155552
If the magnetic field of an electromagnetic wave given as $B_{y}=2 \times 10^{-7} \sin \left(10^{3} x+1.5 \times\right.$ $10^{12}$ t) tesla, the wavelength of the electromagnetic wave is
1 $0.314 \mathrm{~mm}$
2 $0.628 \mathrm{~mm}$
3 $6.28 \mathrm{~mm}$
4 $1.26 \mathrm{~mm}$
5 $0.0628 \mathrm{~mm}$
Explanation:
C Given, $\mathrm{B}_{\mathrm{y}}=2 \times 10^{-7} \sin \left(10^{3} \mathrm{x}+1.5 \times 10^{12} \mathrm{t}\right)$ General equation $\mathrm{B}_{\mathrm{y}}=\mathrm{B}_{\mathrm{o}} \sin (\mathrm{kx}+\omega \mathrm{t})$ On comparing with general equation $\mathrm{k} =10^{3}$ $\mathrm{k} =\frac{2 \pi}{\lambda}$ $\lambda =\frac{2 \pi}{\mathrm{k}} \Rightarrow \lambda=\frac{2 \pi}{10^{3}}$ $\lambda =2 \pi \times 10^{-3}=2 \times 3.14 \times 10^{-3}$ $\lambda =6.28 \times 10^{-3} \mathrm{~m} \sqcup 6.28 \mathrm{~mm}$
Kerala CEE- 2013
Electromagnetic Wave
155555
The electric field of an electromagnetic wave travelling through vacuum is given by the equation $E=E_{0} \sin (k x-\omega t)$. The quantity that is independent of wavelength is
1 $\frac{\mathrm{k}}{\omega}$
2 $\mathrm{k \omega}$
3 $\omega$
4 $\mathrm{k}$
5 $\mathrm{k}^{2} \omega$
Explanation:
A $\mathrm{E} =\mathrm{E}_{\mathrm{o}} \sin (\mathrm{kx}-\omega \mathrm{t})$ $\mathrm{k} =\frac{2 \pi}{\lambda}$ $\omega =2 \pi \mathrm{f}$ $\frac{\mathrm{k}}{\omega} =\frac{2 \pi / \lambda}{2 \pi \mathrm{f}}=\frac{1}{\mathrm{f} \lambda}=\frac{1}{\mathrm{c}} \quad(\because \mathrm{c}=\mathrm{f} \lambda)$ $\frac{\mathrm{k}}{\omega} =\frac{1}{\mathrm{c}}$ $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s} \text { is a constant value }$
Kerala CEE - 2011
Electromagnetic Wave
155556
Electromagnetic waves of frequencies higher than $9 \sqrt{2} \mathrm{MHz}$ are found to be reflected by the ionosphere on a particular day at a place. The maximum electron density in the ionosphere is
1 $\sqrt{5} \times 10^{12} \mathrm{~m}^{-3}$
2 $\sqrt{2} \times 10^{12} \mathrm{~m}^{-3}$
3 $2 \times 10^{12} \mathrm{~m}^{-3}$
4 $5 \times 10^{12} \mathrm{~m}^{-3}$
5 $3 \times 10^{12} \mathrm{~m}^{-3}$
Explanation:
C Given, Frequency of Electromagnetic wave $\mathrm{f}=9 \sqrt{2} \mathrm{MHz}$ Frequency electromagnetic wave $=9 \sqrt{d_{\max }}$ $\mathrm{d}_{\max }=$ maximum electron density $\therefore \quad 9 \sqrt{2} \times 10^{6}=9 \sqrt{d_{\max }}$ $\sqrt{2} \times 10^{6}=\sqrt{\mathrm{d}_{\max }}$ $2 \times 10^{12}=\mathrm{d}_{\max }$ $\mathrm{d}_{\max }=2 \times 10^{12} \mathrm{~m}^{-3}$
Kerala CEE - 2009
Electromagnetic Wave
155557
The refractive index and the permeability of a medium are respectively 1.5 and $5 \times 10^{-7}$ $\mathrm{Hm}^{-1}$. The relative permittivity of the medium is nearly
