Magnetic Field Calculation Using Ampere's Law

This problem involves finding the magnetic field ($B$) generated by a long straight wire carrying a uniform current ($I$) distributed over its circular cross-section. We need to determine how the magnetic field changes with the radial distance ($r$) from the center of the wire. We will use Ampere's circuital law for this. Ampere's law states that the line integral of the magnetic field ($B$) around a closed loop is proportional to the total electric current ($I_{enc}$) enclosed by that loop:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$

We need to consider two cases based on the radial distance $r$ compared to the wire's radius $R$. The constant $\mu_0$ is the permeability of free space.

Magnetic Field Outside the Wire ($r > R$)

First, let's consider the region outside the wire, where the radial distance $r$ is greater than the radius $R$ of the wire ($r > R$).

• Amperian Loop: We choose a circular loop of radius $r$ centered on the axis of the wire. This loop lies entirely outside the wire.
• Symmetry: Due to the symmetry of the wire and the uniform current distribution, the magnetic field ($B$) must be constant in magnitude along this circular loop and tangential to it.
• Applying Ampere's Law: The integral becomes $B \times (2\pi r)$, since the path length is the circumference of the loop.
• Enclosed Current ($I_{enc}$): For a loop outside the wire, the enclosed current is the total current carried by the wire, which is $I$.
• Result: Applying Ampere's law, we get:

  $B \cdot (2\pi r) = \mu_0 I$

  Solving for $B$, we find:

  $B = \frac{\mu_0 I}{2\pi r}$

  This formula applies for $r > R$. Notice that the magnetic field decreases as $1/r$ outside the wire.

Magnetic Field Inside the Wire ($r \le R$)

Next, let's consider the region inside the wire, where the radial distance $r$ is less than or equal to the radius $R$ ($r \le R$).

• Amperian Loop: We choose a circular loop of radius $r$ centered on the axis of the wire, lying entirely within the wire.
• Symmetry: Similar to the outside case, the magnetic field ($B$) is constant in magnitude along this loop and tangential to it.
• Applying Ampere's Law: The integral is again $B \times (2\pi r)$.
• Enclosed Current ($I_{enc}$): This is the crucial part. Since the current $I$ is uniformly distributed over the cross-sectional area $A = \pi R^2$, the current density ($J$) is constant:

  $J = \frac{I}{A} = \frac{I}{\pi R^2}$

  The current enclosed by the Amperian loop of radius $r$ is the current density multiplied by the area of the loop ($A_{enc} = \pi r^2$):

  $I_{enc} = J \times A_{enc} = \left( \frac{I}{\pi R^2} \right) \times (\pi r^2) = \frac{I r^2}{R^2}$

• Result: Substituting this enclosed current into Ampere's law:

  $B \cdot (2\pi r) = \mu_0 \left( \frac{I r^2}{R^2} \right)$

  Solving for $B$, we get:

  $B = \frac{\mu_0 I r^2}{2\pi r R^2} = \frac{\mu_0 I r}{2\pi R^2}$

  This formula applies for $r \le R$. The magnetic field increases linearly with $r$ inside the wire, reaching its maximum value at the surface ($r=R$).

Summary of Magnetic Field

Combining the results for both regions:

• Inside the wire ($r \le R$): $B = \frac{\mu_0 I r}{2\pi R^2}$
• Outside the wire ($r > R$): $B = \frac{\mu_0 I}{2\pi r}$

Comparing this with the given options, the correct description is:

$B = \frac{\mu_0 I r}{2\pi R^2}$ for $r \le R$ and $B = \frac{\mu_0 I}{2\pi r}$ for $r > R$.