A current carrying circular coil of radius 'R' produces magnetic field B₁ at an axial point P at a distance 'x' from its centre and B₂ at point Q placed at its centre respectively . If B₂ = 8B₁ , the value of x is

Step-by-step solution

Magnetic field at an axial point P at a distance x from the centre of a circular coil of radius R is given by: B₁ = ₀ I R^2 2(R^2 + x^2)^ 3/2 Magnetic field at the centre Q of the coil is given by: B₂ = ₀ I 2R Given that B₂ = 8B₁ , substituting the expressions for B₁ and B₂ : ₀ I 2R = 8 ( ₀ I R^2 2(R^2 + x^2)^ 3/2 ) 1 R = 8R^2 (R^2 + x^2)^ 3/2 (R^2 + x^2)^ 3/2 = 8R^3 Taking the cube root on both sides: (R^2 + x^2)^ 1/2 = 2R Squaring both sides: R^2 + x^2 = 4R^2 x^2 = 3R^2 x = 3 R Answer: 3 ,R