NEET Test Series from KOTA - 10 Papers In MS WORD
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Electromagnetic Wave
155616
If a charged particle moves in a gravity free space with uniform velocity, then which of the following is not possible $(\overrightarrow{\mathbf{E}}=\text { electric field, } \overrightarrow{\mathbf{B}}=\text { magnetic field })$
B Option (a) If $=\overrightarrow{\mathrm{E}}=0, \overrightarrow{\mathrm{B}}=0$ $\overrightarrow{\mathrm{F}}_{\text {net }}=0$ Option (c) If $\overrightarrow{\mathrm{E}}=0, \overrightarrow{\mathrm{B}} \neq 0$ and $\theta$ between $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{V}}$ is $0^{\circ}$ or $180^{\circ}$ $\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \times 0+\mathrm{qVB} \sin \left(0^{\circ} \text { or } 180^{\circ}\right)=0$ Option (d) If $\overrightarrow{\mathrm{E}} \neq 0, \overrightarrow{\mathrm{B}} \neq 0$ but $\mathrm{q} \overrightarrow{\mathrm{E}}=-\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})$ $\because$ both forces are equal and opposite so, $\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=0$ But for option (b) $\overrightarrow{\mathrm{B}}=0$ and $\mathrm{E} \neq 0$ $\therefore \quad \overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}$ And this force will change the velocity $\vec{V}$ of charge hence it is not possible.
AP EAMCET-24.04.2017
Electromagnetic Wave
155617
A light beam travelling in the $x$-direction is described by the electric field $E_{y}=300 \mathrm{~V} / \mathrm{m}$ $\sin \omega\left(t-\frac{x}{c}\right)$. The maximum magnetic field is:
1 $300 \mathrm{~T}$
2 $3 \times 10^{-6} \mathrm{~T}$
3 $10^{-6} \mathrm{~T}$
4 $100 \mathrm{~T}$
Explanation:
C Given that, $\mathrm{E}_{\mathrm{y}}=300 \mathrm{~V} / \mathrm{m}$ The maximum magnetic field is, $\mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{c}}$ $\mathrm{B}_{0}=\frac{300}{3 \times 10^{8}}=10^{-6} \mathrm{~T}$ The maximum magnetic field is $10^{-6} \mathrm{~T}$.
Assam CEE-2017
Electromagnetic Wave
155618
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is
1 $3 \mathrm{~V} / \mathrm{m}$
2 $6 \mathrm{~V} / \mathrm{m}$
3 $9 \mathrm{~V} / \mathrm{m}$
4 $12 \mathrm{~V} / \mathrm{m}$
Explanation:
B Magnetic field (B) $=20 \mathrm{nT}=20 \times 10^{-9} \mathrm{~T}$ Velocity of EM wave $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know that, $\because \quad \mathrm{c} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{E} =\mathrm{cB}=3 \times 10^{8} \times 20 \times 10^{-9}$ $\mathrm{E} =60 \times 10^{-1}$ $\mathrm{E} =6 \mathrm{~V} / \mathrm{m}$
BITSAT-2009
Electromagnetic Wave
155620
An electromagnetic wave passes through space and its equation is given by $E=E_{0} \sin (\omega t-k x)$ where $E$ is electric field. Energy density of electromagnetic wave in space is
A Given that, $\mathrm{E}=\mathrm{E}_{0} \sin (\omega \mathrm{t}-\mathrm{kx})$ Energy density of EM wave, $=\varepsilon_{0} \times\left(\mathrm{E}_{\mathrm{rms}}^{2}\right)$ $=\varepsilon_{0}\left(\frac{\mathrm{E}_{0}}{\sqrt{2}}\right)^{2}$ $=\varepsilon_{0} \frac{\mathrm{E}_{0}^{2}}{2}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}$
155616
If a charged particle moves in a gravity free space with uniform velocity, then which of the following is not possible $(\overrightarrow{\mathbf{E}}=\text { electric field, } \overrightarrow{\mathbf{B}}=\text { magnetic field })$
B Option (a) If $=\overrightarrow{\mathrm{E}}=0, \overrightarrow{\mathrm{B}}=0$ $\overrightarrow{\mathrm{F}}_{\text {net }}=0$ Option (c) If $\overrightarrow{\mathrm{E}}=0, \overrightarrow{\mathrm{B}} \neq 0$ and $\theta$ between $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{V}}$ is $0^{\circ}$ or $180^{\circ}$ $\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \times 0+\mathrm{qVB} \sin \left(0^{\circ} \text { or } 180^{\circ}\right)=0$ Option (d) If $\overrightarrow{\mathrm{E}} \neq 0, \overrightarrow{\mathrm{B}} \neq 0$ but $\mathrm{q} \overrightarrow{\mathrm{E}}=-\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})$ $\because$ both forces are equal and opposite so, $\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=0$ But for option (b) $\overrightarrow{\mathrm{B}}=0$ and $\mathrm{E} \neq 0$ $\therefore \quad \overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}$ And this force will change the velocity $\vec{V}$ of charge hence it is not possible.
