NEET Test Series from KOTA - 10 Papers In MS WORD
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Electromagnetic Wave
155539
If ' $c$ ' is the speed of electromagnetic waves in vacuum, then their speed in a medium of dielectric constant ' $K$ ' and relative permittivity ' $\mu_{\mathrm{r}}$ ' is
C Speed of EM wave in vacuum, $\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}=\mathrm{c}$ Speed of EM wave in medium $\frac{1}{\sqrt{\mu \varepsilon}}=\mathrm{v}$ Dividing equation (i) \& (ii), we get - $\frac{\mathrm{c}}{\mathrm{v}}=\frac{1 / \sqrt{\mu_{0} \varepsilon_{0}}}{1 / \sqrt{\varepsilon \mu}}$ $\frac{\mathrm{c}}{\mathrm{v}}=\frac{\sqrt{\varepsilon \mu}}{\sqrt{\mu_{0} \varepsilon_{0}}}=\frac{\sqrt{\varepsilon_{0} \varepsilon_{\mathrm{r}} \mu_{0} \mu_{\mathrm{r}}}}{\sqrt{\mu_{0} \varepsilon_{0}}}$ $\frac{\mathrm{c}}{\mathrm{v}}=\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{r}}} \quad\left(\because \varepsilon_{\mathrm{r}}=\mathrm{K}\right)$ $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\mathrm{K} \mu_{\mathrm{r}}}}$
Kerala CEE 2005
Electromagnetic Wave
155541
The electric field portion of an electromagnetic wave is given by (all variables in SI units ) $E=10^{-4} \sin \left(6 \times 10^{5} t-0.01 x\right)$. The frequency (f) and the speed $(v)$ of electromagnetic wave are
155543
In an electromagnetic wave, the electric and magnetic fields are $100 \mathrm{~V} / \mathrm{m}$ and $0.265 \mathrm{~A} / \mathrm{m}$. The maximum energy flow is
1 $26.5 \mathrm{~W} / \mathrm{m}^{2}$
2 $36.5 \mathrm{~W} / \mathrm{m}^{2}$
3 $46.7 \mathrm{~W} / \mathrm{m}^{2}$
4 $765 \mathrm{~W} / \mathrm{m}^{2}$
Explanation:
A Given, Electric field $\left(\mathrm{E}_{0}\right)=100 \mathrm{~V} / \mathrm{m}$ Magnetic field $\left(\mathrm{B}_{0}\right)=0.265 \mathrm{~A} / \mathrm{m}$ Maximum energy flows $\mathrm{E}_{\max } =\mathrm{E}_{0} \times \mathrm{B}_{0}$ $=100 \times 0.265$ $\mathrm{E}_{\max } =26.5 \mathrm{~W} / \mathrm{m}^{2}$
Manipal UGET-2012
Electromagnetic Wave
155550
In a plane electromagnetic wave, the electric field of amplitude $1 \mathrm{Vm}^{-1}$ varies with time in free space. The average energy density of magnetic field is (in $\mathbf{J m}^{-3}$ )
1 $8.86 \times 10^{-12}$
2 $4.43 \times 10^{-12}$
3 $17.72 \times 10^{-12}$
4 $2.21 \times 10^{-12}$
5 $1.11 \times 10^{-12}$
Explanation:
D Given, $\quad \text { Electric field }(\mathrm{E})=1 \mathrm{~V} / \mathrm{m}$ $\because \quad \quad \text { Energy density }(\mathrm{U})=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^{2}$ $\mathrm{U}=\frac{1}{2} \times 8.85 \times 10^{-12} \times(1)^{2} \quad\left(\because \varepsilon_{0}=8.85 \times 10^{-12}\right)$ $\mathrm{U}=4.43 \times 10^{-12}$ So, the average energy density of magnetic field $\mathrm{U}=\frac{4.43 \times 10^{-12}}{2}$ $\mathrm{U}=2.2162 \times 10^{-12}$ $\mathrm{U} \square 2.21 \times 10^{-12} \mathrm{~J} / \mathrm{m}^{3}$
