From Ampere's circuital law, for a long straight wire of circular cross-section carrying a steady current, the

Question

From Ampere's circuital law, for a long straight wire of circular cross-section carrying a steady current, the variation of the magnetic field inside and outside the region of the wire is:

Accepted Answer

1.  **Inside the Wire ($r < a$):** Assuming the current $I$ is uniformly distributed over the cross-section of radius $a$, the current enclosed by a loop of radius $r$ is $I_{enclosed} = I(r^2/a^2)$. Applying Ampere's Law: $B(2\pi r) = \mu_0 I(r^2/a^2) \implies B = \left(\frac{\mu_0 I}{2\pi a^2}\right)r$. Thus, the magnetic field is directly proportional to the distance ($B \propto r$), representing a linear increase.
2.  **Outside the Wire ($r > a$):** The current enclosed is the total current $I$. Applying Ampere's Law: $B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$. Thus, the magnetic field is inversely proportional to the distance ($B \propto 1/r$).
3.  **Conclusion:** The field increases linearly up to the boundary and then decreases with $1/r$ dependence outside.

Step-by-Step Solution

Practice All PHYSICS Questions