AP EAMCET-24.04.2017
Electromagnetic Wave
155617
A light beam travelling in the $x$-direction is described by the electric field $E_{y}=300 \mathrm{~V} / \mathrm{m}$ $\sin \omega\left(t-\frac{x}{c}\right)$. The maximum magnetic field is:
1 $300 \mathrm{~T}$
2 $3 \times 10^{-6} \mathrm{~T}$
3 $10^{-6} \mathrm{~T}$
4 $100 \mathrm{~T}$
Explanation:
C Given that, $\mathrm{E}_{\mathrm{y}}=300 \mathrm{~V} / \mathrm{m}$ The maximum magnetic field is, $\mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{c}}$ $\mathrm{B}_{0}=\frac{300}{3 \times 10^{8}}=10^{-6} \mathrm{~T}$ The maximum magnetic field is $10^{-6} \mathrm{~T}$.
Assam CEE-2017
Electromagnetic Wave
155618
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is
1 $3 \mathrm{~V} / \mathrm{m}$
2 $6 \mathrm{~V} / \mathrm{m}$
3 $9 \mathrm{~V} / \mathrm{m}$
4 $12 \mathrm{~V} / \mathrm{m}$
Explanation:
B Magnetic field (B) $=20 \mathrm{nT}=20 \times 10^{-9} \mathrm{~T}$ Velocity of EM wave $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know that, $\because \quad \mathrm{c} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{E} =\mathrm{cB}=3 \times 10^{8} \times 20 \times 10^{-9}$ $\mathrm{E} =60 \times 10^{-1}$ $\mathrm{E} =6 \mathrm{~V} / \mathrm{m}$
BITSAT-2009
Electromagnetic Wave
155620
An electromagnetic wave passes through space and its equation is given by $E=E_{0} \sin (\omega t-k x)$ where $E$ is electric field. Energy density of electromagnetic wave in space is
A Given that, $\mathrm{E}=\mathrm{E}_{0} \sin (\omega \mathrm{t}-\mathrm{kx})$ Energy density of EM wave, $=\varepsilon_{0} \times\left(\mathrm{E}_{\mathrm{rms}}^{2}\right)$ $=\varepsilon_{0}\left(\frac{\mathrm{E}_{0}}{\sqrt{2}}\right)^{2}$ $=\varepsilon_{0} \frac{\mathrm{E}_{0}^{2}}{2}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}$
155616
If a charged particle moves in a gravity free space with uniform velocity, then which of the following is not possible $(\overrightarrow{\mathbf{E}}=\text { electric field, } \overrightarrow{\mathbf{B}}=\text { magnetic field })$
B Option (a) If $=\overrightarrow{\mathrm{E}}=0, \overrightarrow{\mathrm{B}}=0$ $\overrightarrow{\mathrm{F}}_{\text {net }}=0$ Option (c) If $\overrightarrow{\mathrm{E}}=0, \overrightarrow{\mathrm{B}} \neq 0$ and $\theta$ between $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{V}}$ is $0^{\circ}$ or $180^{\circ}$ $\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \times 0+\mathrm{qVB} \sin \left(0^{\circ} \text { or } 180^{\circ}\right)=0$ Option (d) If $\overrightarrow{\mathrm{E}} \neq 0, \overrightarrow{\mathrm{B}} \neq 0$ but $\mathrm{q} \overrightarrow{\mathrm{E}}=-\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})$ $\because$ both forces are equal and opposite so, $\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=0$ But for option (b) $\overrightarrow{\mathrm{B}}=0$ and $\mathrm{E} \neq 0$ $\therefore \quad \overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}$ And this force will change the velocity $\vec{V}$ of charge hence it is not possible.
AP EAMCET-24.04.2017
Electromagnetic Wave
155617
A light beam travelling in the $x$-direction is described by the electric field $E_{y}=300 \mathrm{~V} / \mathrm{m}$ $\sin \omega\left(t-\frac{x}{c}\right)$. The maximum magnetic field is:
1 $300 \mathrm{~T}$
2 $3 \times 10^{-6} \mathrm{~T}$
3 $10^{-6} \mathrm{~T}$
4 $100 \mathrm{~T}$
Explanation:
C Given that, $\mathrm{E}_{\mathrm{y}}=300 \mathrm{~V} / \mathrm{m}$ The maximum magnetic field is, $\mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{c}}$ $\mathrm{B}_{0}=\frac{300}{3 \times 10^{8}}=10^{-6} \mathrm{~T}$ The maximum magnetic field is $10^{-6} \mathrm{~T}$.
