The figure shows a long cylindrical straight wire of radius \( a \) carrying steady current \( (I) \). The current \( (I) \) is uniformly distributed across the cross-section of the wire. The ratio of the magnetic fields at points \( r = \displaystyle\frac{a}{3} \) and \( r = \displaystyle\frac{a}{2} \) respectively is: Here \( r \) is the distance of point from the axis of cylinder.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2 : 3
Both points are inside the wire. By Ampere's circuital law, a circle of radius \( r \lt a \) encloses only the fraction \( r^{2}/a^{2} \) of the current.
\( \displaystyle B\,(2\pi r) = \mu_0 I\,\frac{r^{2}}{a^{2}} \;\Rightarrow\; B = \frac{\mu_0 I\,r}{2\pi a^{2}} \;\Rightarrow\; B \propto r \)
\( \displaystyle \frac{B_1}{B_2} = \frac{a/3}{a/2} = \frac{2}{3} \)
So the ratio is \( 2 : 3 \).
Note: NTA picked this question from an NCERT (2026-27) example and added the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Ampere’s Circuital Law – “… Example 4.7 Figure 4.13 shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a.”
Ampere’s Circuital Law – “… Using Ampere’s law,
\( \displaystyle B(2\pi r) = \frac{\mu_0 I r^2}{a^2} \)
\( \displaystyle B = \left(\frac{\mu_0 I}{2\pi a^2}\right) r \qquad [4.15(b)] \qquad\qquad B \propto r \quad (r \lt a) \)”