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NEET PHYSICSMedium

An electromagnetic wave is moving along negative zz (−z-z) direction and at any instant of time, at a point, its electric field vector is 3j^ V/m3\hat j \text{ V/m}. The corresponding magnetic field at that point and instant will be: (Take c=3×108 ms−1c=3\times10^8 \text{ ms}^{-1})

A

10i^ nT10\hat i \text{ nT}

B

−10i^ nT-10\hat i \text{ nT}

C

i^ nT\hat i \text{ nT}

D

−i^ nT-\hat i \text{ nT}

Step-by-Step Solution

  1. Magnitude Calculation: The magnitude of the magnetic field BB is related to the electric field EE and the speed of light cc by the relation B=E/cB = E/c. Given E=3 V/mE = 3 \text{ V/m} and c=3×108 m/sc = 3 \times 10^8 \text{ m/s}: B=33×108=10−8 TB = \frac{3}{3 \times 10^8} = 10^{-8} \text{ T} Converting to nanoTesla (1 nT=10−9 T1 \text{ nT} = 10^{-9} \text{ T}): B=10×10−9 T=10 nTB = 10 \times 10^{-9} \text{ T} = 10 \text{ nT}
  2. Direction Calculation: The direction of propagation of an electromagnetic wave is given by the vector cross product of the electric field direction and the magnetic field direction: v^=E^×B^\hat{v} = \hat{E} \times \hat{B}.
  • Given direction of propagation v^=−k^\hat{v} = -\hat{k} (negative z-direction).
  • Given direction of electric field E^=j^\hat{E} = \hat{j}.
  • We need to find B^\hat{B} such that j^×B^=−k^\hat{j} \times \hat{B} = -\hat{k}.
  • Using the cyclic properties of unit vectors (i^×j^=k^\hat{i} \times \hat{j} = \hat{k}, j^×k^=i^\hat{j} \times \hat{k} = \hat{i}, k^×i^=j^\hat{k} \times \hat{i} = \hat{j}), we know that j^×i^=−k^\hat{j} \times \hat{i} = -\hat{k}.
  • Therefore, the direction of the magnetic field must be i^\hat{i}.
  1. Conclusion: The magnetic field vector is 10i^ nT10\hat{i} \text{ nT}.
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