The electric current in a circular coil of 2 turns produces a magnetic induction B₁ at its centre. The coil is unwound and in rewound into a circular coil of 5 tuns and the same current produces a magnetic induction B₂ at its centre. The ratio of B₂/B₁ is

Solution

Physics sir · 3 steps

The wire length stays fixed, so more turns means a smaller radius, and both effects raise B.

1. Field at the centre. For a coil of N turns and radius r carrying current I, the field at the centre is B = (μ₀ N I)/2r. So B grows with N and also grows as r shrinks.
2. Same wire length. Rewinding does not change the wire, so 2 × 2π r₁ = 5 × 2π r₂. That gives r₂ = 2/5 r₁, which means r₁/r₂ = 5/2.

   B₂/B₁ = N₂/N₁ × r₁/r₂ = 5/2 × 5/2 = 25/4

3. So. The ratio is B₂/B₁ = 25/4, which is option (b).

The idea to keep. When a fixed wire is rewound, N r stays constant, so r ∝ 1/N and the centre field B ∝ N/r goes as N².

Answer from the NTA final answer key. Solution by PrepWiser.