155539
If ' $c$ ' is the speed of electromagnetic waves in vacuum, then their speed in a medium of dielectric constant ' $K$ ' and relative permittivity ' $\mu_{\mathrm{r}}$ ' is
C Speed of EM wave in vacuum, $\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}=\mathrm{c}$ Speed of EM wave in medium $\frac{1}{\sqrt{\mu \varepsilon}}=\mathrm{v}$ Dividing equation (i) \& (ii), we get - $\frac{\mathrm{c}}{\mathrm{v}}=\frac{1 / \sqrt{\mu_{0} \varepsilon_{0}}}{1 / \sqrt{\varepsilon \mu}}$ $\frac{\mathrm{c}}{\mathrm{v}}=\frac{\sqrt{\varepsilon \mu}}{\sqrt{\mu_{0} \varepsilon_{0}}}=\frac{\sqrt{\varepsilon_{0} \varepsilon_{\mathrm{r}} \mu_{0} \mu_{\mathrm{r}}}}{\sqrt{\mu_{0} \varepsilon_{0}}}$ $\frac{\mathrm{c}}{\mathrm{v}}=\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{r}}} \quad\left(\because \varepsilon_{\mathrm{r}}=\mathrm{K}\right)$ $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\mathrm{K} \mu_{\mathrm{r}}}}$
Kerala CEE 2005
Electromagnetic Wave
155541
The electric field portion of an electromagnetic wave is given by (all variables in SI units ) $E=10^{-4} \sin \left(6 \times 10^{5} t-0.01 x\right)$. The frequency (f) and the speed $(v)$ of electromagnetic wave are
155543
In an electromagnetic wave, the electric and magnetic fields are $100 \mathrm{~V} / \mathrm{m}$ and $0.265 \mathrm{~A} / \mathrm{m}$. The maximum energy flow is
1 $26.5 \mathrm{~W} / \mathrm{m}^{2}$
2 $36.5 \mathrm{~W} / \mathrm{m}^{2}$
3 $46.7 \mathrm{~W} / \mathrm{m}^{2}$
4 $765 \mathrm{~W} / \mathrm{m}^{2}$
Explanation:
A Given, Electric field $\left(\mathrm{E}_{0}\right)=100 \mathrm{~V} / \mathrm{m}$ Magnetic field $\left(\mathrm{B}_{0}\right)=0.265 \mathrm{~A} / \mathrm{m}$ Maximum energy flows $\mathrm{E}_{\max } =\mathrm{E}_{0} \times \mathrm{B}_{0}$ $=100 \times 0.265$ $\mathrm{E}_{\max } =26.5 \mathrm{~W} / \mathrm{m}^{2}$
Manipal UGET-2012
Electromagnetic Wave
155550
In a plane electromagnetic wave, the electric field of amplitude $1 \mathrm{Vm}^{-1}$ varies with time in free space. The average energy density of magnetic field is (in $\mathbf{J m}^{-3}$ )
1 $8.86 \times 10^{-12}$
2 $4.43 \times 10^{-12}$
3 $17.72 \times 10^{-12}$
4 $2.21 \times 10^{-12}$
5 $1.11 \times 10^{-12}$
Explanation:
D Given, $\quad \text { Electric field }(\mathrm{E})=1 \mathrm{~V} / \mathrm{m}$ $\because \quad \quad \text { Energy density }(\mathrm{U})=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^{2}$ $\mathrm{U}=\frac{1}{2} \times 8.85 \times 10^{-12} \times(1)^{2} \quad\left(\because \varepsilon_{0}=8.85 \times 10^{-12}\right)$ $\mathrm{U}=4.43 \times 10^{-12}$ So, the average energy density of magnetic field $\mathrm{U}=\frac{4.43 \times 10^{-12}}{2}$ $\mathrm{U}=2.2162 \times 10^{-12}$ $\mathrm{U} \square 2.21 \times 10^{-12} \mathrm{~J} / \mathrm{m}^{3}$
