Q56 / 75 · JEE Main 2026 (06 April Shift 2) · Physics
For an electromagnetic wave propagating through vacuum, $\vec{k}$, $\vec{E}$ and $\omega$ represent propagation vector, electric field and angular frequency, respectively. The magnetic field associated with this wave is represented by :
- A$\dfrac{\vec{E} \times \vec{k}}{\omega}$
- Correct answerB$\dfrac{\vec{k} \times \vec{E}}{\omega}$
- C$\omega(\vec{E} \times \vec{k})$
- D$\omega(\vec{k} \times \vec{E})$
Solution
For an electromagnetic wave, the direction of wave propagation $\hat{k}$ is along the direction of $\vec{E} \times \vec{B}$.
Since $\vec{E}$, $\vec{B}$, and $\vec{k}$ form a mutually perpendicular right-handed system, the direction of the magnetic field is given by $\hat{B} = \hat{k} \times \hat{E}$.
The magnitudes of the electric and magnetic fields are related by $B = \dfrac{E}{c}$.
Using the relation for the speed of light $c = \dfrac{\omega}{k}$, we get $B = \dfrac{kE}{\omega}$.
Combining the magnitude and direction gives:
$\vec{B} = B\hat{B} = \dfrac{kE}{\omega} (\hat{k} \times \hat{E})$
Since $\vec{k} = k\hat{k}$ and $\vec{E} = E\hat{E}$, this simplifies to:
$\vec{B} = \dfrac{\vec{k} \times \vec{E}}{\omega}$
Answer: $\dfrac{\vec{k} \times \vec{E}}{\omega}$
Since $\vec{E}$, $\vec{B}$, and $\vec{k}$ form a mutually perpendicular right-handed system, the direction of the magnetic field is given by $\hat{B} = \hat{k} \times \hat{E}$.
The magnitudes of the electric and magnetic fields are related by $B = \dfrac{E}{c}$.
Using the relation for the speed of light $c = \dfrac{\omega}{k}$, we get $B = \dfrac{kE}{\omega}$.
Combining the magnitude and direction gives:
$\vec{B} = B\hat{B} = \dfrac{kE}{\omega} (\hat{k} \times \hat{E})$
Since $\vec{k} = k\hat{k}$ and $\vec{E} = E\hat{E}$, this simplifies to:
$\vec{B} = \dfrac{\vec{k} \times \vec{E}}{\omega}$
Answer: $\dfrac{\vec{k} \times \vec{E}}{\omega}$
Hence, the correct answer is option: 'B'.