The ratio of magnetic field at the centre of a current carrying coil of radius r to the magnetic field at distance r from the centre of coil on its axis is √x: 1. The value of x is __________

Solution

Physics sir · 4 steps

Put the axial point at a distance r and the whole ratio becomes (2r²)^(3/2) divided by r³.

1. Field at centre. For a circular loop of radius r carrying current I, Biot-Savart gives B_c = (μ₀ I)/2r at the centre.
2. Field on axis. At a distance d from the centre along the axis, Bₐ = (μ₀ I r²)/2(r² + d²)^(3/2). Here d = r, so the bracket becomes (2r²)^(3/2) = 2√2 r³.

   B_c/Bₐ = (μ₀ I / 2r)/(μ₀ I r² / (2 × 2√2 r³)) = (2r²)^(3/2)/r³ = 2√2

3. Match the form. Write 2√2 as a single root: 2√2 = √(4 × 2) = √8. So the ratio is √8 : 1.
4. So. Comparing with √x : 1 gives x = 8.

The common slip: 4. Raising only r² to the power 3/2 and forgetting the 2 makes (2r²)^(3/2) = 2r³. That gives a ratio of 2 = √4, so x = 4 instead of 8.

The idea to keep. The axial field of a loop is the centre field times r³/(r²+d²)^(3/2). At d = r that factor is 1/2√2, a value worth remembering.

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