Question:
If the magnetic field in a plane electromagnetic wave is given by
\( \overrightarrow B \) = 3 \( \times \) 10-8 sin(1.6 \( \times \) 103x + 48 \( \times \) 1010t)\( \widehat j \) T, then what will be expression for electric field ?
If the magnetic field in a plane electromagnetic wave is given by
\( \overrightarrow B \) = 3 \( \times \) 10-8 sin(1.6 \( \times \) 103x + 48 \( \times \) 1010t)\( \widehat j \) T, then what will be expression for electric field ?
\( \overrightarrow B \) = 3 \( \times \) 10-8 sin(1.6 \( \times \) 103x + 48 \( \times \) 1010t)\( \widehat j \) T, then what will be expression for electric field ?
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In an EM wave \(E_0=cB_0\), and E, B and the propagation direction are mutually perpendicular with \(\vec E\times\vec B\) along the direction of travel.
Updated On: Oct 31, 2026
- \( \overrightarrow E \) = (9sin(1.6 \( \times \) 103x + 48 \( \times \) 1010t)\( \widehat k \) V/m)
- \( \overrightarrow E \) = (60sin(1.6 \( \times \) 103x + 48 \( \times \) 1010t)\( \widehat k \) V/m)
- \( \overrightarrow E \) = (3 \( \times \) 10-8 sin(1.6 \( \times \) 103x + 48 \( \times \) 1010t)\( \widehat i \) V/m)
- \( \overrightarrow E \) = (3 \( \times \) 10-8 sin(1.6 \( \times \) 103x + 48 \( \times \) 1010t)\( \widehat j \) V/m)
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The Correct Option is A
Solution and Explanation
\(B_0=3\times10^{-8}\) T. Wave speed \(c=\dfrac{\omega}{k}=\dfrac{48\times10^{10}}{1.6\times10^{3}}=3\times10^8\) m/s.
\(E_0=cB_0=3\times10^8\times3\times10^{-8}=9\) V/m.
The phase \(kx+\omega t\) means the wave travels along \(-\hat x\). Since \(\vec E\times\vec B\) is along the propagation direction and \(\vec B\parallel\hat j\): \(\hat k\times\hat j=-\hat i\), so \(\vec E\) is along \(\hat k\).
\(\vec E=9\sin\left(1.6\times10^3x+48\times10^{10}t\right)\hat k\) V/m.
Answer: (A)
\(E_0=cB_0=3\times10^8\times3\times10^{-8}=9\) V/m.
The phase \(kx+\omega t\) means the wave travels along \(-\hat x\). Since \(\vec E\times\vec B\) is along the propagation direction and \(\vec B\parallel\hat j\): \(\hat k\times\hat j=-\hat i\), so \(\vec E\) is along \(\hat k\).
\(\vec E=9\sin\left(1.6\times10^3x+48\times10^{10}t\right)\hat k\) V/m.
Answer: (A)
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