155562
A plane electromagnetic wave propagating in $x$ $(-)$ direction as a wave function (in SI units) is given as $\psi(x, t)=10^{3} \sin \pi\left(3 \times 10^{6} x-9 \times 10^{14} t\right)$. The speed of the wave is
1 $3 \times 10^{6} \mathrm{~m} / \mathrm{s}$
2 $3 \times 10^{7} \mathrm{~m} / \mathrm{s}$
3 $3 \times 10^{8} \mathrm{~m} / \mathrm{s}$
4 $9 \times 10^{14} \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, $\psi(\mathrm{x}, \mathrm{t})=10^{3} \sin \pi\left(3 \times 10^{6} \mathrm{x}-9 \times 10^{14} \mathrm{t}\right)$ General equation, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin (\mathrm{kx}-\omega \mathrm{t})$ On comparison equation (i) and (ii) we get- $\mathrm{k} =3 \times 10^{6} \mathrm{~m}^{-1}$ $\omega =9 \times 10^{14} \mathrm{rad} / \mathrm{sec}$ $\therefore \quad \mathrm{v} =\frac{\omega}{\mathrm{K}}=\frac{9 \times 10^{14}}{3 \times 10^{6}}$ $\mathrm{v} =3 \times 10^{8} \mathrm{~m} / \mathrm{s}$
JCECE-2013
Electromagnetic Wave
155566
An electromagnetic wave of frequency $v=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with relative permittivity $\in=4.0$. Then
1 Wavelength is doubled and frequency becomes half
2 Wavelength is halved and frequency remains unchanged
3 Wavelength is frequency both remain unchanged
4 Wavelength is doubled and frequency unchanged.
Explanation:
B Velocity wave in medium $\mathrm{v}_{\text {medium }}=\frac{1}{\sqrt{\mu_{\mathrm{o}} \mu_{\mathrm{r}} \varepsilon_{\mathrm{o}} \varepsilon_{\mathrm{r}}}}$ For air, $\mu_{\mathrm{r}}=1$ $\mathrm{v}_{\text {medium }}=\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}} 4}}=\frac{1}{2} \frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}$ We know that, $c=\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}$ (in vacuum) $\mathrm{v}_{\text {medium }}=\frac{\mathrm{c}}{2}$ $\frac{\lambda_{\text {medium }}}{\lambda_{\text {air }}}=\frac{\mathrm{v}_{\text {medium }}}{\mathrm{v}_{\text {air }}}=\frac{\mathrm{c} / 2}{\mathrm{c}}=\frac{1}{2}$ $\lambda_{\text {medium }}=\frac{\lambda_{\text {air }}}{2}$ Wavelength is halved and frequency remains unchanged.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B Oscillating charge between dipolar electric field produces EM wave. $\therefore$ Assertion is correct. Electromagnetic wave is produce by a accelerated charge particle. Reason is also correct. But it is not correct explanation of assertion.
AIIMS-2007
Electromagnetic Wave
155573
The magnetic field amplitude of an electromagnetic wave is $2 \times 10^{-7} \mathrm{~T}$. Its electric field amplitude, if the wave is travelling in free space is :
1 $6 \mathrm{Vm}^{-1}$
2 $60 \mathrm{Vm}^{-1}$
3 $\frac{1}{6} \mathrm{Vm}^{-1}$
4 none of these
Explanation:
B Given, Magnetic field $\mathrm{B}_{\mathrm{o}}=2 \times 10^{-7} \mathrm{~T}$ We know that, $\mathrm{E}_{\mathrm{o}}=\mathrm{cB}_{\mathrm{o}}$ $\mathrm{E}_{\mathrm{o}}=3 \times 10^{8} \times 2 \times 10^{-7}$ $\mathrm{E}_{\mathrm{o}}=6 \times 10=60 \mathrm{~V} / \mathrm{m}$
BCECE-2006
Electromagnetic Wave
155577
Find the value of magnetic field between plates of capacitor at a distance $1 \mathrm{~m}$ from centre, where electric field varies by $10^{10} \mathrm{~V} / \mathrm{m}$ per second.
