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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2026 · 21 January · Morning Shift · Physics · Electromagnetic Waves

The electric field in a plane electromagnetic wave is given by : E_...

The electric field in a plane electromagnetic wave is given by :
Ey=69sin⁡[0.6×103x−1.8×1011t]V/mE_{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.

Options

  1. ABz=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_{z}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} x-1.8 \times 10^{11} t\right]
  2. BBz=2.3×10−7sin⁡[0.6×103x+1.8×1011t]B_{z}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} x+1.8 \times 10^{11} t\right]
  3. CBy=69sin⁡[0.6×103x+1.8×1011t]B_{y}=69 \sin \left[0.6 \times 10^{3} x+1.8 \times 10^{11} t\right]
  4. DBy=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_{y}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} x-1.8 \times 10^{11} t\right]

Correct answer

A. Bz=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_{z}=2.3 \times 10^{-7} \sin \left[0.6 \times 10^{3} x-1.8 \times 10^{11} t\right]

Step-by-step solution

Given Ey=69sin⁡[0.6×103x−1.8×1011t]E_y = 69 \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t] V/m
Wave propagates along +x (form kx−ωtkx - \omega t), E⃗\vec{E} is along y.
For EM wave: E⃗×B⃗\vec{E} \times \vec{B} = direction of propagation
j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, so B⃗\vec{B} is along z-direction.
B0=E0c=693×108=2.3×10−7B_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.
Bz=2.3×10−7sin⁡[0.6×103x−1.8×1011t]B_z = 2.3 \times 10^{-7} \sin[0.6 \times 10^3 x - 1.8 \times 10^{11} t] T
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