Magnetic field at large distances on the axis of a current carrying circular loop depends on distance from the centre as
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → inversely proportional to the cube of distance
The field on the axis of a circular loop of radius \( R \) carrying current \( I \), at a distance \( x \) from its centre, follows from the Biot-Savart law.
\( \displaystyle B = \frac{\mu_0 I R^{2}}{2\left(R^{2} + x^{2}\right)^{3/2}} \)
At large distances (\( x \gg R \)), \( R^{2} \) can be neglected beside \( x^{2} \), so \( (R^{2} + x^{2})^{3/2} \approx x^{3} \).
\( \displaystyle B \approx \frac{\mu_0 I R^{2}}{2x^{3}} \;\Rightarrow\; B \propto \frac{1}{x^{3}} \)
NCERT (2026-27):
Circular Current Loop as a Magnetic Dipole – “… We shall show that the magnetic field (at large distances) due to current in a circular current loop is very similar in behaviour to the electric field of an electric dipole. … For x >> R, we may drop the R2 term in the denominator. Thus,
\( \displaystyle B = \frac{\mu_0 I R^2}{2x^3} \)”