1 $5.56 \times 10^{-8} \mathrm{~T}$
2 $5.56 \times 10^{-3} \mathrm{~T}$
3 $5.56 \mu \mathrm{T}$
4 $5.55 \mathrm{~T}$
Explanation:
A Given, $\frac{\mathrm{dE}}{\mathrm{dt}}=10^{10} \mathrm{~V} / \mathrm{m}$ Magnetic field between plates of capacitor $\mathrm{B} =\frac{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}{2} \cdot \frac{\mathrm{dE}}{\mathrm{dt}}$ $\therefore \quad \mathrm{B} =\frac{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12} \times 10^{10}}{2}$ $\mathrm{~B} =5.56 \times 10^{-8} \mathrm{~T}$
155562
A plane electromagnetic wave propagating in $x$ $(-)$ direction as a wave function (in SI units) is given as $\psi(x, t)=10^{3} \sin \pi\left(3 \times 10^{6} x-9 \times 10^{14} t\right)$. The speed of the wave is
1 $3 \times 10^{6} \mathrm{~m} / \mathrm{s}$
2 $3 \times 10^{7} \mathrm{~m} / \mathrm{s}$
3 $3 \times 10^{8} \mathrm{~m} / \mathrm{s}$
4 $9 \times 10^{14} \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, $\psi(\mathrm{x}, \mathrm{t})=10^{3} \sin \pi\left(3 \times 10^{6} \mathrm{x}-9 \times 10^{14} \mathrm{t}\right)$ General equation, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin (\mathrm{kx}-\omega \mathrm{t})$ On comparison equation (i) and (ii) we get- $\mathrm{k} =3 \times 10^{6} \mathrm{~m}^{-1}$ $\omega =9 \times 10^{14} \mathrm{rad} / \mathrm{sec}$ $\therefore \quad \mathrm{v} =\frac{\omega}{\mathrm{K}}=\frac{9 \times 10^{14}}{3 \times 10^{6}}$ $\mathrm{v} =3 \times 10^{8} \mathrm{~m} / \mathrm{s}$
JCECE-2013
Electromagnetic Wave
155566
An electromagnetic wave of frequency $v=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with relative permittivity $\in=4.0$. Then
1 Wavelength is doubled and frequency becomes half
2 Wavelength is halved and frequency remains unchanged
3 Wavelength is frequency both remain unchanged
4 Wavelength is doubled and frequency unchanged.
Explanation:
B Velocity wave in medium $\mathrm{v}_{\text {medium }}=\frac{1}{\sqrt{\mu_{\mathrm{o}} \mu_{\mathrm{r}} \varepsilon_{\mathrm{o}} \varepsilon_{\mathrm{r}}}}$ For air, $\mu_{\mathrm{r}}=1$ $\mathrm{v}_{\text {medium }}=\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}} 4}}=\frac{1}{2} \frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}$ We know that, $c=\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}$ (in vacuum) $\mathrm{v}_{\text {medium }}=\frac{\mathrm{c}}{2}$ $\frac{\lambda_{\text {medium }}}{\lambda_{\text {air }}}=\frac{\mathrm{v}_{\text {medium }}}{\mathrm{v}_{\text {air }}}=\frac{\mathrm{c} / 2}{\mathrm{c}}=\frac{1}{2}$ $\lambda_{\text {medium }}=\frac{\lambda_{\text {air }}}{2}$ Wavelength is halved and frequency remains unchanged.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B Oscillating charge between dipolar electric field produces EM wave. $\therefore$ Assertion is correct. Electromagnetic wave is produce by a accelerated charge particle. Reason is also correct. But it is not correct explanation of assertion.
