times {10^{ - 8}}\;\,T\)

3 \(\frac{1}{2}\;\,T\)

4 \(\frac{1}{2} \times {10^{ - 8}}\;\,T\)

KCET - 2014

PHXI15:WAVES

358875 An electromagnetic wave in vacuum has the electric and magnetic field \(\overrightarrow E \) and \(\overrightarrow B \) which are always perpendicular to each other. The direction of polarisation is given by \(\vec{X}\) and that of wave propagation by \(\overrightarrow k \). Then

1 \({\rm{\vec X||\vec B}}\,\,{\rm{and}}\,\,{\rm{\vec k||\vec B \times \vec E}}\)

2 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

3 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

4 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}}\)

PHXI15:WAVES

358876 A parallel-plate capacitor made of circular plates, each of radius \({R=6.0 {~cm}}\), has a capacitance \(C = 100\,pF\). The capacitor is connected to a \(230\,V\,a.c.\) supply with a (angular) frequency of \(300\,rad/s.\) Determine the amplitude of \({B}\) (in \({10^{-11} {~T}}\) ) at a point \(3.0\,cm\) from the axis between the plates.
supporting img

1 1.63

2 5.98

3 3.65

4 7.24

Explanation:

We are given that, \(C = 100\,pF = 100 \times {10^{ - 12}}\;F = {10^{ - 10}}\;F,\)
\({V_{rms}} = 230\,\;V,\omega = 300\,rad/s.\)
Capacitive reactance, \({X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{300 \times 10^{-10}}=\dfrac{10^{8}}{3} \Omega}\)
If \({I_{{rms}}}\) is the \(rms\) value of the conduction current,
\(I_{{rms}}=\dfrac{V_{{rms}}}{X_{C}}=\dfrac{230}{10^{8} / 3}=6.9 \times 10^{-6} {~A}=6.9 \mu {A}\)
The conduction current is equal to the displacement even in case of \(ac\).
Let \({I_{0}}\) be the peak value (amplitude) of the conduction current \((I)\) Clearly, \({I_{{rms}}=\dfrac{I_{0}}{\sqrt{2}}}\)
Or \({\quad I_{0}=I_{\text {rms }} \times \sqrt{2}=(6.9 \mu A) \times 1.414=9.76 \mu {A}}\)
Obviously, peak value (amplitude) of \({I_{d}}\) is also the same, \(i.e.,\) \({9.76 \mu {A}}\).
Consider a circular path of radius \({3.0 {~cm}(=0.03 {~m})}\) parallel to the plates of the capacitor and with its centre \({(O)}\) on the axis of the plates.
According to the Ampere's law,
\(\oint {\vec B} \cdot \overrightarrow {d\ell } = {\mu _0}\left( {I + {I_d}} \right){\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
Now \({\oint \vec{B} \cdot \overrightarrow{d \ell}=B \times}\) circumference of the circle of radius \(0.03\;m = B(2\pi \times 0.03)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
As there is no conduction current in the plates,
\(I = 0{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
Further, \({I_{d}=\dfrac{9.76(\mu {A})}{\pi(0.06)^{2}} \times \pi(0.03)^{2}}\)
\( = 2.44\,\mu A = 2.44 \times {10^{ - 6}}\;A\quad \quad (4)\)
From Eqs. (1), (2), (3) and (4),
\(B(2\pi \times 0.03) = {\mu _0}\left( {2.44 \times {{10}^{ - 6}}} \right)\)
Or \(B = \left( {\frac{{{\mu _0}}}{{2\pi }}} \right)\frac{{2.44 \times {{10}^{ - 6}}}}{{0.03}}\)
\( = \left( {2 \times {{10}^{ - 7}}} \right)\left( {81.3 \times {{10}^{ - 6}}} \right)\)
\( = 1.63 \times {10^{ - 11}}\;T\)

PHXI15:WAVES

358877 The magnetic field in travelling EM wave has a peak value of \(20nT\). The peak value of electric field strength is

1 \(6\;V/m\)

2 \(3\;V/m\)

3 \(9\;V/m\)

4 \(12\;V/m\)

PHXI15:WAVES

358873 A plane electromagnetic wave of frequency 25 \(MHz\) travels in free space along the \(x\)-direction. At a perpendicular point in space and time, \(E = 6.3j\,V/m\). What is \(B\) at that point?

1 \(5.1 \times {10^{ - 8}}\,\hat kT\)

2 \(4.1 \times {10^{ - 8}}\,\hat kT\)

3 \(3.1 \times {10^{ - 8}}\,\hat kT\)

4 \(2.1 \times {10^{ - 8}}\,\hat kT\)

PHXI15:WAVES

358874 A plane electromagnetic wave of frequency 20 \(MHz\) travels through a space along \(x\) direction. If the electric field vector at a certain point in space is \(6\;\,V\;{m^{ - 1}}\), what is the magnetic field vector at that point?

