Question

A coaxial cable consists of an inner wire of radius 'a' surrounded by an outer shell of inner and outer radii 'b' and 'c' respectively. The inner wire carries an electric current \(i_0\), which is distributed uniformly across cross-sectional area. The outer shell carries an equal current in opposite direction and distributed uniformly. What will be the ratio of the magnetic field at a distance x from the axis when (i) x \(<\) a and (ii) a \(<\) x \(<\) b?

• A. \( \frac{x^2}{a^2} \)
• B. \( \frac{a^2}{x^2} \)
• C. \( \frac{x^2}{b^2 - a^2} \)
• D. \( \frac{b^2 - a^2}{x^2} \)

Hint:

For problems involving Ampere's law with uniform current distribution, the enclosed current is the key. Inside a solid conductor (\(x<a\)), the enclosed current is proportional to the area, \(I_{\text{enc}} \propto x^2\). Outside a conductor (\(x>a\)), the enclosed current is the total current of that conductor. Mastering this concept is essential for magnetism problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the ratio of the magnetic field expressions in two different regions of a coaxial cable: inside the inner conductor (\(x<a\)) and between the inner and outer conductors (\(a<x<b\)). We will use Ampere's Law to find the magnetic field in each region.
Step 2: Key Formula or Approach:
Ampere's Circuital Law: The line integral of the magnetic field \(\vec{B}\) around a closed loop is proportional to the total current \(I_{\text{enc}}\) enclosed by the loop.
\[ \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}} \] For a cylindrical wire, due to symmetry, this simplifies to \( B(2\pi x) = \mu_0 I_{\text{enc}} \), where \(x\) is the radial distance from the axis.
Step 3: Detailed Explanation:
Case (i): Magnetic field at \( x<a \) (inside the inner wire)
Let's call the magnetic field in this region \( B_1 \). We apply Ampere's law for a circular loop of radius \(x\).
The current \(i_0\) is distributed uniformly over the cross-sectional area \( \pi a^2 \). The current density is \( J = \frac{i_0}{\pi a^2} \).
The current enclosed by the loop of radius \(x\) is:
\[ I_{\text{enc}} = J \times (\text{Area of loop}) = \left(\frac{i_0}{\pi a^2}\right) \times (\pi x^2) = i_0 \frac{x^2}{a^2} \] Applying Ampere's Law:
\( B_1 (2\pi x) = \mu_0 I_{\text{enc}} = \mu_0 \left( i_0 \frac{x^2}{a^2} \right) \)
\[ B_1 = \frac{\mu_0 i_0 x}{2\pi a^2} \] Case (ii): Magnetic field at \( a<x<b \) (between the conductors)
Let's call the magnetic field in this region \( B_2 \). We apply Ampere's law for a circular loop of radius \(x\).
In this region, the loop encloses the entire current \(i_0\) from the inner wire. The current from the outer shell is outside the loop, so it is not included in \(I_{\text{enc}}\).
\[ I_{\text{enc}} = i_0 \] Applying Ampere's Law:
\( B_2 (2\pi x) = \mu_0 i_0 \)
\[ B_2 = \frac{\mu_0 i_0}{2\pi x} \] Finding the Ratio:
The question asks for the ratio of the magnetic field at a distance \(x\) in case (i) to the magnetic field at a distance \(x\) in case (ii). Although a single value of \(x\) cannot be in both regions simultaneously, the question implies finding the ratio of the functional forms of the magnetic field expressions in the two regions.
Ratio = \( \frac{B_1(x)}{B_2(x)} \)
\[ \text{Ratio} = \frac{\frac{\mu_0 i_0 x}{2\pi a^2}}{\frac{\mu_0 i_0}{2\pi x}} \] \[ \text{Ratio} = \left(\frac{\mu_0 i_0 x}{2\pi a^2}\right) \times \left(\frac{2\pi x}{\mu_0 i_0}\right) \] The terms \( \mu_0, i_0, 2\pi \) cancel out.
\[ \text{Ratio} = \frac{x}{a^2} \times x = \frac{x^2}{a^2} \] Step 4: Final Answer:
The ratio of the magnetic field expressions is \( \frac{x^2}{a^2} \).