Case: Read the passage and answer the question.
Consider a circular current-carrying conducting coil.
If the radius of the coil is halved, keeping the current unchanged, what happens to the magnetic field on its axis at far-off points?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → It becomes one-fourth
On the axis of a circular loop of radius \( R \), at a distance \( x \) from its centre, the field is:
\( \displaystyle B=\frac{\mu_0IR^{2}}{2(x^{2}+R^{2})^{3/2}} \)
At far-off points (\( x\gg R \)) the \( R^{2} \) in the denominator is very small next to \( x^{2} \) and can be left out. So, for the same current and point, the field is proportional to \( R^{2} \).
\( \displaystyle B\approx\frac{\mu_0IR^{2}}{2x^{3}} \)
With the radius halved:
\( \begin{aligned} \frac{B'}{B} &= \left(\frac{R/2}{R}\right)^{2} \\ &= \frac{1}{4} \end{aligned} \)
Therefore, the field at far-off axial points becomes one-fourth.
NCERT (2026-27):
Circular Current Loop as a Magnetic Dipole – “… Here, x is the distance along the axis from the centre of the loop. For x >> R, we may drop the R2 term in the denominator. Thus,
\( \displaystyle B=\frac{\mu_0IR^{2}}{2x^{3}} \)”