155552
If the magnetic field of an electromagnetic wave given as $B_{y}=2 \times 10^{-7} \sin \left(10^{3} x+1.5 \times\right.$ $10^{12}$ t) tesla, the wavelength of the electromagnetic wave is
1 $0.314 \mathrm{~mm}$
2 $0.628 \mathrm{~mm}$
3 $6.28 \mathrm{~mm}$
4 $1.26 \mathrm{~mm}$
5 $0.0628 \mathrm{~mm}$
Explanation:
C Given, $\mathrm{B}_{\mathrm{y}}=2 \times 10^{-7} \sin \left(10^{3} \mathrm{x}+1.5 \times 10^{12} \mathrm{t}\right)$ General equation $\mathrm{B}_{\mathrm{y}}=\mathrm{B}_{\mathrm{o}} \sin (\mathrm{kx}+\omega \mathrm{t})$ On comparing with general equation $\mathrm{k} =10^{3}$ $\mathrm{k} =\frac{2 \pi}{\lambda}$ $\lambda =\frac{2 \pi}{\mathrm{k}} \Rightarrow \lambda=\frac{2 \pi}{10^{3}}$ $\lambda =2 \pi \times 10^{-3}=2 \times 3.14 \times 10^{-3}$ $\lambda =6.28 \times 10^{-3} \mathrm{~m} \sqcup 6.28 \mathrm{~mm}$
Kerala CEE- 2013
Electromagnetic Wave
155555
The electric field of an electromagnetic wave travelling through vacuum is given by the equation $E=E_{0} \sin (k x-\omega t)$. The quantity that is independent of wavelength is
1 $\frac{\mathrm{k}}{\omega}$
2 $\mathrm{k \omega}$
3 $\omega$
4 $\mathrm{k}$
5 $\mathrm{k}^{2} \omega$
Explanation:
A $\mathrm{E} =\mathrm{E}_{\mathrm{o}} \sin (\mathrm{kx}-\omega \mathrm{t})$ $\mathrm{k} =\frac{2 \pi}{\lambda}$ $\omega =2 \pi \mathrm{f}$ $\frac{\mathrm{k}}{\omega} =\frac{2 \pi / \lambda}{2 \pi \mathrm{f}}=\frac{1}{\mathrm{f} \lambda}=\frac{1}{\mathrm{c}} \quad(\because \mathrm{c}=\mathrm{f} \lambda)$ $\frac{\mathrm{k}}{\omega} =\frac{1}{\mathrm{c}}$ $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s} \text { is a constant value }$
Kerala CEE - 2011
Electromagnetic Wave
155556
Electromagnetic waves of frequencies higher than $9 \sqrt{2} \mathrm{MHz}$ are found to be reflected by the ionosphere on a particular day at a place. The maximum electron density in the ionosphere is
1 $\sqrt{5} \times 10^{12} \mathrm{~m}^{-3}$
2 $\sqrt{2} \times 10^{12} \mathrm{~m}^{-3}$
3 $2 \times 10^{12} \mathrm{~m}^{-3}$
4 $5 \times 10^{12} \mathrm{~m}^{-3}$
5 $3 \times 10^{12} \mathrm{~m}^{-3}$
Explanation:
C Given, Frequency of Electromagnetic wave $\mathrm{f}=9 \sqrt{2} \mathrm{MHz}$ Frequency electromagnetic wave $=9 \sqrt{d_{\max }}$ $\mathrm{d}_{\max }=$ maximum electron density $\therefore \quad 9 \sqrt{2} \times 10^{6}=9 \sqrt{d_{\max }}$ $\sqrt{2} \times 10^{6}=\sqrt{\mathrm{d}_{\max }}$ $2 \times 10^{12}=\mathrm{d}_{\max }$ $\mathrm{d}_{\max }=2 \times 10^{12} \mathrm{~m}^{-3}$
Kerala CEE - 2009
Electromagnetic Wave
155557
The refractive index and the permeability of a medium are respectively 1.5 and $5 \times 10^{-7}$ $\mathrm{Hm}^{-1}$. The relative permittivity of the medium is nearly
