A circular loop is formed by joining two wire arcs, Arc 1 and Arc 2. Arc 1 subtends an angle of 60^ at the centre of the loop, while Arc 2 subtends 300^ . A steady current enters the loop at one junction of the arcs and leaves at the other junction. It is observed that the magnitude of the magnetic field at the centre due to Arc 1 is exactly twice the magnitude of the magnetic field due to Arc 2. What is the ratio of

Step-by-step solution

The magnetic field at the centre of a circular arc subtending an angle is given by B = ₀ I 2R 360^ . For Arc 1 ( ₁ = 60^ ): B₁ = ₀ I₁ 2R 60^ 360^ = ₀ I₁ 12R For Arc 2 ( ₂ = 300^ ): B₂ = ₀ I₂ 2R 300^ 360^ = 5 ₀ I₂ 12R It is given that B₁ = 2B₂ . Substituting the expressions: ₀ I₁ 12R = 2 ( 5 ₀ I₂ 12R ) I₁ = 10 I₂ I₁ I₂ = 10 Since the two arcs are connected in parallel between the entry and exit junctions, the potential difference across them is the same. V₁ = V₂ I₁ R₁ = I₂ R₂ R₁ R₂ = I₂ I₁ = 1 10 Thus, the ratio of