155546
Light wave is travelling along $\mathbf{y}$ - directions. If the corresponding $\vec{E}$ vector at any time is along the $x$ - axis, the direction of $\vec{B}$ vector at that time is along :
1 $y$ - axis
2 $x$ - axis
3 $+\mathrm{z}-$ axis
4 $-\mathrm{z}$ - axis
Explanation:
C If wave traveling is y-direction \& corresponding $\overrightarrow{\mathrm{E}}$ vector is in $\mathrm{x}$-axis. Then direction of $\overrightarrow{\mathrm{B}}$ vector will be in $\mathrm{z}$-direction. $\therefore \overrightarrow{\mathrm{E}} \& \overrightarrow{\mathrm{B}}$ are perpendicular to each other.
UPSEE - 2005
Electromagnetic Wave
155558
The dielectric constant of air is 1.006. The speed of electromagnetic wave travelling in air is a $\times 10^{8} \mathrm{~ms}^{-1}$, where $a$ is about
1 3
2 3.88
3 2.5
4 3.2
5 2.8
Explanation:
A Given, Dielectric constant of air $\left(\varepsilon_{\mathrm{r}}\right)=1.006 \sqcup 1$ $\because$ Speed of light in a medium $\mathrm{v}=\frac{1}{\sqrt{\varepsilon \mu}}$ $\mathrm{v}=\frac{1}{\sqrt{\varepsilon_{\mathrm{r}} \varepsilon_{\mathrm{o}} \mu_{\mathrm{r}} \mu_{\mathrm{o}}}}$ For air $\mu_{\mathrm{r}}=1$ $\therefore \quad \mathrm{v}=\frac{1}{\sqrt{1 \times 8.85 \times 10^{-12} \times 4 \pi \times 10^{-7}}}$ $\mathrm{a} \times 10^{8}=305078786$ $\mathrm{a} \times 10^{8} \sqcup 3 \times 10^{8} \Rightarrow \mathrm{a}=3$
COMEDK 2016
Electromagnetic Wave
155559
The amplitude of the sinusoidially oscillating electric field of a plane wave is $60 \mathrm{~V} / \mathrm{m}$. Then the amplitude of magnetic field is :
1 $2 \times 10^{2} \mathrm{~T}$
2 $6 \times 10^{7} \mathrm{~T}$
3 $6 \times 10^{2} \mathrm{~T}$
4 $2 \times 10^{-7} \mathrm{~T}$
5 $3 \times 10^{8} \mathrm{~T}$
Explanation:
D Given, Electric field $(\mathrm{E})=60 \mathrm{~V} / \mathrm{m}$ We know, $\mathrm{c} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{B} =\frac{\mathrm{E}}{\mathrm{c}}=\frac{60}{3 \times 10^{8}} \quad\left(\because \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ $\mathrm{B} =20 \times 10^{-8}$ $\mathrm{~B} =2 \times 10^{-7} \mathrm{~T}$
Kerala CEE 2006
Electromagnetic Wave
155561
A plane electromagnetic wave of frequency 50 $\mathrm{MHz}$ travels in free space along the $\mathrm{X}$ direction. At a particular point in space $E=$ $7.2 \hat{\mathbf{j}} \mathrm{V} / \mathrm{m}$. At this point, $B$ is equal to
155546
Light wave is travelling along $\mathbf{y}$ - directions. If the corresponding $\vec{E}$ vector at any time is along the $x$ - axis, the direction of $\vec{B}$ vector at that time is along :
1 $y$ - axis
2 $x$ - axis
3 $+\mathrm{z}-$ axis
4 $-\mathrm{z}$ - axis
Explanation:
C If wave traveling is y-direction \& corresponding $\overrightarrow{\mathrm{E}}$ vector is in $\mathrm{x}$-axis. Then direction of $\overrightarrow{\mathrm{B}}$ vector will be in $\mathrm{z}$-direction. $\therefore \overrightarrow{\mathrm{E}} \& \overrightarrow{\mathrm{B}}$ are perpendicular to each other.
