Verified Solution • 2025–26 Session moderate

QNEET-2022-Ph-520: NEET 2022 Physics - Class 12 NEET-2022-Ph-520 Solution: The magnetic field of a plane electromagnetic wave is given by…

The magnetic field of a plane electromagnetic wave is given by B⃗=3×10−8cos⁡(1.6×103x+48×1010t)j^\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: 9cos⁡(1.6×103x+48×1010t)k^ V/m9\cos(1.6 \times 10^3 x + 48 \times 10^{10} t)\hat{k}\ \text{V/m} 3×10−8cos⁡(1.6×103x+48×1010t)i^ V/m3 \times 10^{-8} \cos(1.6 \times 10^3 x + 48 \times 10^{10} t)\hat{i}\ \text{V/m} 3×10−8sin⁡(1.6×103x+48×1010t)i^ V/m3 \times 10^{-8} \sin(1.6 \times 10^3 x + 48 \times 10^{10} t)\hat{i}\ \text{V/m} 9sin⁡(1.6×103x−48×1010t)k^ V/m9\sin(1.6 \times 10^3 x - 48 \times 10^{10} t)\hat{k}\ \text{V/m}

Multiple Choice Options

  • A 9cos⁡(1.6×103x+48×1010t)k^9\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat k 9 cos ( 1.6 _ 1 0 3 x + 48 _ 1 0 10 t ) k V/m
  • B 3×10−8cos⁡(1.6×103x+48×1010t)i^3 \times {10^{ - 8}}\cos (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat i 3 _ 1 0 _ 8 cos ( 1.6 _ 1 0 3 x + 48 _ 1 0 10 t ) i V/m
  • C 3×10−8sin⁡(1.6×103x+48×1010t)i^3 \times {10^{ - 8}}\sin (1.6 \times {10^3}x + 48 \times {10^{10}}t)\widehat i 3 _ 1 0 _ 8 sin ( 1.6 _ 1 0 3 x + 48 _ 1 0 10 t ) i V/m
  • D 9sin⁡(1.6×103x−48×1010t)k^9\sin (1.6 \times {10^3}x - 48 \times {10^{10}}t)\widehat k 9 sin ( 1.6 _ 1 0 3 x _ 48 _ 1 0 10 t ) k V/m

Step-by-Step Verified Solution

For electromagnetic wave, | B _ | = | E _ | c Here B _ is magnetic field associated with EM wave E _ is electric field associated with EM wave c is the speed of EM wave _ | E _ | = c | B _ | = 3 _ 10 8 _ 3 _ 10 _ 8 cos _ ( 1.6 _ 10 3 x + 48 _ 10 10 t ) V/m Direction can be determined from Poynting vector = E _ _ B _ _ 0 E _ = 9 cos _ ( 1.6 _ 10 3 x + 48 _ 10 10 t ) k ^ V/m

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