91Ó°ÊÓ

Ampere's law states that the closed-line integral of the magnetic field (B) around any closed loop in the direction that is tangent to the loop is equal to \(\mu _{0}\) times the current enclosed by the loop. Mathematically, Ampere's Law can be expressed as: \[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_{0}I_{enclosed} \]

We need to choose an appropriate amperian loop (a closed-loop) for integration. Since the magnetic field due to the current in the thin-walled tube has cylindrical symmetry, we'll choose a circular loop of radius r as the amperian loop. Let's call this loop L.

According to Ampere's law, we can calculate the magnetic field inside the tube as follows: \[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_{0}I_{enclosed} \] Since B is uniform along the tangent to the amperian loop L, we can remove it from the integral: \[ B \oint d\mathbf{l} = \mu_{0}I_{enclosed} \] As loop L has a circumference of 2Ï€r, we have: \[ B(2 \pi r) = \mu_{0}I_{enclosed} \] Now, we need to find the current enclosed by the Amperian loop L: \[ I_{enclosed} = i \] (Had we taken a circular Amperian loop outside the tube, we would have had no current enclosed by the loop, and subsequently a magnetic induction of zero) Now, let's solve for B: \[ B = \frac{\mu_{0}i}{2 \pi r} \] Comparing with the options mentioned earlier, we find that neither of the given options matches the derived value for magnetic induction. The question appears faulty having none of the given options as the correct answer.