The magnetic field of an electromagnetic wave is given by:- B = 1.6 × 10⁻⁶ cos (2 × 10⁷ z + 6 × 10¹⁵ t) (2i + j) Wb/m² The associated electric field will be:- (a) E = 4.8 × 10² cos (2 × 10⁷ z - 6 × 10¹⁵ t) (-2i + j) V/m (b) E = 4.8 × 10² cos (2 × 10⁷ z - 6 × 10¹⁵ t) (2i + j) V/m (c) E = 4.8 × 10² cos (2 × 10⁷ z + 6 × 10¹⁵ t) (i - 2j) V/m (d) E = 4.8 × 10² cos (2 × 10⁷ z + 6 × 10¹⁵ t) (-i + 2j) V/m

Answer: (c). v = 6×10¹⁵2×10⁷ = 3×10⁸ m/s; (kz + ω t) means travel along -z. E₀ = cB₀ = 3×10⁸×1.6×10⁻⁶ = 4.8×10² V/m (per unit of the vector). E = c B× n with n = - k: (2 i + j)×(- k) = 2 j - i Check: (- i + 2 j)×(2 i + j) = -5 k, along -z as required; ( i - 2 j) would give +5 k. So the working gives (d) E = 4.8×10²cos (2×10⁷z + 6×10¹⁵t)(- i + 2 j) V/m.

The magnetic field of an electromagnetic wave is given by:- B = 1.6 ×… | JEE Main 2019 (8 Apr, Shift 2)

Q67JEE Main 2019 · 8 APR · SHIFT 2Medium

Nature of EM Waves & Wave Equation

The magnetic field of an electromagnetic wave is given by :-

$\vec{B} = 1.6 \times 10^{-6} \cos\left(2 \times 10^{7} z + 6 \times 10^{15} t\right) \left(2\hat{i} + \hat{j}\right) \frac{\text{Wb}}{\text{m}^2}$

The associated electric field will be :-

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Answer: (c)

$v = \dfrac{6\times10^{15}}{2\times10^{7}} = 3\times10^{8}$ m/s; $(kz + \omega t)$ means travel along $-z$ .
$E_0 = cB_0 = 3\times10^{8}\times1.6\times10^{-6} = 4.8\times10^{2}$ V/m (per unit of the vector).
$\vec E = c\,\vec B\times\hat n$ with $\hat n = -\hat k$ : $(2\hat i + \hat j)\times(-\hat k) = 2\hat j - \hat i$
Check: $(-\hat i + 2\hat j)\times(2\hat i + \hat j) = -5\hat k$ , along $-z$ as required; $(\hat i - 2\hat j)$ would give $+5\hat k$ .
So the working gives (d) $\vec E = 4.8\times10^{2}\cos(2\times10^{7}z + 6\times10^{15}t)(-\hat i + 2\hat j)$ V/m.

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Nature of EM Waves & Wave Equation question from JEE Main 2019 (8 Apr, Shift 2)

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