155539
If ' $c$ ' is the speed of electromagnetic waves in vacuum, then their speed in a medium of dielectric constant ' $K$ ' and relative permittivity ' $\mu_{\mathrm{r}}$ ' is
C Speed of EM wave in vacuum, $\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}=\mathrm{c}$ Speed of EM wave in medium $\frac{1}{\sqrt{\mu \varepsilon}}=\mathrm{v}$ Dividing equation (i) \& (ii), we get - $\frac{\mathrm{c}}{\mathrm{v}}=\frac{1 / \sqrt{\mu_{0} \varepsilon_{0}}}{1 / \sqrt{\varepsilon \mu}}$ $\frac{\mathrm{c}}{\mathrm{v}}=\frac{\sqrt{\varepsilon \mu}}{\sqrt{\mu_{0} \varepsilon_{0}}}=\frac{\sqrt{\varepsilon_{0} \varepsilon_{\mathrm{r}} \mu_{0} \mu_{\mathrm{r}}}}{\sqrt{\mu_{0} \varepsilon_{0}}}$ $\frac{\mathrm{c}}{\mathrm{v}}=\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{r}}} \quad\left(\because \varepsilon_{\mathrm{r}}=\mathrm{K}\right)$ $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\mathrm{K} \mu_{\mathrm{r}}}}$
Kerala CEE 2005
Electromagnetic Wave
155541
The electric field portion of an electromagnetic wave is given by (all variables in SI units ) $E=10^{-4} \sin \left(6 \times 10^{5} t-0.01 x\right)$. The frequency (f) and the speed $(v)$ of electromagnetic wave are
155543
In an electromagnetic wave, the electric and magnetic fields are $100 \mathrm{~V} / \mathrm{m}$ and $0.265 \mathrm{~A} / \mathrm{m}$. The maximum energy flow is
1 $26.5 \mathrm{~W} / \mathrm{m}^{2}$
2 $36.5 \mathrm{~W} / \mathrm{m}^{2}$
3 $46.7 \mathrm{~W} / \mathrm{m}^{2}$
4 $765 \mathrm{~W} / \mathrm{m}^{2}$
Explanation:
A Given, Electric field $\left(\mathrm{E}_{0}\right)=100 \mathrm{~V} / \mathrm{m}$ Magnetic field $\left(\mathrm{B}_{0}\right)=0.265 \mathrm{~A} / \mathrm{m}$ Maximum energy flows $\mathrm{E}_{\max } =\mathrm{E}_{0} \times \mathrm{B}_{0}$ $=100 \times 0.265$ $\mathrm{E}_{\max } =26.5 \mathrm{~W} / \mathrm{m}^{2}$
Manipal UGET-2012
Electromagnetic Wave
155550
In a plane electromagnetic wave, the electric field of amplitude $1 \mathrm{Vm}^{-1}$ varies with time in free space. The average energy density of magnetic field is (in $\mathbf{J m}^{-3}$ )
1 $8.86 \times 10^{-12}$
2 $4.43 \times 10^{-12}$
3 $17.72 \times 10^{-12}$
4 $2.21 \times 10^{-12}$
5 $1.11 \times 10^{-12}$
Explanation:
D Given, $\quad \text { Electric field }(\mathrm{E})=1 \mathrm{~V} / \mathrm{m}$ $\because \quad \quad \text { Energy density }(\mathrm{U})=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^{2}$ $\mathrm{U}=\frac{1}{2} \times 8.85 \times 10^{-12} \times(1)^{2} \quad\left(\because \varepsilon_{0}=8.85 \times 10^{-12}\right)$ $\mathrm{U}=4.43 \times 10^{-12}$ So, the average energy density of magnetic field $\mathrm{U}=\frac{4.43 \times 10^{-12}}{2}$ $\mathrm{U}=2.2162 \times 10^{-12}$ $\mathrm{U} \square 2.21 \times 10^{-12} \mathrm{~J} / \mathrm{m}^{3}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Electromagnetic Wave
