Learn

Saturday, 31 October

JEE Main 2023 · PhysicsMultiple choiceSingle correctEasyRecall

JEE Main 6 April 2023, Shift 2, Physics Q46

Question 46 of 90 in this shift, Physics question 16 of 30, Section A.

The energy density associated with electric field E⃗\vec{E} and magnetic field B⃗\vec{B} of an electromagnetic wave in free space is given by (ϵ0−\epsilon_0- permittivity of free space, μ0=\mu_0= permeability of free space)
  1. (1)UE=E22ϵ0, UB=μ0B22U_E=\frac{E^2}{2\epsilon_0},\ U_B=\frac{\mu_0 B^2}{2}
  2. (2)UE=ϵ0E22, UB=μ0B22U_E=\frac{\epsilon_0 E^2}{2},\ U_B=\frac{\mu_0 B^2}{2}
  3. (3)UE=E22ϵ0, UB=B22μ0U_E=\frac{E^2}{2\epsilon_0},\ U_B=\frac{B^2}{2\mu_0}
  4. (4)UE=ϵ0E22, UB=B22μ0U_E=\frac{\epsilon_0 E^2}{2},\ U_B=\frac{B^2}{2\mu_0}Official answer

Official answer

Option 4

NTA final key.

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.