JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
The magnetic field in a plane electromagnetic wave is \(\mathrm{B}_y=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 \mathrm{x}+0.5\right.\) \(\left.\times 10^{11} \mathrm{t}\right) \mathrm{T}\). The corresponding electric field will be _______.
- A \(\mathrm{E}_{\mathrm{y}}=1.17 \sin \left(1.5 \times 10^3 \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^1\)
- B \(\mathrm{E}_{\mathrm{z}}=105 \sin \left(1.5 \times 10^3 \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}\)
- C \(\mathrm{E}_z=1.17 \sin \left(1.5 \times 10^5 \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}\)
- D \(\mathrm{E}_{\mathrm{y}}=10.5 \sin \left(1.5 \times 10^3 \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{E}_{\mathrm{z}}=105 \sin \left(1.5 \times 10^3 \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right) \mathrm{Vm}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{E}_0=\mathrm{B}_0 \mathrm{C}\) \(\mathrm{E}_0=3 \times 10^8 \times\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 \mathrm{x}+0.5 \times 10^{11} \mathrm{t}\right)\)…
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