1 \(2\,\;T\)

2 \(2 \times {10^{ - 8}}\;\,T\)

3 \(\frac{1}{2}\;\,T\)

4 \(\frac{1}{2} \times {10^{ - 8}}\;\,T\)

KCET - 2014

PHXI15:WAVES

358875 An electromagnetic wave in vacuum has the electric and magnetic field \(\overrightarrow E \) and \(\overrightarrow B \) which are always perpendicular to each other. The direction of polarisation is given by \(\vec{X}\) and that of wave propagation by \(\overrightarrow k \). Then

1 \({\rm{\vec X||\vec B}}\,\,{\rm{and}}\,\,{\rm{\vec k||\vec B \times \vec E}}\)

2 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

3 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

4 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}}\)

PHXI15:WAVES

358876 A parallel-plate capacitor made of circular plates, each of radius \({R=6.0 {~cm}}\), has a capacitance \(C = 100\,pF\). The capacitor is connected to a \(230\,V\,a.c.\) supply with a (angular) frequency of \(300\,rad/s.\) Determine the amplitude of \({B}\) (in \({10^{-11} {~T}}\) ) at a point \(3.0\,cm\) from the axis between the plates.
supporting img

1 1.63

2 5.98

3 3.65

4 7.24

Explanation:

We are given that, \(C = 100\,pF = 100 \times {10^{ - 12}}\;F = {10^{ - 10}}\;F,\)
\({V_{rms}} = 230\,\;V,\omega = 300\,rad/s.\)
Capacitive reactance, \({X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{300 \times 10^{-10}}=\dfrac{10^{8}}{3} \Omega}\)
If \({I_{{rms}}}\) is the \(rms\) value of the conduction current,
\(I_{{rms}}=\dfrac{V_{{rms}}}{X_{C}}=\dfrac{230}{10^{8} / 3}=6.9 \times 10^{-6} {~A}=6.9 \mu {A}\)
The conduction current is equal to the displacement even in case of \(ac\).
Let \({I_{0}}\) be the peak value (amplitude) of the conduction current \((I)\) Clearly, \({I_{{rms}}=\dfrac{I_{0}}{\sqrt{2}}}\)
Or \({\quad I_{0}=I_{\text {rms }} \times \sqrt{2}=(6.9 \mu A) \times 1.414=9.76 \mu {A}}\)
Obviously, peak value (amplitude) of \({I_{d}}\) is also the same, \(i.e.,\) \({9.76 \mu {A}}\).
Consider a circular path of radius \({3.0 {~cm}(=0.03 {~m})}\) parallel to the plates of the capacitor and with its centre \({(O)}\) on the axis of the plates.
According to the Ampere's law,
\(\oint {\vec B} \cdot \overrightarrow {d\ell } = {\mu _0}\left( {I + {I_d}} \right){\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
Now \({\oint \vec{B} \cdot \overrightarrow{d \ell}=B \times}\) circumference of the circle of radius \(0.03\;m = B(2\pi \times 0.03)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
As there is no conduction current in the plates,
\(I = 0{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
Further, \({I_{d}=\dfrac{9.76(\mu {A})}{\pi(0.06)^{2}} \times \pi(0.03)^{2}}\)
\( = 2.44\,\mu A = 2.44 \times {10^{ - 6}}\;A\quad \quad (4)\)
From Eqs. (1), (2), (3) and (4),
\(B(2\pi \times 0.03) = {\mu _0}\left( {2.44 \times {{10}^{ - 6}}} \right)\)
Or \(B = \left( {\frac{{{\mu _0}}}{{2\pi }}} \right)\frac{{2.44 \times {{10}^{ - 6}}}}{{0.03}}\)
\( = \left( {2 \times {{10}^{ - 7}}} \right)\left( {81.3 \times {{10}^{ - 6}}} \right)\)
\( = 1.63 \times {10^{ - 11}}\;T\)

PHXI15:WAVES

358877 The magnetic field in travelling EM wave has a peak value of \(20nT\). The peak value of electric field strength is

1 \(6\;V/m\)

2 \(3\;V/m\)

3 \(9\;V/m\)

4 \(12\;V/m\)

PHXI15:WAVES

358873 A plane electromagnetic wave of frequency 25 \(MHz\) travels in free space along the \(x\)-direction. At a perpendicular point in space and time, \(E = 6.3j\,V/m\). What is \(B\) at that point?

