A circular wire loop of radius R is placed in the x-y plane centered at origin O. A square loop of side 'a' (a << R) of single turn is placed with its plane parallel to the x-y plane and at a distance of z = √3 R . The mutual inductance between the loops is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{\mu_0 a^{2}}{16R} \)
The mutual inductance is the flux through one loop per unit current in the other, \( M = \Phi/I \). Let a current I flow in the large circular loop. Because \( a \ll R \), its field is nearly uniform over the small square and equal to the field on the axis at \( z = \sqrt{3}R \).
\( \displaystyle B = \frac{\mu_0 I R^{2}}{2\left(R^{2} + z^{2}\right)^{3/2}} \)
\( \displaystyle R^{2} + z^{2} = R^{2} + 3R^{2} = 4R^{2} \Rightarrow \left(4R^{2}\right)^{3/2} = 8R^{3} \)
\( \displaystyle B = \frac{\mu_0 I R^{2}}{2 \times 8R^{3}} = \frac{\mu_0 I}{16R} \)
The square is parallel to the x-y plane, so the axial field is normal to its surface and the flux is B times its area:
\( \displaystyle \Phi = B a^{2} = \frac{\mu_0 I a^{2}}{16R} \)
\( \displaystyle M = \frac{\Phi}{I} = \frac{\mu_0 a^{2}}{16R} \)
Shortcut: Mutual inductance has the dimensions of \( \mu_0 \) times a length. \( \mu_0 R/8a^{2} \) and \( \mu_0 R/4a \) fail this check, leaving Options (1) and (4); the numerical factor from the axial field then decides.
Note: The question does not state that the centre of the square loop lies on the axis of the circular loop (the z-axis). This is implied by giving the position of the square as z = √3R, and the question is solved on that reading.