A long straight thin-walled tube and a long straight solid wire have the same radius R and carry equal steady currents I. In the wire the current is uniformly distributed over its cross-section. At a distance R/2 from the axis, the magnitudes of the magnetic field due to the tube and due to the wire are, respectively,
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → zero and \( \mu_0 I/4\pi R \)
Take an Amperian loop, a circle of radius \( r=R/2 \) concentric with the axis, in each case. By symmetry, the field has the same magnitude all round the loop and is tangential to it. So Ampere’s circuital law gives \( B(2\pi r)=\mu_0 I_e \), where \( I_e \) is the current enclosed by the loop.
For the thin-walled tube, all the current flows in the wall at radius R, outside the loop. So \( I_e=0 \) and the field is zero.
For the solid wire, the loop encloses only the part of the current that flows inside radius \( R/2 \).
The current enclosed in the wire is:
\( \begin{aligned} I_e &= I\,\frac{\pi(R/2)^{2}}{\pi R^{2}} \\ &= \frac{I}{4} \end{aligned} \)
The field due to the wire at \( r=R/2 \) is then:
\( \begin{aligned} B &= \frac{\mu_0 I_e}{2\pi r} \\ &= \frac{\mu_0 (I/4)}{2\pi (R/2)} \\ &= \frac{\mu_0 I}{4\pi R} \end{aligned} \)
Hence, the fields due to the tube and the wire are zero and \( \mu_0 I/4\pi R \), respectively.
NCERT (2026-27):
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,”
\( \displaystyle I_e=I\left(\frac{\pi r^{2}}{\pi a^{2}}\right)=\frac{Ir^{2}}{a^{2}} \)
“Using Ampere’s law,”
\( \displaystyle B(2\pi r)=\mu_0\frac{Ir^{2}}{a^{2}} \)
\( \displaystyle B=\left(\frac{\mu_0 I}{2\pi a^{2}}\right)r \qquad [4.15(b)] \)
\( B\propto r \quad (r\lt a) \)