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Saturday, 31 October

JEE Main 2024 · PhysicsMultiple choiceSingle correctEasyApplication

JEE Main 4 April 2024, Shift 1, Physics Q44

Question 44 of 90 in this shift, Physics question 14 of 30, Section A.

The electric field in an electromagnetic wave is given by E⃗=i^ 40cos⁡ω(t−zc)N C−1\vec{E} = \hat{i}\ 40\cos\omega\left(t - \frac{z}{c}\right)\mathrm{N\,C^{-1}}. The magnetic field induction of this wave is (in SI unit) :
  1. (1)B⃗=j^ 40cos⁡ω(t−zc)\vec{B} = \hat{j}\ 40\cos\omega\left(t - \frac{z}{c}\right)
  2. (2)B⃗=k^ 40ccos⁡ω(t−zc)\vec{B} = \hat{k}\ \frac{40}{c}\cos\omega\left(t - \frac{z}{c}\right)
  3. (3)B⃗=j^ 40ccos⁡ω(t−zc)\vec{B} = \hat{j}\ \frac{40}{c}\cos\omega\left(t - \frac{z}{c}\right)Official answer
  4. (4)B⃗=i^ 40ccos⁡ω(t−zc)\vec{B} = \hat{i}\ \frac{40}{c}\cos\omega\left(t - \frac{z}{c}\right)

Official answer

Option 3

NTA final key.

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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.