A current carrying circular coil of radius 'R' produces magnetic field 'B'₁ at an axial point P at a distance 'x' from its centre and B₂ at point Q placed at its centre ,respectively. If B₂ = 27 , B₁ , the value of 'x' is

Step-by-step solution

The 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 The magnetic field at the centre Q of the coil is given by: B₂ = ₀ I 2R Given that B₂ = 27 B₁ , substituting the expressions for B₁ and B₂ : ₀ I 2R = 27 ( ₀ I R^2 2(R^2 + x^2)^ 3/2 ) 1 R = 27 R^2 (R^2 + x^2)^ 3/2 (R^2 + x^2)^ 3/2 = 27 R^3 Taking the cube root on both sides: (R^2 + x^2)^ 1/2 = 3R Squaring both sides: R^2 + x^2 = 9R^2 x^2 = 8R^2 x = 2 2 R Answer: 2 2 , R