JEE Main202325 Jan 2023Morning ShiftPhysicsMagnetic Effects of CurrentActual
Match List I with List II List – I (Current configuration) List – II (Magnetic field at point O ) A I. B 0 = μ 0 I 4 π r [ π + 2 ] B II. B 0 = μ 0 4 I r C III. B 0 = μ 0 I 2 π r [ π - 1 ] D IV. B 0 = μ 0 I 4 π r [ π + 1 ] Choose the correct answer from the option given below:
Options
- AA-III, B-IV, C-I, D-II
- BA-I, B-III, C-IV, D-II
- CA-III, B-I, C-IV, D-II
- DA-II, B-I, C-IV, D-III
Correct answer
C. A-III, B-I, C-IV, D-II
Step-by-step solution
(A) Here net magnetic field will be sum of magnetic field due to straight wires ab ,   de and due to loop bcd . So, B ab = μ 0 4 π I r = B de (directed out of the plane in both cases) B b c d = μ 0 4 π I r ( 2 π ) (in the plane) Thus, magnetic field at O is B O = - μ 0 4 π I r + μ 0 4 π I r ( 2 π ) - μ 0 4 π I r B O = μ 0 2 π I r ( π - 1 ) (III) (B) Here net magnetic field will be sum of magnetic field due to straight wires ab ,