Q66 / 75 · JEE Main 2026 (21 January Shift 1) · Physics
The electric field in a plane electromagnetic wave is given by :
$E_{y}=69 \sin \left[0.6 \times 10^{3} x-1.8 \times 10^{11} t\right] \mathrm{V} / \mathrm{m}$.
The expression for magnetic field associated with this electromagnetic wave is $\_\_\_\_$ T.
$E_{y}=69 \sin \left[0.6 \times 10^{3} x-1.8 \times 10^{11} t\right] \mathrm{V} / \mathrm{m}$.
The expression for magnetic field associated with this electromagnetic wave is $\_\_\_\_$ T.
- Correct answerA$B_{z}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} x-1.8 \times 10^{11} t\right]$
- B$B_{z}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} x+1.8 \times 10^{11} t\right]$
- C$B_{y}=69 \sin \left[0.6 \times 10^{3} x+1.8 \times 10^{11} t\right]$
- D$B_{y}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} x-1.8 \times 10^{11} t\right]$
Solution
Given $E_y = 69 \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t]$ V/m
Wave propagates along +x (form $kx - \omega t$), $\vec{E}$ is along y.
For EM wave: $\vec{E} \times \vec{B}$ = direction of propagation
$\hat{j} \times \hat{k} = \hat{i}$, so $\vec{B}$ is along z-direction.
$B_0 = \frac{E_0}{c} = \frac{69}{3 \times 10^8} = 2.3 \times 10^{-7}$ T
E and B are in phase for plane EM wave.
$B_z = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t]$ T
Wave propagates along +x (form $kx - \omega t$), $\vec{E}$ is along y.
For EM wave: $\vec{E} \times \vec{B}$ = direction of propagation
$\hat{j} \times \hat{k} = \hat{i}$, so $\vec{B}$ is along z-direction.
$B_0 = \frac{E_0}{c} = \frac{69}{3 \times 10^8} = 2.3 \times 10^{-7}$ T
E and B are in phase for plane EM wave.
$B_z = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t]$ T
Hence, the correct answer is option: 'A'.