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If the magnetic field of a plane electromagnetic wave is given by (The speed of light =3× 10^8 m/s) B=100× 10^-6 sin⁡[2π × 2× 10^15(t- x )] then the…

JEE Main 2019 — Physics Electromagnetism

2019mcqeasy

If the magnetic field of a plane electromagnetic wave is given by (The speed of light =3×108m/s=3\times {10}^{8} m/s=3×108m/s)

B=100×10−6sin⁡[2π×2×1015(t−xc)]B=100\times {10}^{-6}\mathrm{sin}⁡[2\pi \times 2\times {10}^{15}(t-\frac{x}{c})]B=100×10−6sin⁡[2π×2×1015(t−cx​)] then the maximum electric field associated with it is:

Official previous-year question

Held on 10 Jan 2019 · Verified 6 Jul 2026.

Options

  1. A

    3×104NC−13\times {10}^{4} N{C}^{-1}3×104NC−1

  2. B

    4×104NC−14\times {10}^{4} N{C}^{-1}4×104NC−1

  3. C

    4.5×104NC−14.5\times {10}^{4} N{C}^{-1}4.5×104NC−1

  4. D

    6×104NC−16\times {10}^{4} N{C}^{-1}6×104NC−1

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