Question

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

• A. Uniform and remains constant for both the regions
• B. A linearly increasing function of distance up to the boundary of the wire and then linearly decreasing for the outside region
• C. A linearly increasing function of distance r up to the boundary of the wire and then decreasing one with 1/r dependence for the outside region
• D. A linearly decreasing function of distance up to the boundary of the wire and then a linearly increasing one for the outside region

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Ampere's Circuital Law (\(\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{encl}\)) is used to find the magnetic field distribution for symmetric current configurations like a long thick wire.
Key Formula or Approach:
Let \(R\) be the radius of the wire and \(r\) be the distance from the axis.
1. Inside (\(r<R\)): Current enclosed \(I' = I \times \frac{\pi r^2}{\pi R^2}\).
2. Outside (\(r \ge R\)): Total current \(I\) is enclosed.
Step 2: Detailed Explanation:
Case 1: Inside the wire (\(r<R\)):
Applying Ampere's law:
\[ B(2\pi r) = \mu_0 \left( I \frac{r^2}{R^2} \right) \]
\[ B = \frac{\mu_0 I}{2\pi R^2} r \]
Since all other terms are constant, \(B \propto r\). The field increases linearly from the center.

Case 2: Outside the wire (\(r \ge R\)):
Applying Ampere's law:
\[ B(2\pi r) = \mu_0 I \]
\[ B = \frac{\mu_0 I}{2\pi r} \]
Here, \(B \propto \frac{1}{r}\). The field decreases as the reciprocal of the distance.
Step 3: Final Answer:
The field increases linearly inside the wire and decreases as \(1/r\) outside the wire.