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The electric field associated with an electromagnetic wave travelling in vacuum is given by E0 (3y + 4z + t) i, where 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 = 3 108 ms-1.]
- The wave is travelling in - 15 (3 j + 4 k ) direction.
- The magnitude of the wave vector is 0.5 m-1.
- The value of is 1.5 109 rad s-1.
- The magnetic field associated with this wave is given by E0c (3y + 4z + t) (4 j - 3 k ).
Answer
(1) The wave is travelling in - 15 (3 j + 4 k ) direction., (3) The value of is 1.5 109 rad s-1.
Solution
Phase k r + t with k = 3 j + 4 k: the wave moves along - k, unit vector n = - 15(3 j + 4 k) — (A) correct.
| k| = 9 + 16 = 5 m-1 — (B) incorrect.
= ck = 3 108 5 = 1.5 109 rad s-1 — (C) correct.
B = 1c n E = E05c (3y + 4z + t) [-(3 j + 4 k) i ] = - E05c (3y + 4z + t)(4 j - 3 k); magnitude E0/c, so (D) (amplitude 5E0/c) is incorrect.
Answer: (A), (C)
JEE Advanced 2026 Paper 1