The magnetic field of a plane electromagnetic wave is given by B = 3× 10⁻⁸cos (1.6× 10³x + 48× 10¹⁰t)j, then the associated electric field will be: (a) 9cos (1.6× 10³x + 48× 10¹⁰t)k V/m (b) 3× 10⁻⁸cos (1.6× 10³x + 48× 10¹⁰t)i V/m (c) 3× 10⁻⁸sin (1.6× 10³x + 48× 10¹⁰t)i V/m (d) 9sin (1.6× 10³x - 48× 10¹⁰t)k V/m

Answer: (a). v = ω/k = 48×10¹⁰1.6×10³ = 3×10⁸ m/s; wave moves along -x. E₀ = cB₀ = 3×10⁸×3×10⁻⁸ = 9 V/m Need E× j = - i: k× j = - i, so E is along k with the same phase. So the answer is (a) 9cos (1.6×10³x+48×10¹⁰t) k V/m.

The magnetic field of a plane electromagnetic wave is given by B = 3×… | NEET 2022 (Phase 2)

Q37NEET 2022 · PHASE 2Medium

Nature of EM Waves & Wave Equation

The magnetic field of a plane electromagnetic wave is given by

$\vec{B} = 3\times 10^{-8}\cos(1.6\times 10^{3}x + 48\times 10^{10}t)\hat{j}$ , then the associated electric field will be:

Show answer and solution

Answer: (a)

$v = \omega/k = \dfrac{48\times10^{10}}{1.6\times10^{3}} = 3\times10^{8}$ m/s; wave moves along $-x$ .
$E_0 = cB_0 = 3\times10^{8}\times3\times10^{-8} = 9$ V/m
Need $\hat E\times\hat j = -\hat i$ : $\hat k\times\hat j = -\hat i$ , so $\vec E$ is along $\hat k$ with the same phase.
So the answer is (a) $9\cos(1.6\times10^{3}x+48\times10^{10}t)\hat k$ V/m.

Report a mistake

Nature of EM Waves & Wave Equation question from NEET 2022 (Phase 2)

More questions on Nature of EM Waves & Wave Equation