Question:

The electric field in an electromagnetic wave is given by \[ \vec{E} = \hat{i} 40 \cos \omega \left( t - \frac{z}{c} \right) \, \text{NC}^{-1}.\]The magnetic field induction of this wave is (in SI unit):

Updated On: Nov 26, 2024
  • \( \vec{B} = \hat{i} \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right) \)
  • \( \vec{B} = \hat{j} 40 \cos \omega \left( t - \frac{z}{c} \right) \)
  • \( \vec{B} = \hat{k} \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right) \)
  • \( \vec{B} = \hat{j} \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right) \)
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The Correct Option is D

Solution and Explanation

Given the electric field of the electromagnetic wave: 
\[ \vec{E} = \hat{i} 40 \cos \omega \left( t - \frac{z}{c} \right) \, \text{NC}^{-1} \] 
In an electromagnetic wave, the magnetic field \(\vec{B}\) is perpendicular to both the electric field \(\vec{E}\) and the direction of propagation. 

Since \(\vec{E}\) is along the \(\hat{i}\)-direction and the wave propagates along the \(\hat{k}\)-direction, the magnetic field \(\vec{B}\) must be along the \(\hat{j}\)-direction. 

The relationship between the magnitudes of the electric and magnetic fields in an electromagnetic wave is given by:
\[ B = \frac{E}{c} \] 

Substituting the given electric field magnitude: 
\[ B = \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right) \] 
Thus, the magnetic field is: 
\[ \vec{B} = \hat{j} \frac{40}{c} \cos \omega \left( t - \frac{z}{c} \right) \]

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