A long coaxial cable has a thin central wire and a thin outer metal tube of radius 2 cm around the same axis. The central wire carries a current of 4 A. The tube carries 4 A back in the opposite direction. The magnitude of the magnetic field at a distance of 1 cm from the axis is (take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(8\times10^{-5}\ \mathrm{T}\)
Take an Amperian loop as a circle of radius 1 cm around the axis. By symmetry, \(B\) is tangential and has the same size at every point of this circle. Ampere’s law uses only the current enclosed by the loop. The tube lies outside this circle, so only the 4 A of the central wire is enclosed.
\(I_e=4\ \mathrm{A}\)
\(r=1\ \mathrm{cm}=0.01\ \mathrm{m}\)
\( \begin{aligned} B(2\pi r) &= \mu_0 I_e \\ B &= \frac{\mu_0 I_e}{2\pi r} \\ &= \frac{4\pi\times10^{-7}\times4}{2\pi\times0.01} \\ &= 8\times10^{-5}\ \mathrm{T} \end{aligned} \)
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.,”
\( \oint \vec{B}\cdot d\vec{l}=\mu_0 I \)\( \qquad [4.13(a)] \)
Summary – “7. The magnitude of the magnetic field at a distance R from a long, straight wire carrying a current I is given by:”
\( B=\frac{\mu_0 I}{2\pi R} \)