A plane EM wave is propagating along x direction. It has a wavelength of 4 mm. If electric field is in y direction with the maximum magnitude of 60 V m⁻¹, the equation for magnetic field is: (a) Bz = 2× 10⁻⁷sin [π/2(x - 3× 10⁸t)]k T (b) Bz = 2× 10⁻⁷sin [π/2× 10³(x - 3× 10⁸t)]k T (c) Bz = 60sin [π/2(x - 3× 10⁸t)]k T (d) Bx = 60sin [π/2(x - 3× 10⁸t)]i T

Answer: (b). k = (2π)/(λ) = 2π4×10⁻³ = π/2×10³ m ⁻¹ B₀ = 603×10⁸ = 2×10⁻⁷ T; y× z = x, so B along z. Bz = B₀sin [k(x - ct)] So the answer is (b) Bz = 2×10⁻⁷sin [π/2×10³(x - 3×10⁸t)] k T.

A plane EM wave is propagating along x direction. It has a wavelength… | JEE Main 2024 (9 Apr, Shift 1)

Q16JEE Main 2024 · 9 APR · SHIFT 1Medium

Nature of EM Waves & Wave Equation

A plane EM wave is propagating along $x$ direction. It has a wavelength of $4\text{ mm}$ . If electric field is in $y$ direction with the maximum magnitude of $60\text{ V m}^{-1}$ , the equation for magnetic field is:

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Answer: (b)

$k = \dfrac{2\pi}{\lambda} = \dfrac{2\pi}{4\times10^{-3}} = \dfrac{\pi}{2}\times10^{3}$ m $^{-1}$
$B_0 = \dfrac{60}{3\times10^{8}} = 2\times10^{-7}$ T; $\hat y\times\hat z = \hat x$ , so $\vec B$ along $z$ .
$B_z = B_0\sin[k(x - ct)]$
So the answer is (b) $B_z = 2\times10^{-7}\sin[\tfrac{\pi}{2}\times10^{3}(x - 3\times10^{8}t)]\hat k$ T.

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