Assam CEE-2017
Electromagnetic Wave
155618
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is
1 $3 \mathrm{~V} / \mathrm{m}$
2 $6 \mathrm{~V} / \mathrm{m}$
3 $9 \mathrm{~V} / \mathrm{m}$
4 $12 \mathrm{~V} / \mathrm{m}$
Explanation:
B Magnetic field (B) $=20 \mathrm{nT}=20 \times 10^{-9} \mathrm{~T}$ Velocity of EM wave $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know that, $\because \quad \mathrm{c} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{E} =\mathrm{cB}=3 \times 10^{8} \times 20 \times 10^{-9}$ $\mathrm{E} =60 \times 10^{-1}$ $\mathrm{E} =6 \mathrm{~V} / \mathrm{m}$
BITSAT-2009
Electromagnetic Wave
155620
An electromagnetic wave passes through space and its equation is given by $E=E_{0} \sin (\omega t-k x)$ where $E$ is electric field. Energy density of electromagnetic wave in space is
A Given that, $\mathrm{E}=\mathrm{E}_{0} \sin (\omega \mathrm{t}-\mathrm{kx})$ Energy density of EM wave, $=\varepsilon_{0} \times\left(\mathrm{E}_{\mathrm{rms}}^{2}\right)$ $=\varepsilon_{0}\left(\frac{\mathrm{E}_{0}}{\sqrt{2}}\right)^{2}$ $=\varepsilon_{0} \frac{\mathrm{E}_{0}^{2}}{2}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}$
155616
If a charged particle moves in a gravity free space with uniform velocity, then which of the following is not possible $(\overrightarrow{\mathbf{E}}=\text { electric field, } \overrightarrow{\mathbf{B}}=\text { magnetic field })$
B Option (a) If $=\overrightarrow{\mathrm{E}}=0, \overrightarrow{\mathrm{B}}=0$ $\overrightarrow{\mathrm{F}}_{\text {net }}=0$ Option (c) If $\overrightarrow{\mathrm{E}}=0, \overrightarrow{\mathrm{B}} \neq 0$ and $\theta$ between $\overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{V}}$ is $0^{\circ}$ or $180^{\circ}$ $\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \times 0+\mathrm{qVB} \sin \left(0^{\circ} \text { or } 180^{\circ}\right)=0$ Option (d) If $\overrightarrow{\mathrm{E}} \neq 0, \overrightarrow{\mathrm{B}} \neq 0$ but $\mathrm{q} \overrightarrow{\mathrm{E}}=-\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})$ $\because$ both forces are equal and opposite so, $\overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}+\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})=0$ But for option (b) $\overrightarrow{\mathrm{B}}=0$ and $\mathrm{E} \neq 0$ $\therefore \quad \overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{q} \overrightarrow{\mathrm{E}}$ And this force will change the velocity $\vec{V}$ of charge hence it is not possible.
AP EAMCET-24.04.2017
Electromagnetic Wave
155617
A light beam travelling in the $x$-direction is described by the electric field $E_{y}=300 \mathrm{~V} / \mathrm{m}$ $\sin \omega\left(t-\frac{x}{c}\right)$. The maximum magnetic field is:
1 $300 \mathrm{~T}$
2 $3 \times 10^{-6} \mathrm{~T}$
3 $10^{-6} \mathrm{~T}$
4 $100 \mathrm{~T}$
Explanation:
C Given that, $\mathrm{E}_{\mathrm{y}}=300 \mathrm{~V} / \mathrm{m}$ The maximum magnetic field is, $\mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{c}}$ $\mathrm{B}_{0}=\frac{300}{3 \times 10^{8}}=10^{-6} \mathrm{~T}$ The maximum magnetic field is $10^{-6} \mathrm{~T}$.
Assam CEE-2017
Electromagnetic Wave
155618
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is
1 $3 \mathrm{~V} / \mathrm{m}$
2 $6 \mathrm{~V} / \mathrm{m}$
3 $9 \mathrm{~V} / \mathrm{m}$
4 $12 \mathrm{~V} / \mathrm{m}$
Explanation:
B Magnetic field (B) $=20 \mathrm{nT}=20 \times 10^{-9} \mathrm{~T}$ Velocity of EM wave $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know that, $\because \quad \mathrm{c} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{E} =\mathrm{cB}=3 \times 10^{8} \times 20 \times 10^{-9}$ $\mathrm{E} =60 \times 10^{-1}$ $\mathrm{E} =6 \mathrm{~V} / \mathrm{m}$
BITSAT-2009
Electromagnetic Wave
155620
An electromagnetic wave passes through space and its equation is given by $E=E_{0} \sin (\omega t-k x)$ where $E$ is electric field. Energy density of electromagnetic wave in space is
A Given that, $\mathrm{E}=\mathrm{E}_{0} \sin (\omega \mathrm{t}-\mathrm{kx})$ Energy density of EM wave, $=\varepsilon_{0} \times\left(\mathrm{E}_{\mathrm{rms}}^{2}\right)$ $=\varepsilon_{0}\left(\frac{\mathrm{E}_{0}}{\sqrt{2}}\right)^{2}$ $=\varepsilon_{0} \frac{\mathrm{E}_{0}^{2}}{2}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2}$
