Question

The ratio of magnetic field due to coil at centre and at distance of R from the centre on the axis passing through the centre and perpendicular to the plane of ring is : 1x (R is the radius of coil), find the value of x.

Answer 1

The magnetic field at a distance \( x \) along the axis of a circular loop of radius \( R \) carrying current \( I \) is: \[ B_{\text{axis}} = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}}. \] At the center of the loop (\( x = 0 \)): \[ B_{\text{center}} = \frac{\mu_0 I}{2R}. \] At \( x = R \), substitute \( x = R \) into \( B_{\text{axis}} \): \[ B_{\text{axis}} = \frac{\mu_0 I R^2}{2 (R^2 + R^2)^{3/2}} = \frac{\mu_0 I R^2}{2 (2R^2)^{3/2}} = \frac{\mu_0 I}{2 \sqrt{8} R}. \] The ratio is: \[ \frac{B_{\text{center}}}{B_{\text{axis}}} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I}{2 \sqrt{8} R}} = \sqrt{8}. \] Final Answer: The value of \( x \) is: \[ \boxed{8}. \]