Question

A circular wire loop of radius \(R\) is placed in the \(x\)-\(y\) plane centered at the origin. A square loop of side \(a\;(a\ll R)\) of single turn is placed with its plane parallel to the \(x\)-\(y\) plane and at a distance \(z=\sqrt{3}\,R\). The mutual inductance between the loops is

• A. \[ \frac{\mu_0a^2}{4R} \]
• B. \[ \frac{\mu_0R}{8a^2} \]
• C. \[ \frac{\mu_0R}{4a} \]
• D. \[ \frac{\mu_0a^2}{16R} \]

Hint:

For a circular loop, magnetic field on its axis is \[ B= \frac{\mu_0IR^2} {2(R^2+z^2)^{3/2}} \] When a small loop \((a\ll R)\) is placed in this field, \[ \Phi \approx BA \] and \[ M=\frac{\Phi}{I} \] This approximation is frequently used in mutual inductance problems.

Answer 1

The Correct Option is D

Step 1:Calculate the magnetic field at the location of the square loop. Given, \[ z=\sqrt3\,R \] Hence, \[ R^2+z^2 = R^2+3R^2 = 4R^2 \] Therefore, \[ B= \frac{\mu_0IR^2} {2(4R^2)^{3/2}} \] \[ B= \frac{\mu_0IR^2} {2(8R^3)} \] \[ B= \frac{\mu_0I}{16R} \]

Step 2: Find the magnetic flux through the square loop. Area of square loop: \[ A=a^2 \] Therefore, \[ \Phi=BA \] \[ \Phi= \frac{\mu_0I}{16R}\,a^2 \] \[ \Phi= \frac{\mu_0Ia^2}{16R} \]

Step 3: Calculate mutual inductance. \[ M=\frac{\Phi}{I} \] \[ M= \frac{\mu_0a^2}{16R} \]

Step 4: State the answer. \[ { M=\frac{\mu_0a^2}{16R} } \] Hence, the correct option is \[ {(D)} \]