- 1(current)
Q. No.Due to the presence of an electromagnetic wave whose electric component is given by \(E=100 \sin (\omega t-k x)~\text{NC}^{-1},\) a cylinder of length \(200~\text{cm}\) holds certain amount of electromagnetic energy inside it. If another cylinder of the same length but half the diameter the previous one holds the same amount of electromagnetic energy, the magnitude of the electric field of the corresponding electromagnetic wave should be modified as:
1. \(25 \sin (\omega t-\mathrm{kx})~\text {NC}^{-1}\)
2. \(50 \sin (\omega t-k x) ~\text N C^{-1}\)
3. \(200 \sin (\omega t-\mathrm{kx})~\text {NC}^{-1}\)
4. \(400 \sin (\omega t-k x)~\text {NC}^{-1}\)
1. \(25 \sin (\omega t-\mathrm{kx})~\text {NC}^{-1}\)
2. \(50 \sin (\omega t-k x) ~\text N C^{-1}\)
3. \(200 \sin (\omega t-\mathrm{kx})~\text {NC}^{-1}\)
4. \(400 \sin (\omega t-k x)~\text {NC}^{-1}\)
Subtopic: Generation of EM Waves |
54%
Level 3: 35%-60%
Please attempt this question first.
Hints
The magnetic field of an electromagnetic wave is given by \(\vec{B}=\left(\dfrac{\sqrt{3}}{2} \hat{\imath}+\dfrac{1}{2} \hat{\jmath}\right) 30 \sin \left[\omega\left(t-\dfrac{z}{c}\right)\right] \text { (S.I. Units). } \)The corresponding electric field in S.I. units is:
1. \(\vec{E}=\left(\dfrac{\sqrt{3}}{2} \hat{\imath}-\dfrac{1}{2} \hat{\jmath}\right) 30 c \sin \left[\omega\left(t+\dfrac{z}{c}\right)\right] ~\)
2. \(\vec{{E}}=\left(\dfrac{1}{2} \hat{{i}}-\dfrac{\sqrt{3}}{2} \hat{{j}}\right) 30{c\sin}\left[\omega\left({t}-\dfrac{{z}}{{c}}\right)\right] ~\)
3. \(\vec{E}=\left(\dfrac{1}{2} \hat{\imath}+\dfrac{\sqrt{3}}{2} \hat{\jmath}\right) 30 c \sin \left[\omega\left(t+\dfrac{z}{c}\right)\right] ~\)
4. \(\vec{E}=\left(\dfrac{3}{4} \hat{\imath}+\dfrac{1}{4} \hat{\jmath}\right) 30 c \cos \left[\omega\left(t-\dfrac{z}{c}\right)\right]~\)
1. \(\vec{E}=\left(\dfrac{\sqrt{3}}{2} \hat{\imath}-\dfrac{1}{2} \hat{\jmath}\right) 30 c \sin \left[\omega\left(t+\dfrac{z}{c}\right)\right] ~\)
2. \(\vec{{E}}=\left(\dfrac{1}{2} \hat{{i}}-\dfrac{\sqrt{3}}{2} \hat{{j}}\right) 30{c\sin}\left[\omega\left({t}-\dfrac{{z}}{{c}}\right)\right] ~\)
3. \(\vec{E}=\left(\dfrac{1}{2} \hat{\imath}+\dfrac{\sqrt{3}}{2} \hat{\jmath}\right) 30 c \sin \left[\omega\left(t+\dfrac{z}{c}\right)\right] ~\)
4. \(\vec{E}=\left(\dfrac{3}{4} \hat{\imath}+\dfrac{1}{4} \hat{\jmath}\right) 30 c \cos \left[\omega\left(t-\dfrac{z}{c}\right)\right]~\)
Subtopic: Generation of EM Waves |
88%
Level 1: 80%+
Please attempt this question first.
Hints
The electric field of an electromagnetic wave in free space is
\( \vec{{E}}=57 \cos \left[7.5 \times 10^6 {t}-5 \times 10^{-3}(3{x}+4{y})\right](4 \hat{\imath}-3 \hat{\jmath})~\text {N/C} \) The associated magnetic field in Tesla is
\( \vec{{E}}=57 \cos \left[7.5 \times 10^6 {t}-5 \times 10^{-3}(3{x}+4{y})\right](4 \hat{\imath}-3 \hat{\jmath})~\text {N/C} \) The associated magnetic field in Tesla is
| 1. | \( \vec{{B}}=\dfrac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 {t}-5 \times 10^{-3}(3 {x}+4 {y})\right](5 \hat{{k}})\) |
| 2. | \( \vec{{B}}=-\dfrac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 {t}-5 \times 10^{-3}(3 \mathrm{x}+4 {y})\right](5 \hat{{k}}) \) |
| 3. | \( \vec{{B}}=-\dfrac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 {t}-5 \times 10^{-3}(3 {x}+4 {y})\right](\hat{{k}}) \) |
| 4. | \( \vec{B}=\dfrac{57}{3 \times 10^8} \cos \left[7.5 \times 10^6 t-5 \times 10^{-3}(3 x+4 y)\right](\hat{k}) \) |
Subtopic: Generation of EM Waves |
Please attempt this question first.
Hints
Please attempt this question first.
- 1(current)
Q. No.
© 2026 GoodEd Technologies Pvt. Ltd.