A long straight wire carries a steady current. The magnetic field at a distance of 5 cm from the wire is \(2\times10^{-5}\ \mathrm{T}\). The value of \(\oint \vec{B}\cdot d\vec{l}\) around a circle of radius 10 cm, with the wire along its axis, is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2\pi\times10^{-6}\ \mathrm{T\,m}\)
By Ampere’s law, \( \oint \vec{B}\cdot d\vec{l} \) around a closed loop is \( \mu_0 \) times the current through the loop. Both circles go round the same wire, so the integral is the same for both. We can find it from the 5 cm circle, where \( B \) is known.
On a circle round the wire, \( \vec{B} \) is along the circle and has the same size at every point. So the integral is \( B\times2\pi r \).
Here \( r=5\ \mathrm{cm}=0.05\ \mathrm{m} \).
\( \begin{aligned} \oint \vec{B}\cdot d\vec{l} &= B\times2\pi r \\ &= 2\times10^{-5}\times2\pi\times0.05 \\ &= 2\pi\times10^{-6}\ \mathrm{T\,m} \end{aligned} \)
Therefore, the integral around the 10 cm circle is also \( 2\pi\times10^{-6}\ \mathrm{T\,m} \). On that circle the field is half as large, but the circle is twice as long.
NCERT (2026-27):
Ampere’s Circuital Law – “… Ampere’s law states that this integral is equal to μ0 times the total current passing through the surface, i.e.,”
\( \oint \vec{B}\cdot d\vec{l}=\mu_0 I \)\( \qquad [4.13(a)] \)
Ampere’s Circuital Law – “… It implies that the field at every point on a circle of radius r, (with the wire along the axis), is same in magnitude. … (ii) The field direction at any point on this circle is tangential to it.”