Step 1: Understanding the Concept:
According to Ampere’s Circuital Law, the magnetic field produced by a current-carrying wire depends on the distance from its central axis.
For a solid cylindrical wire of radius \(a\) with a uniformly distributed current \(I\):
1. Inside the wire (\(r<a\)): Only the current enclosed by the Amperean loop of radius \(r\) contributes to the field. This current is \(I' = I \times \frac{\text{Area of loop}}{\text{Total area}} = I \frac{r^2}{a^2}\).
2. Outside the wire (\(r \geq a\)): The entire current \(I\) is enclosed by any Amperean loop of radius \(r \geq a\).
Step 2: Key Formula or Approach:
The formulas for the magnetic field (\(B\)) at a distance \(r\) from the axis are:
Inside the wire (\(r<a\)): \[ B_{in} = \frac{\mu_0 I r}{2 \pi a^2} \]
Outside the wire (\(r>a\)): \[ B_{out} = \frac{\mu_0 I}{2 \pi r} \]
Step 3: Detailed Explanation:
Let's calculate the magnetic field at the two given points.
Point 1: \(r_1 = a/4\)
Since \(a/4<a\), this point is inside the wire. We use the formula for \(B_{in}\):
\[ B_1 = \frac{\mu_0 I (a/4)}{2 \pi a^2} = \frac{\mu_0 I a}{8 \pi a^2} = \frac{\mu_0 I}{8 \pi a} \]
Point 2: \(r_2 = 3a\)
Since \(3a>a\), this point is outside the wire. We use the formula for \(B_{out}\):
\[ B_2 = \frac{\mu_0 I}{2 \pi (3a)} = \frac{\mu_0 I}{6 \pi a} \]
Calculating the ratio \(B_1 : B_2\):
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{8 \pi a}}{\frac{\mu_0 I}{6 \pi a}} \]
The constant terms (\(\mu_0, I, \pi, a\)) cancel out:
\[ \frac{B_1}{B_2} = \frac{1/8}{1/6} = \frac{6}{8} \]
Simplifying the ratio:
\[ \frac{B_1}{B_2} = \frac{3}{4} \]
Step 4: Final Answer:
The ratio of the magnetic fields at \(a/4\) and \(3a\) is 3 : 4.