AIIMS-2007
Electromagnetic Wave
155573
The magnetic field amplitude of an electromagnetic wave is $2 \times 10^{-7} \mathrm{~T}$. Its electric field amplitude, if the wave is travelling in free space is :
1 $6 \mathrm{Vm}^{-1}$
2 $60 \mathrm{Vm}^{-1}$
3 $\frac{1}{6} \mathrm{Vm}^{-1}$
4 none of these
Explanation:
B Given, Magnetic field $\mathrm{B}_{\mathrm{o}}=2 \times 10^{-7} \mathrm{~T}$ We know that, $\mathrm{E}_{\mathrm{o}}=\mathrm{cB}_{\mathrm{o}}$ $\mathrm{E}_{\mathrm{o}}=3 \times 10^{8} \times 2 \times 10^{-7}$ $\mathrm{E}_{\mathrm{o}}=6 \times 10=60 \mathrm{~V} / \mathrm{m}$
BCECE-2006
Electromagnetic Wave
155577
Find the value of magnetic field between plates of capacitor at a distance $1 \mathrm{~m}$ from centre, where electric field varies by $10^{10} \mathrm{~V} / \mathrm{m}$ per second.
1 $5.56 \times 10^{-8} \mathrm{~T}$
2 $5.56 \times 10^{-3} \mathrm{~T}$
3 $5.56 \mu \mathrm{T}$
4 $5.55 \mathrm{~T}$
Explanation:
A Given, $\frac{\mathrm{dE}}{\mathrm{dt}}=10^{10} \mathrm{~V} / \mathrm{m}$ Magnetic field between plates of capacitor $\mathrm{B} =\frac{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}{2} \cdot \frac{\mathrm{dE}}{\mathrm{dt}}$ $\therefore \quad \mathrm{B} =\frac{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12} \times 10^{10}}{2}$ $\mathrm{~B} =5.56 \times 10^{-8} \mathrm{~T}$
155562
A plane electromagnetic wave propagating in $x$ $(-)$ direction as a wave function (in SI units) is given as $\psi(x, t)=10^{3} \sin \pi\left(3 \times 10^{6} x-9 \times 10^{14} t\right)$. The speed of the wave is
1 $3 \times 10^{6} \mathrm{~m} / \mathrm{s}$
2 $3 \times 10^{7} \mathrm{~m} / \mathrm{s}$
3 $3 \times 10^{8} \mathrm{~m} / \mathrm{s}$
4 $9 \times 10^{14} \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, $\psi(\mathrm{x}, \mathrm{t})=10^{3} \sin \pi\left(3 \times 10^{6} \mathrm{x}-9 \times 10^{14} \mathrm{t}\right)$ General equation, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin (\mathrm{kx}-\omega \mathrm{t})$ On comparison equation (i) and (ii) we get- $\mathrm{k} =3 \times 10^{6} \mathrm{~m}^{-1}$ $\omega =9 \times 10^{14} \mathrm{rad} / \mathrm{sec}$ $\therefore \quad \mathrm{v} =\frac{\omega}{\mathrm{K}}=\frac{9 \times 10^{14}}{3 \times 10^{6}}$ $\mathrm{v} =3 \times 10^{8} \mathrm{~m} / \mathrm{s}$
JCECE-2013
Electromagnetic Wave
155566
An electromagnetic wave of frequency $v=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with relative permittivity $\in=4.0$. Then
1 Wavelength is doubled and frequency becomes half
2 Wavelength is halved and frequency remains unchanged
3 Wavelength is frequency both remain unchanged
4 Wavelength is doubled and frequency unchanged.
Explanation:
B Velocity wave in medium $\mathrm{v}_{\text {medium }}=\frac{1}{\sqrt{\mu_{\mathrm{o}} \mu_{\mathrm{r}} \varepsilon_{\mathrm{o}} \varepsilon_{\mathrm{r}}}}$ For air, $\mu_{\mathrm{r}}=1$ $\mathrm{v}_{\text {medium }}=\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}} 4}}=\frac{1}{2} \frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}$ We know that, $c=\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}$ (in vacuum) $\mathrm{v}_{\text {medium }}=\frac{\mathrm{c}}{2}$ $\frac{\lambda_{\text {medium }}}{\lambda_{\text {air }}}=\frac{\mathrm{v}_{\text {medium }}}{\mathrm{v}_{\text {air }}}=\frac{\mathrm{c} / 2}{\mathrm{c}}=\frac{1}{2}$ $\lambda_{\text {medium }}=\frac{\lambda_{\text {air }}}{2}$ Wavelength is halved and frequency remains unchanged.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B Oscillating charge between dipolar electric field produces EM wave. $\therefore$ Assertion is correct. Electromagnetic wave is produce by a accelerated charge particle. Reason is also correct. But it is not correct explanation of assertion.