UPSEE - 2005
Electromagnetic Wave
155558
The dielectric constant of air is 1.006. The speed of electromagnetic wave travelling in air is a $\times 10^{8} \mathrm{~ms}^{-1}$, where $a$ is about
1 3
2 3.88
3 2.5
4 3.2
5 2.8
Explanation:
A Given, Dielectric constant of air $\left(\varepsilon_{\mathrm{r}}\right)=1.006 \sqcup 1$ $\because$ Speed of light in a medium $\mathrm{v}=\frac{1}{\sqrt{\varepsilon \mu}}$ $\mathrm{v}=\frac{1}{\sqrt{\varepsilon_{\mathrm{r}} \varepsilon_{\mathrm{o}} \mu_{\mathrm{r}} \mu_{\mathrm{o}}}}$ For air $\mu_{\mathrm{r}}=1$ $\therefore \quad \mathrm{v}=\frac{1}{\sqrt{1 \times 8.85 \times 10^{-12} \times 4 \pi \times 10^{-7}}}$ $\mathrm{a} \times 10^{8}=305078786$ $\mathrm{a} \times 10^{8} \sqcup 3 \times 10^{8} \Rightarrow \mathrm{a}=3$
COMEDK 2016
Electromagnetic Wave
155559
The amplitude of the sinusoidially oscillating electric field of a plane wave is $60 \mathrm{~V} / \mathrm{m}$. Then the amplitude of magnetic field is :
1 $2 \times 10^{2} \mathrm{~T}$
2 $6 \times 10^{7} \mathrm{~T}$
3 $6 \times 10^{2} \mathrm{~T}$
4 $2 \times 10^{-7} \mathrm{~T}$
5 $3 \times 10^{8} \mathrm{~T}$
Explanation:
D Given, Electric field $(\mathrm{E})=60 \mathrm{~V} / \mathrm{m}$ We know, $\mathrm{c} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{B} =\frac{\mathrm{E}}{\mathrm{c}}=\frac{60}{3 \times 10^{8}} \quad\left(\because \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ $\mathrm{B} =20 \times 10^{-8}$ $\mathrm{~B} =2 \times 10^{-7} \mathrm{~T}$
Kerala CEE 2006
Electromagnetic Wave
155561
A plane electromagnetic wave of frequency 50 $\mathrm{MHz}$ travels in free space along the $\mathrm{X}$ direction. At a particular point in space $E=$ $7.2 \hat{\mathbf{j}} \mathrm{V} / \mathrm{m}$. At this point, $B$ is equal to
155546
Light wave is travelling along $\mathbf{y}$ - directions. If the corresponding $\vec{E}$ vector at any time is along the $x$ - axis, the direction of $\vec{B}$ vector at that time is along :
1 $y$ - axis
2 $x$ - axis
3 $+\mathrm{z}-$ axis
4 $-\mathrm{z}$ - axis
Explanation:
C If wave traveling is y-direction \& corresponding $\overrightarrow{\mathrm{E}}$ vector is in $\mathrm{x}$-axis. Then direction of $\overrightarrow{\mathrm{B}}$ vector will be in $\mathrm{z}$-direction. $\therefore \overrightarrow{\mathrm{E}} \& \overrightarrow{\mathrm{B}}$ are perpendicular to each other.
UPSEE - 2005
Electromagnetic Wave
155558
The dielectric constant of air is 1.006. The speed of electromagnetic wave travelling in air is a $\times 10^{8} \mathrm{~ms}^{-1}$, where $a$ is about
1 3
2 3.88
3 2.5
4 3.2
5 2.8
Explanation:
A Given, Dielectric constant of air $\left(\varepsilon_{\mathrm{r}}\right)=1.006 \sqcup 1$ $\because$ Speed of light in a medium $\mathrm{v}=\frac{1}{\sqrt{\varepsilon \mu}}$ $\mathrm{v}=\frac{1}{\sqrt{\varepsilon_{\mathrm{r}} \varepsilon_{\mathrm{o}} \mu_{\mathrm{r}} \mu_{\mathrm{o}}}}$ For air $\mu_{\mathrm{r}}=1$ $\therefore \quad \mathrm{v}=\frac{1}{\sqrt{1 \times 8.85 \times 10^{-12} \times 4 \pi \times 10^{-7}}}$ $\mathrm{a} \times 10^{8}=305078786$ $\mathrm{a} \times 10^{8} \sqcup 3 \times 10^{8} \Rightarrow \mathrm{a}=3$
COMEDK 2016
Electromagnetic Wave
155559
