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Saturday, 31 October

JEE Main 2022 · PhysicsMultiple choiceSingle correctMediumApplication

JEE Main 28 June 2022, Shift 2, Physics Q45

Question 45 of 90 in this shift, Physics question 15 of 30, Section A.

An EM wave propagating in xx-direction has a wavelength of 8 mm. The electric field vibrating yy-direction has maximum magnitude of 60 Vm−1^{-1}. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum :
  1. (1)Ey=60sin⁡[π4×103 (x−3×108t)]j^E_y = 60\sin\left[\frac{\pi}{4}\times10^{3}\,(x-3\times10^{8}t)\right]\hat{j} Vm−1^{-1} Bz=2sin⁡[π4×103 (x−3×108t)]k^B_z = 2\sin\left[\frac{\pi}{4}\times10^{3}\,(x-3\times10^{8}t)\right]\hat{k} T
  2. (2)Ey=60sin⁡[π4×103 (x−3×108t)]j^E_y = 60\sin\left[\frac{\pi}{4}\times10^{3}\,(x-3\times10^{8}t)\right]\hat{j} Vm−1^{-1} Bz=2×10−7sin⁡[π4×103 (x−3×108t)]k^B_z = 2\times10^{-7}\sin\left[\frac{\pi}{4}\times10^{3}\,(x-3\times10^{8}t)\right]\hat{k} TOfficial answer
  3. (3)Ey=2×10−7sin⁡[π4×103 (x−3×108t)]j^E_y = 2\times10^{-7}\sin\left[\frac{\pi}{4}\times10^{3}\,(x-3\times10^{8}t)\right]\hat{j} Vm−1^{-1} Bz=60sin⁡[π4×103 (x−3×108t)]k^B_z = 60\sin\left[\frac{\pi}{4}\times10^{3}\,(x-3\times10^{8}t)\right]\hat{k} T
  4. (4)Ey=2×10−7sin⁡[π4×104 (x−4×108t)]j^E_y = 2\times10^{-7}\sin\left[\frac{\pi}{4}\times10^{4}\,(x-4\times10^{8}t)\right]\hat{j} Vm−1^{-1} Bz=60sin⁡[π4×104 (x−4×108t)]k^B_z = 60\sin\left[\frac{\pi}{4}\times10^{4}\,(x-4\times10^{8}t)\right]\hat{k} T

Official answer

Option 2

NTA final key.

Chapter
Electromagnetic Waves
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.