AIIMS-2007
Electromagnetic Wave
155573
The magnetic field amplitude of an electromagnetic wave is $2 \times 10^{-7} \mathrm{~T}$. Its electric field amplitude, if the wave is travelling in free space is :
1 $6 \mathrm{Vm}^{-1}$
2 $60 \mathrm{Vm}^{-1}$
3 $\frac{1}{6} \mathrm{Vm}^{-1}$
4 none of these
Explanation:
B Given, Magnetic field $\mathrm{B}_{\mathrm{o}}=2 \times 10^{-7} \mathrm{~T}$ We know that, $\mathrm{E}_{\mathrm{o}}=\mathrm{cB}_{\mathrm{o}}$ $\mathrm{E}_{\mathrm{o}}=3 \times 10^{8} \times 2 \times 10^{-7}$ $\mathrm{E}_{\mathrm{o}}=6 \times 10=60 \mathrm{~V} / \mathrm{m}$
BCECE-2006
Electromagnetic Wave
155577
Find the value of magnetic field between plates of capacitor at a distance $1 \mathrm{~m}$ from centre, where electric field varies by $10^{10} \mathrm{~V} / \mathrm{m}$ per second.
1 $5.56 \times 10^{-8} \mathrm{~T}$
2 $5.56 \times 10^{-3} \mathrm{~T}$
3 $5.56 \mu \mathrm{T}$
4 $5.55 \mathrm{~T}$
Explanation:
A Given, $\frac{\mathrm{dE}}{\mathrm{dt}}=10^{10} \mathrm{~V} / \mathrm{m}$ Magnetic field between plates of capacitor $\mathrm{B} =\frac{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}{2} \cdot \frac{\mathrm{dE}}{\mathrm{dt}}$ $\therefore \quad \mathrm{B} =\frac{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12} \times 10^{10}}{2}$ $\mathrm{~B} =5.56 \times 10^{-8} \mathrm{~T}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Electromagnetic Wave
155562
A plane electromagnetic wave propagating in $x$ $(-)$ direction as a wave function (in SI units) is given as $\psi(x, t)=10^{3} \sin \pi\left(3 \times 10^{6} x-9 \times 10^{14} t\right)$. The speed of the wave is
1 $3 \times 10^{6} \mathrm{~m} / \mathrm{s}$
2 $3 \times 10^{7} \mathrm{~m} / \mathrm{s}$
3 $3 \times 10^{8} \mathrm{~m} / \mathrm{s}$
4 $9 \times 10^{14} \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, $\psi(\mathrm{x}, \mathrm{t})=10^{3} \sin \pi\left(3 \times 10^{6} \mathrm{x}-9 \times 10^{14} \mathrm{t}\right)$ General equation, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin (\mathrm{kx}-\omega \mathrm{t})$ On comparison equation (i) and (ii) we get- $\mathrm{k} =3 \times 10^{6} \mathrm{~m}^{-1}$ $\omega =9 \times 10^{14} \mathrm{rad} / \mathrm{sec}$ $\therefore \quad \mathrm{v} =\frac{\omega}{\mathrm{K}}=\frac{9 \times 10^{14}}{3 \times 10^{6}}$ $\mathrm{v} =3 \times 10^{8} \mathrm{~m} / \mathrm{s}$
JCECE-2013
Electromagnetic Wave
155566
An electromagnetic wave of frequency $v=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with relative permittivity $\in=4.0$. Then