The amplitude of the sinusoidially oscillating electric field of a plane wave is $60 \mathrm{~V} / \mathrm{m}$. Then the amplitude of magnetic field is :
1 $2 \times 10^{2} \mathrm{~T}$
2 $6 \times 10^{7} \mathrm{~T}$
3 $6 \times 10^{2} \mathrm{~T}$
4 $2 \times 10^{-7} \mathrm{~T}$
5 $3 \times 10^{8} \mathrm{~T}$
Explanation:
D Given, Electric field $(\mathrm{E})=60 \mathrm{~V} / \mathrm{m}$ We know, $\mathrm{c} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{B} =\frac{\mathrm{E}}{\mathrm{c}}=\frac{60}{3 \times 10^{8}} \quad\left(\because \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ $\mathrm{B} =20 \times 10^{-8}$ $\mathrm{~B} =2 \times 10^{-7} \mathrm{~T}$
Kerala CEE 2006
Electromagnetic Wave
155561
A plane electromagnetic wave of frequency 50 $\mathrm{MHz}$ travels in free space along the $\mathrm{X}$ direction. At a particular point in space $E=$ $7.2 \hat{\mathbf{j}} \mathrm{V} / \mathrm{m}$. At this point, $B$ is equal to
155546
Light wave is travelling along $\mathbf{y}$ - directions. If the corresponding $\vec{E}$ vector at any time is along the $x$ - axis, the direction of $\vec{B}$ vector at that time is along :
1 $y$ - axis
2 $x$ - axis
3 $+\mathrm{z}-$ axis
4 $-\mathrm{z}$ - axis
Explanation:
C If wave traveling is y-direction \& corresponding $\overrightarrow{\mathrm{E}}$ vector is in $\mathrm{x}$-axis. Then direction of $\overrightarrow{\mathrm{B}}$ vector will be in $\mathrm{z}$-direction. $\therefore \overrightarrow{\mathrm{E}} \& \overrightarrow{\mathrm{B}}$ are perpendicular to each other.
UPSEE - 2005
Electromagnetic Wave
155558
The dielectric constant of air is 1.006. The speed of electromagnetic wave travelling in air is a $\times 10^{8} \mathrm{~ms}^{-1}$, where $a$ is about
1 3
2 3.88
3 2.5
4 3.2
5 2.8
Explanation:
A Given, Dielectric constant of air $\left(\varepsilon_{\mathrm{r}}\right)=1.006 \sqcup 1$ $\because$ Speed of light in a medium $\mathrm{v}=\frac{1}{\sqrt{\varepsilon \mu}}$ $\mathrm{v}=\frac{1}{\sqrt{\varepsilon_{\mathrm{r}} \varepsilon_{\mathrm{o}} \mu_{\mathrm{r}} \mu_{\mathrm{o}}}}$ For air $\mu_{\mathrm{r}}=1$ $\therefore \quad \mathrm{v}=\frac{1}{\sqrt{1 \times 8.85 \times 10^{-12} \times 4 \pi \times 10^{-7}}}$ $\mathrm{a} \times 10^{8}=305078786$ $\mathrm{a} \times 10^{8} \sqcup 3 \times 10^{8} \Rightarrow \mathrm{a}=3$
COMEDK 2016
Electromagnetic Wave
155559
The amplitude of the sinusoidially oscillating electric field of a plane wave is $60 \mathrm{~V} / \mathrm{m}$. Then the amplitude of magnetic field is :
1 $2 \times 10^{2} \mathrm{~T}$
2 $6 \times 10^{7} \mathrm{~T}$
3 $6 \times 10^{2} \mathrm{~T}$
4 $2 \times 10^{-7} \mathrm{~T}$
5 $3 \times 10^{8} \mathrm{~T}$
Explanation:
D Given, Electric field $(\mathrm{E})=60 \mathrm{~V} / \mathrm{m}$ We know, $\mathrm{c} =\frac{\mathrm{E}}{\mathrm{B}}$ $\mathrm{B} =\frac{\mathrm{E}}{\mathrm{c}}=\frac{60}{3 \times 10^{8}} \quad\left(\because \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$ $\mathrm{B} =20 \times 10^{-8}$ $\mathrm{~B} =2 \times 10^{-7} \mathrm{~T}$
Kerala CEE 2006
Electromagnetic Wave
155561
A plane electromagnetic wave of frequency 50 $\mathrm{MHz}$ travels in free space along the $\mathrm{X}$ direction. At a particular point in space $E=$ $7.2 \hat{\mathbf{j}} \mathrm{V} / \mathrm{m}$. At this point, $B$ is equal to
