NTA Abhyas JEE Main2020PhysicsMagnetic Effects of CurrentPractice
The magnetic field at P in the arrangement shown is
Options
- Aμ 0 i 2 π d 1 - 1 2 ⊗
- B2 μ 0 i 2 π d ⊗
- Cμ 0 i 2 π d ⊗
- Dμ 0 i 2 2 π d 1 + 1 2 ⊗
Correct answer
A. μ 0 i 2 π d 1 - 1 2 ⊗
Step-by-step solution
x d = sin 4 5 °   ⇒ x = d 2 we can extend the wire and then subtract the field of the extended part. B 1 = μ 0 4 π I d / 2 sin 0 ° + sin 9 0 ° - μ 0 4 π I d / 2 sin 0 ° + sin 4 5 ° = μ 0 4 π 2 I d 1 - 1 2 = μ 0 I 2 2  π d 1 - 1 2 ⊗ B 1 ⊗ , B 2 ⊗ ⇒ B = B 1 + B 2 ⊕ = 2 B 1 ⊗ = μ 0 I 2  π d 1 - 1 2 ⊗