Question

Two points A and B on the axis of a circular current loop are at distances of 4 cm and \( 3\sqrt{3} \) cm from the center of the loop. If the ratio of the induced magnetic fields at points A and B is 216:125, the radius of the loop is:

• A. \( 3 \) cm
• B. \( 4 \) cm
• C. \( 5 \) cm
• D. \( 6 \) cm

Hint:

The magnetic field at a point on the axis of a circular loop is given by \( B = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \).
- To find the radius, use the given ratio of fields and solve for \( R \).

Answer 1

The Correct Option is A

The magnetic field at a point on the axis of a circular current loop is given by: \[ B = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}} \] where: - \( R \) is the radius of the loop, - \( x \) is the distance from the center along the axis. 1. Ratio of Magnetic Fields at A and B: \[ \frac{B_A}{B_B} = \frac{(R^2 + x_B^2)^{3/2}}{(R^2 + x_A^2)^{3/2}} \] Given \( x_A = 4 \) cm, \( x_B = 3\sqrt{3} \) cm, and \( \frac{B_A}{B_B} = \frac{216}{125} \), we equate: \[ \left( R^2 + (3\sqrt{3})^2 \right)^{3/2} = \frac{125}{216} \left( R^2 + 4^2 \right)^{3/2} \] 2. Solving for \( R \): \[ \left( R^2 + 27 \right)^{3/2} = \frac{125}{216} \left( R^2 + 16 \right)^{3/2} \] Taking cube roots and solving, \[ R = 3 \text{ cm} \] Thus, the correct answer is \(\boxed{3 \text{ cm}}\).