1 Wavelength is doubled and frequency becomes half
2 Wavelength is halved and frequency remains unchanged
3 Wavelength is frequency both remain unchanged
4 Wavelength is doubled and frequency unchanged.
Explanation:
B Velocity wave in medium $\mathrm{v}_{\text {medium }}=\frac{1}{\sqrt{\mu_{\mathrm{o}} \mu_{\mathrm{r}} \varepsilon_{\mathrm{o}} \varepsilon_{\mathrm{r}}}}$ For air, $\mu_{\mathrm{r}}=1$ $\mathrm{v}_{\text {medium }}=\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}} 4}}=\frac{1}{2} \frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}$ We know that, $c=\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}$ (in vacuum) $\mathrm{v}_{\text {medium }}=\frac{\mathrm{c}}{2}$ $\frac{\lambda_{\text {medium }}}{\lambda_{\text {air }}}=\frac{\mathrm{v}_{\text {medium }}}{\mathrm{v}_{\text {air }}}=\frac{\mathrm{c} / 2}{\mathrm{c}}=\frac{1}{2}$ $\lambda_{\text {medium }}=\frac{\lambda_{\text {air }}}{2}$ Wavelength is halved and frequency remains unchanged.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B Oscillating charge between dipolar electric field produces EM wave. $\therefore$ Assertion is correct. Electromagnetic wave is produce by a accelerated charge particle. Reason is also correct. But it is not correct explanation of assertion.
AIIMS-2007
Electromagnetic Wave
155573
The magnetic field amplitude of an electromagnetic wave is $2 \times 10^{-7} \mathrm{~T}$. Its electric field amplitude, if the wave is travelling in free space is :
1 $6 \mathrm{Vm}^{-1}$
2 $60 \mathrm{Vm}^{-1}$
3 $\frac{1}{6} \mathrm{Vm}^{-1}$
4 none of these
Explanation:
B Given, Magnetic field $\mathrm{B}_{\mathrm{o}}=2 \times 10^{-7} \mathrm{~T}$ We know that, $\mathrm{E}_{\mathrm{o}}=\mathrm{cB}_{\mathrm{o}}$ $\mathrm{E}_{\mathrm{o}}=3 \times 10^{8} \times 2 \times 10^{-7}$ $\mathrm{E}_{\mathrm{o}}=6 \times 10=60 \mathrm{~V} / \mathrm{m}$
BCECE-2006
Electromagnetic Wave
155577
Find the value of magnetic field between plates of capacitor at a distance $1 \mathrm{~m}$ from centre, where electric field varies by $10^{10} \mathrm{~V} / \mathrm{m}$ per second.
1 $5.56 \times 10^{-8} \mathrm{~T}$
2 $5.56 \times 10^{-3} \mathrm{~T}$
3 $5.56 \mu \mathrm{T}$
4 $5.55 \mathrm{~T}$
Explanation:
A Given, $\frac{\mathrm{dE}}{\mathrm{dt}}=10^{10} \mathrm{~V} / \mathrm{m}$ Magnetic field between plates of capacitor $\mathrm{B} =\frac{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}{2} \cdot \frac{\mathrm{dE}}{\mathrm{dt}}$ $\therefore \quad \mathrm{B} =\frac{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12} \times 10^{10}}{2}$ $\mathrm{~B} =5.56 \times 10^{-8} \mathrm{~T}$
155562
A plane electromagnetic wave propagating in $x$ $(-)$ direction as a wave function (in SI units) is given as $\psi(x, t)=10^{3} \sin \pi\left(3 \times 10^{6} x-9 \times 10^{14} t\right)$. The speed of the wave is
1 $3 \times 10^{6} \mathrm{~m} / \mathrm{s}$
2 $3 \times 10^{7} \mathrm{~m} / \mathrm{s}$
3 $3 \times 10^{8} \mathrm{~m} / \mathrm{s}$
4 $9 \times 10^{14} \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, $\psi(\mathrm{x}, \mathrm{t})=10^{3} \sin \pi\left(3 \times 10^{6} \mathrm{x}-9 \times 10^{14} \mathrm{t}\right)$ General equation, $\mathrm{E}=\mathrm{E}_{\mathrm{o}} \sin (\mathrm{kx}-\omega \mathrm{t})$ On comparison equation (i) and (ii) we get- $\mathrm{k} =3 \times 10^{6} \mathrm{~m}^{-1}$ $\omega =9 \times 10^{14} \mathrm{rad} / \mathrm{sec}$ $\therefore \quad \mathrm{v} =\frac{\omega}{\mathrm{K}}=\frac{9 \times 10^{14}}{3 \times 10^{6}}$ $\mathrm{v} =3 \times 10^{8} \mathrm{~m} / \mathrm{s}$
JCECE-2013
Electromagnetic Wave
155566
An electromagnetic wave of frequency $v=3.0 \mathrm{MHz}$ passes from vacuum into a dielectric medium with relative permittivity $\in=4.0$. Then