155552
If the magnetic field of an electromagnetic wave given as $B_{y}=2 \times 10^{-7} \sin \left(10^{3} x+1.5 \times\right.$ $10^{12}$ t) tesla, the wavelength of the electromagnetic wave is
1 $0.314 \mathrm{~mm}$
2 $0.628 \mathrm{~mm}$
3 $6.28 \mathrm{~mm}$
4 $1.26 \mathrm{~mm}$
5 $0.0628 \mathrm{~mm}$
Explanation:
C Given, $\mathrm{B}_{\mathrm{y}}=2 \times 10^{-7} \sin \left(10^{3} \mathrm{x}+1.5 \times 10^{12} \mathrm{t}\right)$ General equation $\mathrm{B}_{\mathrm{y}}=\mathrm{B}_{\mathrm{o}} \sin (\mathrm{kx}+\omega \mathrm{t})$ On comparing with general equation $\mathrm{k} =10^{3}$ $\mathrm{k} =\frac{2 \pi}{\lambda}$ $\lambda =\frac{2 \pi}{\mathrm{k}} \Rightarrow \lambda=\frac{2 \pi}{10^{3}}$ $\lambda =2 \pi \times 10^{-3}=2 \times 3.14 \times 10^{-3}$ $\lambda =6.28 \times 10^{-3} \mathrm{~m} \sqcup 6.28 \mathrm{~mm}$
Kerala CEE- 2013
Electromagnetic Wave
155555
The electric field of an electromagnetic wave travelling through vacuum is given by the equation $E=E_{0} \sin (k x-\omega t)$. The quantity that is independent of wavelength is
1 $\frac{\mathrm{k}}{\omega}$
2 $\mathrm{k \omega}$
3 $\omega$
4 $\mathrm{k}$
5 $\mathrm{k}^{2} \omega$
Explanation:
A $\mathrm{E} =\mathrm{E}_{\mathrm{o}} \sin (\mathrm{kx}-\omega \mathrm{t})$ $\mathrm{k} =\frac{2 \pi}{\lambda}$ $\omega =2 \pi \mathrm{f}$ $\frac{\mathrm{k}}{\omega} =\frac{2 \pi / \lambda}{2 \pi \mathrm{f}}=\frac{1}{\mathrm{f} \lambda}=\frac{1}{\mathrm{c}} \quad(\because \mathrm{c}=\mathrm{f} \lambda)$ $\frac{\mathrm{k}}{\omega} =\frac{1}{\mathrm{c}}$ $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s} \text { is a constant value }$
Kerala CEE - 2011
Electromagnetic Wave
155556
Electromagnetic waves of frequencies higher than $9 \sqrt{2} \mathrm{MHz}$ are found to be reflected by the ionosphere on a particular day at a place. The maximum electron density in the ionosphere is
1 $\sqrt{5} \times 10^{12} \mathrm{~m}^{-3}$
2 $\sqrt{2} \times 10^{12} \mathrm{~m}^{-3}$
3 $2 \times 10^{12} \mathrm{~m}^{-3}$
4 $5 \times 10^{12} \mathrm{~m}^{-3}$
5 $3 \times 10^{12} \mathrm{~m}^{-3}$
Explanation:
C Given, Frequency of Electromagnetic wave $\mathrm{f}=9 \sqrt{2} \mathrm{MHz}$ Frequency electromagnetic wave $=9 \sqrt{d_{\max }}$ $\mathrm{d}_{\max }=$ maximum electron density $\therefore \quad 9 \sqrt{2} \times 10^{6}=9 \sqrt{d_{\max }}$ $\sqrt{2} \times 10^{6}=\sqrt{\mathrm{d}_{\max }}$ $2 \times 10^{12}=\mathrm{d}_{\max }$ $\mathrm{d}_{\max }=2 \times 10^{12} \mathrm{~m}^{-3}$
Kerala CEE - 2009
Electromagnetic Wave
155557
The refractive index and the permeability of a medium are respectively 1.5 and $5 \times 10^{-7}$ $\mathrm{Hm}^{-1}$. The relative permittivity of the medium is nearly
