Question:

The electric field of a plane electromagnetic wave is given by
\(\vec{E} = E_0(\hat{x} + \hat{y})\sin(kz - \omega t)\)
Its magnetic field will be given by :

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The wave travels along \(+z\) and \(\vec{E}\times\vec{B}\) must point along the propagation direction, so \(\vec{B} = \frac{1}{c}\hat{k}\times\vec{E}\), in phase with \(\vec{E}\).
Updated On: Nov 5, 2026
  • \(\frac{E_0}{c}(\hat{x} + \hat{y})\sin(kz - \omega t)\)
  • \(\frac{E_0}{c}(\hat{x} - \hat{y})\sin(kz - \omega t)\)
  • \(\frac{E_0}{c}(\hat{x} - \hat{y})\cos(kz - \omega t)\)
  • \(\frac{E_0}{c}(-\hat{x} + \hat{y})\sin(kz - \omega t)\)
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The Correct Option is D

Solution and Explanation

Phase \(kz - \omega t\) means propagation along \(+z\). For an EM wave, \(\vec{B} = \frac{1}{c}\hat{k} \times \vec{E}\), in phase with E.
\(\hat{z} \times (\hat{x} + \hat{y}) = \hat{y} - \hat{x}\) (since \(\hat{z}\times\hat{x} = \hat{y}\) and \(\hat{z}\times\hat{y} = -\hat{x}\)).
So \(\vec{B} = \frac{E_0}{c}(-\hat{x} + \hat{y})\sin(kz - \omega t)\).
Check: \(\vec{E}\times\vec{B} \propto (\hat{x}+\hat{y})\times(-\hat{x}+\hat{y}) = \hat{z} + \hat{z} = 2\hat{z}\), along propagation. Option (2) has the opposite sign and would give \(\vec{E}\times\vec{B}\) along \(-z\).
Answer: (4) \(\frac{E_0}{c}(-\hat{x} + \hat{y})\sin(kz - \omega t)\)
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