A long straight solid wire of radius 3 mm carries a steady current of 18 A. The current is spread uniformly over its cross-section. A circle of radius 1 mm is drawn inside the wire, with its centre on the axis of the wire and its plane perpendicular to the wire. The value of \(\oint \vec{B}\cdot d\vec{l}\) around this circle is (take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2.5\times10^{-6}\ \mathrm{T\,m}\)
Ampere’s law uses only the current that passes through the loop. This circle lies inside the wire, so it encloses only a part of the current. When the current is uniform, this part is in the ratio of the two areas, \( I_e=Ir^{2}/a^{2} \).
Given:
\( I=18\ \mathrm{A} \)
\( a=3\ \mathrm{mm} \)
\( r=1\ \mathrm{mm} \)
\( \begin{aligned} I_e &= \frac{Ir^{2}}{a^{2}} \\ &= 18\times\left(\frac{1}{3}\right)^{2} \\ &= 2\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} \oint \vec{B}\cdot d\vec{l} &= \mu_0 I_e \\ &= 4\pi\times10^{-7}\times2 \\ &= 2.5\times10^{-6}\ \mathrm{T\,m} \end{aligned} \)
Therefore, only 2 A of the 18 A counts for this circle.
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.,”
\(\oint \vec{B}\cdot d\vec{l}=\mu_0 I \qquad [4.13(a)] \)
Example 4.7 – “Now the current enclosed Ie is not I, but is less than this value. Since the current distribution is uniform, the current enclosed is,”
\( I_e=I\left(\frac{\pi r^{2}}{\pi a^{2}}\right)=\frac{Ir^{2}}{a^{2}} \)