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

Ampere’s circuital law for a long straight wire states that the line integral of the magnetic field B around a closed path is equal to µ₀ times the total current I enclosed by the path. Mathematically, this is expressed as:

$$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I$$

We consider the two cases:

1. Inside the wire: For a point at a distance r from the center of the wire where \( r < R \) (R being the radius of the wire), the magnetic field is determined by the fraction of the current enclosed by the circle of radius r. If the current density is uniform, the enclosed current at radius r is \( I_{\text{enclosed}} = I \cdot \frac{\pi r^2}{\pi R^2} = I \cdot \frac{r^2}{R^2} \). Applying Ampere’s law: $$ B \cdot 2\pi r = \mu_0 \cdot I \cdot \frac{r^2}{R^2} $$ $$ B = \frac{\mu_0 I r}{2\pi R^2} $$ This implies B increases linearly with r inside the wire.
2. Outside the wire: For a point at a distance r where \( r \geq R \), the entire current I is enclosed within the loop. By Ampere’s law: $$ B \cdot 2\pi r = \mu_0 I $$ $$ B = \frac{\mu_0 I}{2\pi r} $$ Hence, B decreases with 1/r outside the wire.

Thus, the correct description of the variation of magnetic field is that it is a linearly increasing function of distance r up to the boundary of the wire and then decreases with a 1/r dependence for the outside region.