1 Wavelength is doubled and frequency becomes half
2 Wavelength is halved and frequency remains unchanged
3 Wavelength is frequency both remain unchanged
4 Wavelength is doubled and frequency unchanged.
Explanation:
B Velocity wave in medium $\mathrm{v}_{\text {medium }}=\frac{1}{\sqrt{\mu_{\mathrm{o}} \mu_{\mathrm{r}} \varepsilon_{\mathrm{o}} \varepsilon_{\mathrm{r}}}}$ For air, $\mu_{\mathrm{r}}=1$ $\mathrm{v}_{\text {medium }}=\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}} 4}}=\frac{1}{2} \frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}$ We know that, $c=\frac{1}{\sqrt{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}}$ (in vacuum) $\mathrm{v}_{\text {medium }}=\frac{\mathrm{c}}{2}$ $\frac{\lambda_{\text {medium }}}{\lambda_{\text {air }}}=\frac{\mathrm{v}_{\text {medium }}}{\mathrm{v}_{\text {air }}}=\frac{\mathrm{c} / 2}{\mathrm{c}}=\frac{1}{2}$ $\lambda_{\text {medium }}=\frac{\lambda_{\text {air }}}{2}$ Wavelength is halved and frequency remains unchanged.
1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.
3 If the Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
5 If the Assertion is incorrect but the Reason is correct.
Explanation:
B Oscillating charge between dipolar electric field produces EM wave. $\therefore$ Assertion is correct. Electromagnetic wave is produce by a accelerated charge particle. Reason is also correct. But it is not correct explanation of assertion.
AIIMS-2007
Electromagnetic Wave
155573
The magnetic field amplitude of an electromagnetic wave is $2 \times 10^{-7} \mathrm{~T}$. Its electric field amplitude, if the wave is travelling in free space is :
1 $6 \mathrm{Vm}^{-1}$
2 $60 \mathrm{Vm}^{-1}$
3 $\frac{1}{6} \mathrm{Vm}^{-1}$
4 none of these
Explanation:
B Given, Magnetic field $\mathrm{B}_{\mathrm{o}}=2 \times 10^{-7} \mathrm{~T}$ We know that, $\mathrm{E}_{\mathrm{o}}=\mathrm{cB}_{\mathrm{o}}$ $\mathrm{E}_{\mathrm{o}}=3 \times 10^{8} \times 2 \times 10^{-7}$ $\mathrm{E}_{\mathrm{o}}=6 \times 10=60 \mathrm{~V} / \mathrm{m}$
BCECE-2006
Electromagnetic Wave
155577
Find the value of magnetic field between plates of capacitor at a distance $1 \mathrm{~m}$ from centre, where electric field varies by $10^{10} \mathrm{~V} / \mathrm{m}$ per second.
1 $5.56 \times 10^{-8} \mathrm{~T}$
2 $5.56 \times 10^{-3} \mathrm{~T}$
3 $5.56 \mu \mathrm{T}$
4 $5.55 \mathrm{~T}$
Explanation:
A Given, $\frac{\mathrm{dE}}{\mathrm{dt}}=10^{10} \mathrm{~V} / \mathrm{m}$ Magnetic field between plates of capacitor $\mathrm{B} =\frac{\mu_{\mathrm{o}} \varepsilon_{\mathrm{o}}}{2} \cdot \frac{\mathrm{dE}}{\mathrm{dt}}$ $\therefore \quad \mathrm{B} =\frac{4 \pi \times 10^{-7} \times 8.85 \times 10^{-12} \times 10^{10}}{2}$ $\mathrm{~B} =5.56 \times 10^{-8} \mathrm{~T}$