The ratio of the magnetic field at the center of a circular coil to the magnetic field at a distance \( x \) from the center of the circular coil is:
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For calculating the magnetic field on the axis of a circular coil, use the formula for field at a point along the axis, and use symmetry to relate it to the field at the center.
The problem requires calculating the ratio of the magnetic field at the center of a circular coil to the magnetic field at a distance \( x \) from the center.
The magnetic field at the coil's center is given by:
\[ B_{\text{center}} = \frac{\mu_0 I}{2R} \]
where \( \mu_0 \) is the permeability of free space, \( I \) is the coil current, and \( R \) is the coil radius.
The magnetic field on the axis at distance \( x \) from the center is:
\[ B_x = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \]
The ratio \( \frac{B_{\text{center}}}{B_x} \) is computed as:
\[ \frac{B_{\text{center}}}{B_x} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}}} \]
Simplifying this yields:
\[ \frac{B_{\text{center}}}{B_x} = \frac{(R^2 + x^2)^{3/2}}{R^3} \]
Substituting \( x = \frac{3}{4}R \):
\[ \frac{B_{\text{center}}}{B_x} = \frac{(R^2 + \left(\frac{3}{4}R\right)^2)^{3/2}}{R^3} = \frac{(R^2 + \frac{9}{16}R^2)^{3/2}}{R^3} \]
Further simplification leads to:
\[ \frac{B_{\text{center}}}{B_x} = \frac{( \frac{25}{16} R^2 )^{3/2}}{R^3} = \frac{\frac{125}{64} R^3}{R^3} = \frac{125}{64} \]
The problem statement incorrectly concludes that the ratio is \( \frac{3}{4} \). The derived ratio \( \frac{B_{\text{center}}}{B_x} \) is \( \frac{125}{64} \). The value \( \frac{3}{4} \) corresponds to \( \frac{x}{R} \).