155552
If the magnetic field of an electromagnetic wave given as $B_{y}=2 \times 10^{-7} \sin \left(10^{3} x+1.5 \times\right.$ $10^{12}$ t) tesla, the wavelength of the electromagnetic wave is
1 $0.314 \mathrm{~mm}$
2 $0.628 \mathrm{~mm}$
3 $6.28 \mathrm{~mm}$
4 $1.26 \mathrm{~mm}$
5 $0.0628 \mathrm{~mm}$
Explanation:
C Given, $\mathrm{B}_{\mathrm{y}}=2 \times 10^{-7} \sin \left(10^{3} \mathrm{x}+1.5 \times 10^{12} \mathrm{t}\right)$ General equation $\mathrm{B}_{\mathrm{y}}=\mathrm{B}_{\mathrm{o}} \sin (\mathrm{kx}+\omega \mathrm{t})$ On comparing with general equation $\mathrm{k} =10^{3}$ $\mathrm{k} =\frac{2 \pi}{\lambda}$ $\lambda =\frac{2 \pi}{\mathrm{k}} \Rightarrow \lambda=\frac{2 \pi}{10^{3}}$ $\lambda =2 \pi \times 10^{-3}=2 \times 3.14 \times 10^{-3}$ $\lambda =6.28 \times 10^{-3} \mathrm{~m} \sqcup 6.28 \mathrm{~mm}$
Kerala CEE- 2013
Electromagnetic Wave
155555
The electric field of an electromagnetic wave travelling through vacuum is given by the equation $E=E_{0} \sin (k x-\omega t)$. The quantity that is independent of wavelength is
1 $\frac{\mathrm{k}}{\omega}$
2 $\mathrm{k \omega}$
3 $\omega$
4 $\mathrm{k}$
5 $\mathrm{k}^{2} \omega$
Explanation:
A $\mathrm{E} =\mathrm{E}_{\mathrm{o}} \sin (\mathrm{kx}-\omega \mathrm{t})$ $\mathrm{k} =\frac{2 \pi}{\lambda}$ $\omega =2 \pi \mathrm{f}$ $\frac{\mathrm{k}}{\omega} =\frac{2 \pi / \lambda}{2 \pi \mathrm{f}}=\frac{1}{\mathrm{f} \lambda}=\frac{1}{\mathrm{c}} \quad(\because \mathrm{c}=\mathrm{f} \lambda)$ $\frac{\mathrm{k}}{\omega} =\frac{1}{\mathrm{c}}$ $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s} \text { is a constant value }$
Kerala CEE - 2011
Electromagnetic Wave
155556
Electromagnetic waves of frequencies higher than $9 \sqrt{2} \mathrm{MHz}$ are found to be reflected by the ionosphere on a particular day at a place. The maximum electron density in the ionosphere is
1 $\sqrt{5} \times 10^{12} \mathrm{~m}^{-3}$
2 $\sqrt{2} \times 10^{12} \mathrm{~m}^{-3}$
3 $2 \times 10^{12} \mathrm{~m}^{-3}$
4 $5 \times 10^{12} \mathrm{~m}^{-3}$
5 $3 \times 10^{12} \mathrm{~m}^{-3}$
Explanation:
C Given, Frequency of Electromagnetic wave $\mathrm{f}=9 \sqrt{2} \mathrm{MHz}$ Frequency electromagnetic wave $=9 \sqrt{d_{\max }}$ $\mathrm{d}_{\max }=$ maximum electron density $\therefore \quad 9 \sqrt{2} \times 10^{6}=9 \sqrt{d_{\max }}$ $\sqrt{2} \times 10^{6}=\sqrt{\mathrm{d}_{\max }}$ $2 \times 10^{12}=\mathrm{d}_{\max }$ $\mathrm{d}_{\max }=2 \times 10^{12} \mathrm{~m}^{-3}$
Kerala CEE - 2009
Electromagnetic Wave
155557
The refractive index and the permeability of a medium are respectively 1.5 and $5 \times 10^{-7}$ $\mathrm{Hm}^{-1}$. The relative permittivity of the medium is nearly
