Electromagnetic Waves

JEE Main level, single correct, medium

4 of 165

Question 4 of 165

For an electromagnetic wave propagating through vacuum, k⃗\vec{k}, E⃗\vec{E} and ω\omega represent propagation vector, electric field and angular frequency, respectively. The magnetic field associated with this wave is represented by :

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Answer: (B) k⃗×E⃗ω\dfrac{\vec{k} \times \vec{E}}{\omega}

For an electromagnetic wave, the direction of wave propagation k^\hat{k} is along the direction of E⃗×B⃗\vec{E} \times \vec{B}.

Since E⃗\vec{E}, B⃗\vec{B}, and k⃗\vec{k} form a mutually perpendicular right-handed system, the direction of the magnetic field is given by B^=k^×E^\hat{B} = \hat{k} \times \hat{E}.

The magnitudes of the electric and magnetic fields are related by B=EcB = \dfrac{E}{c}.

Using the relation for the speed of light c=ωkc = \dfrac{\omega}{k}, we get B=kEωB = \dfrac{kE}{\omega}.

Combining the magnitude and direction gives:

B⃗=BB^=kEω(k^×E^)\vec{B} = B\hat{B} = \dfrac{kE}{\omega} (\hat{k} \times \hat{E})

Since k⃗=kk^\vec{k} = k\hat{k} and E⃗=EE^\vec{E} = E\hat{E}, this simplifies to:

B⃗=k⃗×E⃗ω\vec{B} = \dfrac{\vec{k} \times \vec{E}}{\omega}

Answer: k⃗×E⃗ω\dfrac{\vec{k} \times \vec{E}}{\omega}

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