Question

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:

• A. \( \frac{x}{R} = \frac{3}{4} \)
• B. \( \frac{x}{R} = \frac{4}{3} \)
• C. \( \frac{x}{R} = 1 \)
• D. \( \frac{x}{R} = 2 \)

Hint:

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.

Answer 1

The Correct Option is A

We are given a circular coil, and we are asked to find the ratio of the magnetic field at the center of the coil to the magnetic field at a distance \( x \) from the center. The magnetic field at the center of a circular coil is given by the formula: \[ B_{\text{center}} = \frac{\mu_0 I}{2R} \] Where: - \( \mu_0 \) is the permeability of free space, - \( I \) is the current through the coil, - \( R \) is the radius of the coil. The magnetic field at a point on the axis of a circular coil at a distance \( x \) from the center is given by the formula: \[ B_x = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \] To find the ratio of the magnetic field at the center to the magnetic field at a distance \( x \), we calculate: \[ \frac{B_{\text{center}}}{B_x} = \frac{\frac{\mu_0 I}{2R}}{\frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}}} \] Simplifying the expression: \[ \frac{B_{\text{center}}}{B_x} = \frac{(R^2 + x^2)^{3/2}}{R^3} \] At \( x = \frac{3}{4}R \), substitute this value into the equation: \[ \frac{B_{\text{center}}}{B_x} = \frac{(R^2 + \left(\frac{3}{4}R\right)^2)^{3/2}}{R^3} = \frac{(R^2 + \frac{9}{16}R^2)^{3/2}}{R^3} \] This simplifies to: \[ \frac{B_{\text{center}}}{B_x} = \frac{( \frac{25}{16} R^2 )^{3/2}}{R^3} = \frac{\frac{125}{64} R^3}{R^3} = \frac{125}{64} = \frac{3}{4} \] Thus, the correct ratio is \( \frac{x}{R} = \frac{3}{4} \).