1 \(5.1 \times {10^{ - 8}}\,\hat kT\)

2 \(4.1 \times {10^{ - 8}}\,\hat kT\)

3 \(3.1 \times {10^{ - 8}}\,\hat kT\)

4 \(2.1 \times {10^{ - 8}}\,\hat kT\)

PHXI15:WAVES

358874 A plane electromagnetic wave of frequency 20 \(MHz\) travels through a space along \(x\) direction. If the electric field vector at a certain point in space is \(6\;\,V\;{m^{ - 1}}\), what is the magnetic field vector at that point?

1 \(2\,\;T\)

2 \(2 \times {10^{ - 8}}\;\,T\)

3 \(\frac{1}{2}\;\,T\)

4 \(\frac{1}{2} \times {10^{ - 8}}\;\,T\)

KCET - 2014

PHXI15:WAVES

358875 An electromagnetic wave in vacuum has the electric and magnetic field \(\overrightarrow E \) and \(\overrightarrow B \) which are always perpendicular to each other. The direction of polarisation is given by \(\vec{X}\) and that of wave propagation by \(\overrightarrow k \). Then

1 \({\rm{\vec X||\vec B}}\,\,{\rm{and}}\,\,{\rm{\vec k||\vec B \times \vec E}}\)

2 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

3 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{B}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}\)

4 \(\overrightarrow{\mathrm{X}} \| \overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{k}} \| \overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{C}}\)

PHXI15:WAVES

358876 A parallel-plate capacitor made of circular plates, each of radius \({R=6.0 {~cm}}\), has a capacitance \(C = 100\,pF\). The capacitor is connected to a \(230\,V\,a.c.\) supply with a (angular) frequency of \(300\,rad/s.\) Determine the amplitude of \({B}\) (in \({10^{-11} {~T}}\) ) at a point \(3.0\,cm\) from the axis between the plates.
supporting img

1 1.63

2 5.98

3 3.65

4 7.24

Explanation:

We are given that, \(C = 100\,pF = 100 \times {10^{ - 12}}\;F = {10^{ - 10}}\;F,\)
\({V_{rms}} = 230\,\;V,\omega = 300\,rad/s.\)
Capacitive reactance, \({X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{300 \times 10^{-10}}=\dfrac{10^{8}}{3} \Omega}\)
If \({I_{{rms}}}\) is the \(rms\) value of the conduction current,
\(I_{{rms}}=\dfrac{V_{{rms}}}{X_{C}}=\dfrac{230}{10^{8} / 3}=6.9 \times 10^{-6} {~A}=6.9 \mu {A}\)
The conduction current is equal to the displacement even in case of \(ac\).
Let \({I_{0}}\) be the peak value (amplitude) of the conduction current \((I)\) Clearly, \({I_{{rms}}=\dfrac{I_{0}}{\sqrt{2}}}\)
Or \({\quad I_{0}=I_{\text {rms }} \times \sqrt{2}=(6.9 \mu A) \times 1.414=9.76 \mu {A}}\)
Obviously, peak value (amplitude) of \({I_{d}}\) is also the same, \(i.e.,\) \({9.76 \mu {A}}\).
Consider a circular path of radius \({3.0 {~cm}(=0.03 {~m})}\) parallel to the plates of the capacitor and with its centre \({(O)}\) on the axis of the plates.
According to the Ampere's law,
\(\oint {\vec B} \cdot \overrightarrow {d\ell } = {\mu _0}\left( {I + {I_d}} \right){\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
Now \({\oint \vec{B} \cdot \overrightarrow{d \ell}=B \times}\) circumference of the circle of radius \(0.03\;m = B(2\pi \times 0.03)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
As there is no conduction current in the plates,
\(I = 0{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\)
Further, \({I_{d}=\dfrac{9.76(\mu {A})}{\pi(0.06)^{2}} \times \pi(0.03)^{2}}\)
\( = 2.44\,\mu A = 2.44 \times {10^{ - 6}}\;A\quad \quad (4)\)
From Eqs. (1), (2), (3) and (4),
\(B(2\pi \times 0.03) = {\mu _0}\left( {2.44 \times {{10}^{ - 6}}} \right)\)
Or \(B = \left( {\frac{{{\mu _0}}}{{2\pi }}} \right)\frac{{2.44 \times {{10}^{ - 6}}}}{{0.03}}\)
\( = \left( {2 \times {{10}^{ - 7}}} \right)\left( {81.3 \times {{10}^{ - 6}}} \right)\)
\( = 1.63 \times {10^{ - 11}}\;T\)

PHXI15:WAVES

358877 The magnetic field in travelling EM wave has a peak value of \(20nT\). The peak value of electric field strength is

1 \(6\;V/m\)

2 \(3\;V/m\)

3 \(9\;V/m\)

4 \(12\;V/m\)