The magnetic field in a plane electromagnetic wave is By = (3.5× 10⁻⁷)sin (1.5× 10³x + 0.5× 10¹¹t) T. The corresponding electric field will be: (a) Ez = 105sin (1.5× 10³x + 0.5× 10¹¹t) V m⁻¹ (b) Ey = 10.5sin (1.5× 10³x + 0.5× 10¹¹t) V m⁻¹ (c) Ey = 1.17sin (1.5× 10³x + 0.5× 10¹¹t) V m⁻¹ (d) Ez = 1.17sin (1.5× 10³x + 0.5× 10¹¹t) V m⁻¹

Answer: (a). (kx+ω t): wave moves along -x; B along j, so E× j = - i gives E along k ( Ez). E₀ = cB₀ = 3×10⁸×3.5×10⁻⁷ = 105 V/m So the answer is (a) Ez = 105sin (1.5×10³x + 0.5×10¹¹t) V m ⁻¹.

The magnetic field in a plane electromagnetic wave is By = (3.5×… | JEE Main 2024 (9 Apr, Shift 2)

Q15JEE Main 2024 · 9 APR · SHIFT 2Medium

Nature of EM Waves & Wave Equation

The magnetic field in a plane electromagnetic wave is

$B_y = \left(3.5\times 10^{-7}\right)\sin\left(1.5\times 10^{3}x + 0.5\times 10^{11}t\right)\text{ T}$ . The corresponding electric field will be:

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Answer: (a)

$(kx+\omega t)$ : wave moves along $-x$ ; $\vec B$ along $\hat j$ , so $\hat E\times\hat j = -\hat i$ gives $\vec E$ along $\hat k$ ( $E_z$ ).
$E_0 = cB_0 = 3\times10^{8}\times3.5\times10^{-7} = 105$ V/m
So the answer is (a) $E_z = 105\sin(1.5\times10^{3}x + 0.5\times10^{11}t)$ V m $^{-1}$ .

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