The magnetic field of a plane electromagnetic wave is given by: B = 2× 10⁻⁸sin (0.5× 10³x + 1.5× 10¹¹t)j T. The amplitude of the electric field would be: (a) 6 V m⁻¹ along x -axis (b) 3 V m⁻¹ along z -axis (c) 6 V m⁻¹ along z -axis (d) 2× 10⁻⁸ V m⁻¹ along z -axis

Answer: (c). E₀ = cB₀ = 3×10⁸×2×10⁻⁸ = 6 V/m (kx+ω t): wave moves along -x; B along j. Need E× j = - i: k× j = - i, so E is along z. So the answer is (c) 6 V m ⁻¹ along z-axis.

The magnetic field of a plane electromagnetic wave is given by: B =… | JEE Main 2022 (26 Jul, Shift 1)

Q35JEE Main 2022 · 26 JUL · SHIFT 1Medium

Nature of EM Waves & Wave Equation

The magnetic field of a plane electromagnetic wave is given by:

$\vec{B} = 2\times 10^{-8}\sin\left(0.5\times 10^{3}x + 1.5\times 10^{11}t\right)\hat{j}\text{ T}$ .

The amplitude of the electric field would be:

Show answer and solution

Answer: (c)

$E_0 = cB_0 = 3\times10^{8}\times2\times10^{-8} = 6$ V/m
$(kx+\omega t)$ : wave moves along $-x$ ; $\vec B$ along $\hat j$ .
Need $\hat E\times\hat j = -\hat i$ : $\hat k\times\hat j = -\hat i$ , so $\vec E$ is along $z$ .
So the answer is (c) 6 V m $^{-1}$ along z-axis.

Report a mistake

Nature of EM Waves & Wave Equation question from JEE Main 2022 (26 Jul, Shift 1)

More questions on Nature of EM Waves & Wave Equation