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NEET Physics QUESTION #1059
Question 1
The electric field in a plane EM wave is \(E_z=60\cos(5x+1.5\times10^9t)\ \text{V/m}\). The corresponding magnetic field expression is:
  • \(B_z=60\cos(5x+1.5\times10^9t)\ \text{T}\)
  • \(B_y=60\sin(5x+1.5\times10^9t)\ \text{T}\)
  • \(B_y=2\times10^{-7}\cos(5x+1.5\times10^9t)\ \text{T}\)✔️
  • \(B_x=2\times10^{-7}\cos(5x+1.5\times10^9t)\ \text{T}\)
Correct Answer Explanation
Amplitude of B: \(B_0=\frac{E_0}{c}=\frac{60}{3\times10^8}=2\times10^{-7}\ \text{T}\). Wave travels in \(-x\) direction (since \(5x+\omega t\)). E is in z-direction, wave in \(-x\): B must be in y-direction (\(\hat{k}\times(-\hat{i})=\hat{j}\)... checking: \(\vec{E}\times\vec{B}\propto\) direction of propagation \(=-\hat{x}\)). So \(B_y=2\times10^{-7}\cos(5x+1.5\times10^9t)\ \text{T}\).