The magnitude of the magnetic field \( (B) \) on the axis of a circular current \( (I) \) carrying loop of radius \( R \) at a distance \( x \) from its center is given by the expression
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.5\,\mu_{0}IR^{2}(x^{2} + R^{2})^{-3/2} \)
The field on the axis of a circular loop follows from the Biot–Savart law: the components of the field perpendicular to the axis cancel in pairs and the axial components add.
\( \displaystyle B = \frac{\mu_{0}IR^{2}}{2(x^{2}+R^{2})^{3/2}} \)
\( \displaystyle B = 0.5\,\mu_{0}IR^{2}(x^{2}+R^{2})^{-3/2} \)
Shortcut: Limiting case \( x = 0 \): the field at the centre of a loop is \( \mu_{0}I/2R \). Option (2) gives \( 0.5\,\mu_{0}IR^{2}\cdot R^{-3} = \mu_{0}I/2R \). Option (3) gives four times this; Options (1) and (4) increase with x, whereas the field must fall off far from the loop (as \( 1/x^{3} \)).
NCERT (2026-27):
Magnetic Field on the Axis of a Circular Current Loop – “In this section, we shall evaluate the magnetic field due to a circular coil along its axis.”
Magnetic Field on the Axis of a Circular Current Loop – “… Thus, the magnetic field at P due to entire circular loop is
\( \displaystyle \mathbf{B} = B_x \hat{\mathbf{i}} = \frac{\mu_0 I R^2}{2\left(x^2 + R^2\right)^{3/2}}\,\hat{\mathbf{i}} \qquad (4.11) \)”