A current carrying circular coil of radius ' R ' produces magnetic fields ' B ₁ ' and ' B ₂ ' at an axial point ' P ' at a distance ' x ' from its centre and at a point ' Q ' placed at its centre respectively. If B₁= B₂ 8 , the value of ' x ' is

Step-by-step solution

The magnetic field at the center of a circular coil is given by B₂ = ₀ I 2R , while the field at an axial point distance x from the center is B₁ = ₀ I R^2 2(R^2 + x^2)^ 3/2 . Given B₁ = B₂ 8 , substitute both expressions: ₀ I R^2 2(R^2 + x^2)^ 3/2 = 1 8 ₀ I 2R Canceling common factors ₀ I 2 from both sides yields: R^2 (R^2 + x^2)^ 3/2 = 1 8R Cross-multiplying gives 8R^3 = (R^2 + x^2)^ 3/2 . Raising both sides to the 2/3 power: (8R^3)^ 2/3 = [(R^2 + x^2)^ 3/2 ]^ 2/3 (2R)^2 = R^2 + x^2 4R^2 = R^2 + x^2 3R^2 = x^2 x =