Q75 / 90 · JEE Main 2022 (28 Jun Shift 2) · Physics

An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60Vm-1. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum :

  • A

    Ey=60sinπ4×103x-3×108tj ^ V m-1

    Bz=2sinπ4×103x-3×108tk^ T

  • Correct answer
    B

    Ey=60sinπ4×103x-3×108t j^ V m-1

    Bz=2×10-7sinπ4×103x-3×108tk^ T

  • C

    Ey=2×10-7sinπ4×103x-3×108tj^ V m-1

    Bz=60sinπ4×103x-3×108tk^ T

  • D

    Ey=2×10-7sinπ4×104x-4×108tj^ V m-1

    Bz=60sinπ4×104x-4×108tk^ T

Solution

Using the relation E→=cB→, where, c is speed of light in vacuum.

After analysing each option, we get option B is the correct option

as E→B→=602×10-7=3×108 m s-1.

Hence, the correct answer is option: 'B'.

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