KCET2016PhysicsMagnetic Effects of Current
A conducting wire carrying current is arranged as shown. The magnetic field at ' ( O ) '
Options
- A( ₀ I 12 [ 1 R₁ - 1 R₂ ] )
- B( ₀ I 12 [ 1 R₁ + 1 R₂ ] )
- C( ₀ I 6 [ 1 R₁ - 1 R₂ ] )
- D( ₀ I 6 [ 1 R₁ + 1 R₂ ] )
Correct answer
A. ( ₀ I 12 [ 1 R₁ - 1 R₂ ] )
Step-by-step solution
Magnetic field at ( O ) due to inner current carrying wire is [ array l B₁= ₀ 4 2 (1 / 6) I R₁ = ₀ 12 I R₁ B₂= ₀ 4 2 (1 / 6) I R₂ = ₀ 12 I R₂ array ] Therefore, net magnetic field at ( O ) is given as [ B=B₁-B₂= ₀ 12 I R₁ - ₀ 12 I R₂ = ₀ I 12 ( 1 R₁ - 1 R₂ ) ]