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← JEE Advanced 2026 · Paper 1
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JEE Advanced 2026 · Paper 1, Physics question 8

The electric field associated with an electromagnetic wave travelling in vacuum is given by E0sin⁡(3y+4z+ωt)ı^E_0\sin(3y+4z+\omega t)\hat\imath, where ω\omega is the angular frequency. All quantities are in SI units. The correct statement(s) about this wave is/are:
[Given: speed of light in vacuum c=3×108c=3\times10^8 ms−1^{-1}.]

Choose one or more options
Answer and worked solution

Official answer: A, C

The phase is k⃗⋅r⃗+ωt\vec k\cdot\vec r+\omega t with k⃗=3ȷ^+4k^\vec k=3\hat\jmath+4\hat k.

(A) A phase +ωt+\omega t means motion along −k⃗-\vec k: direction −15(3ȷ^+4k^)-\frac15(3\hat\jmath+4\hat k). True.

(B) ∣k⃗∣=5 m−1|\vec k|=5\ \text{m}^{-1}. False.

(C) ω=c∣k⃗∣=1.5×109\omega=c|\vec k|=1.5\times10^9 rad/s. True.

(D) B⃗=1cn^×E⃗\vec B=\dfrac1c\hat n\times\vec E with n^=−15(3ȷ^+4k^)\hat n=-\frac15(3\hat\jmath+4\hat k):
n^×ı^=−15(3ȷ^×ı^+4k^×ı^)=15(3k^−4ȷ^).\hat n\times\hat\imath=-\tfrac15(3\hat\jmath\times\hat\imath+4\hat k\times\hat\imath)=\tfrac15(3\hat k-4\hat\jmath).
So B⃗=E05csin⁡(3y+4z+ωt)(3k^−4ȷ^)\vec B=\dfrac{E_0}{5c}\sin(3y+4z+\omega t)(3\hat k-4\hat\jmath). The stated field has the wrong sign and is 5 times too large. False.

Topics:Properties of EM waves
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