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← JEE Main 2017 · 9 Apr · Shift 1
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JEE Main 2017 · 9 Apr · Shift 1, Physics question 21

The electric field component of a monochromatic radiation is given by
E⃗=2E0i^cos⁡kzcos⁡ωt\vec E=2E_0\hat i\cos kz\cos\omega t
Its magnetic field B⃗\vec B is then given by :

Choose one option
Answer and worked solution

Official answer: C

2E0cos⁡kzcos⁡ωt=E0[cos⁡(kz−ωt)+cos⁡(kz+ωt)]2E_0\cos kz\cos\omega t=E_0[\cos(kz-\omega t)+\cos(kz+\omega t)]. The wave along +z has B⃗=E0cj^cos⁡(kz−ωt)\vec B=\frac{E_0}c\hat j\cos(kz-\omega t) and the wave along −z has −E0cj^cos⁡(kz+ωt)-\frac{E_0}c\hat j\cos(kz+\omega t). Adding:
B⃗=2E0cj^sin⁡kzsin⁡ωt.\vec B=\frac{2E_0}c\hat j\sin kz\sin\omega t.

Topics:Properties of EM waves
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