A long conducting wire having a current I flowing through it, is bent into a circular coil of N turns. Then it is bent into a circular coil of n turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is :

Solution

Physics sir · 4 steps

The wire length is fixed, so more turns means a smaller radius too. N enters twice, and that gives N squared.

1. Field of a coil. At the centre of a circular coil of N turns and radius r, the field is B = (μ₀ N I)/2r. Both N and r change when the wire is rewound.
2. Length fixes radius. The same wire of length L is used both times, so L = N × 2π r. This gives r = L/(2π N): more turns force a smaller radius.

   B = (μ₀ N I)/2 × (2π N)/L = (π μ₀ I N²)/L

3. Compare the cases. I and L are the same in both cases, so B ∝ N². Then B₁/B₂ = N²/n².
4. So. The ratio is N² : n², option (a).

The idea to keep. When a fixed length of wire is rewound into a coil, the radius is not free. Write r in terms of turns using the length, then put it into the formula for B.

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