For a plane electromagnetic wave, the magnetic field at a point x and time t is B(x,t) = [1.2× 10⁻⁷sin (0.5× 10³x + 1.5× 10¹¹t)k] T The instantaneous electric field E corresponding to B is: (speed of light c = 3 × 108 ms–1) (a) E(x,t) = [36sin (1× 10³x + 1.5× 10¹¹t)i]V/m (b) E(x,t) = [36sin (0.5× 10³x + 1.5× 10¹¹t)k]V/m (c) E(x,t) = [36sin (1× 10³x + 0.5× 10¹¹t)j]V/m (d) E(x,t) = [-36sin (0.5× 10³x + 1.5× 10¹¹t)j]V/m

Answer: (d). v = 1.5×10¹¹0.5×10³ = 3×10⁸ m/s; (kx+ω t) means travel along -x. E₀ = cB₀ = 3×10⁸×1.2×10⁻⁷ = 36 V/m Need E× k = - i: (- j)× k = - i, so E along - j. So the answer is (d) E = -36sin (0.5×10³x + 1.5×10¹¹t) j V/m.

For a plane electromagnetic wave, the magnetic field at a point x and… | JEE Main 2020 (6 Sep, Shift 2)

Q51JEE Main 2020 · 6 SEP · SHIFT 2Medium

Nature of EM Waves & Wave Equation

For a plane electromagnetic wave, the magnetic field at a point x and time t is

$\vec{B}(x,t) = \left[1.2\times 10^{-7}\sin\left(0.5\times 10^{3}x + 1.5\times 10^{11}t\right)\hat{k}\right]$ T

The instantaneous electric field $\vec{E}$ corresponding to $\vec{B}$ is:

(speed of light c = 3 × 108 ms–1)

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Answer: (d)

$v = \dfrac{1.5\times10^{11}}{0.5\times10^{3}} = 3\times10^{8}$ m/s; $(kx+\omega t)$ means travel along $-x$ .
$E_0 = cB_0 = 3\times10^{8}\times1.2\times10^{-7} = 36$ V/m
Need $\hat E\times\hat k = -\hat i$ : $(-\hat j)\times\hat k = -\hat i$ , so $\vec E$ along $-\hat j$ .
So the answer is (d) $\vec E = -36\sin(0.5\times10^{3}x + 1.5\times10^{11}t)\hat j$ V/m.

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Nature of EM Waves & Wave Equation question from JEE Main 2020 (6 Sep, Shift 2)

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