155539
If ' $c$ ' is the speed of electromagnetic waves in vacuum, then their speed in a medium of dielectric constant ' $K$ ' and relative permittivity ' $\mu_{\mathrm{r}}$ ' is
C Speed of EM wave in vacuum, $\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}=\mathrm{c}$ Speed of EM wave in medium $\frac{1}{\sqrt{\mu \varepsilon}}=\mathrm{v}$ Dividing equation (i) \& (ii), we get - $\frac{\mathrm{c}}{\mathrm{v}}=\frac{1 / \sqrt{\mu_{0} \varepsilon_{0}}}{1 / \sqrt{\varepsilon \mu}}$ $\frac{\mathrm{c}}{\mathrm{v}}=\frac{\sqrt{\varepsilon \mu}}{\sqrt{\mu_{0} \varepsilon_{0}}}=\frac{\sqrt{\varepsilon_{0} \varepsilon_{\mathrm{r}} \mu_{0} \mu_{\mathrm{r}}}}{\sqrt{\mu_{0} \varepsilon_{0}}}$ $\frac{\mathrm{c}}{\mathrm{v}}=\sqrt{\mu_{\mathrm{r}} \varepsilon_{\mathrm{r}}} \quad\left(\because \varepsilon_{\mathrm{r}}=\mathrm{K}\right)$ $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\mathrm{K} \mu_{\mathrm{r}}}}$
Kerala CEE 2005
Electromagnetic Wave
155541
The electric field portion of an electromagnetic wave is given by (all variables in SI units ) $E=10^{-4} \sin \left(6 \times 10^{5} t-0.01 x\right)$. The frequency (f) and the speed $(v)$ of electromagnetic wave are
155543
In an electromagnetic wave, the electric and magnetic fields are $100 \mathrm{~V} / \mathrm{m}$ and $0.265 \mathrm{~A} / \mathrm{m}$. The maximum energy flow is
1 $26.5 \mathrm{~W} / \mathrm{m}^{2}$
2 $36.5 \mathrm{~W} / \mathrm{m}^{2}$
3 $46.7 \mathrm{~W} / \mathrm{m}^{2}$
4 $765 \mathrm{~W} / \mathrm{m}^{2}$
Explanation:
A Given, Electric field $\left(\mathrm{E}_{0}\right)=100 \mathrm{~V} / \mathrm{m}$ Magnetic field $\left(\mathrm{B}_{0}\right)=0.265 \mathrm{~A} / \mathrm{m}$ Maximum energy flows $\mathrm{E}_{\max } =\mathrm{E}_{0} \times \mathrm{B}_{0}$ $=100 \times 0.265$ $\mathrm{E}_{\max } =26.5 \mathrm{~W} / \mathrm{m}^{2}$
Manipal UGET-2012
Electromagnetic Wave
155550
In a plane electromagnetic wave, the electric field of amplitude $1 \mathrm{Vm}^{-1}$ varies with time in free space. The average energy density of magnetic field is (in $\mathbf{J m}^{-3}$ )
1 $8.86 \times 10^{-12}$
2 $4.43 \times 10^{-12}$
3 $17.72 \times 10^{-12}$
4 $2.21 \times 10^{-12}$
5 $1.11 \times 10^{-12}$
Explanation:
D Given, $\quad \text { Electric field }(\mathrm{E})=1 \mathrm{~V} / \mathrm{m}$ $\because \quad \quad \text { Energy density }(\mathrm{U})=\frac{1}{2} \varepsilon_{\mathrm{o}} \mathrm{E}^{2}$ $\mathrm{U}=\frac{1}{2} \times 8.85 \times 10^{-12} \times(1)^{2} \quad\left(\because \varepsilon_{0}=8.85 \times 10^{-12}\right)$ $\mathrm{U}=4.43 \times 10^{-12}$ So, the average energy density of magnetic field $\mathrm{U}=\frac{4.43 \times 10^{-12}}{2}$ $\mathrm{U}=2.2162 \times 10^{-12}$ $\mathrm{U} \square 2.21 \times 10^{-12} \mathrm{~J} / \mathrm{m}^{3}$