A co-axial cable consists of an inner wire of radius a surrounded by an outer shell of inner and out

Q: Why is the enclosed current for $x < a$ not $i_0$?

Because the current is uniformly distributed over the entire cross-section of radius $a$. A smaller loop of radius $x$ only encloses a fraction of this area.

Q: What is the direction of the magnetic field?

By the right-hand grip rule, it forms concentric circles around the axis.

Q: Why don't we consider the outer shell for $a < x < b$?

Ampere's law only cares about the current *enclosed* by the loop. The outer shell is outside this loop.

Q: What happens to the magnetic field inside the outer shell ($b < x < c$)?

The enclosed current will be $i_0$ minus the fraction of the return current enclosed by the loop. The field will start decreasing.

Q: Why is the magnetic field zero outside the cable ($x > c$)?

The total enclosed current is $i_0 - i_0 = 0$. Hence, by Ampere's law, the magnetic field is zero.

Question

A co-axial 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) $a < x < b$?

Options:

(A) $\frac{x^2}{a^2}$

(B) $\frac{a^2}{x^2}$

(C) $\frac{x^2}{b^2 - a^2}$

(D) $\frac{b^2 - a^2}{x^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine the cross-section of a co-axial cable. In the center, we have an inner wire of radius $a$, surrounded by an outer shell with inner and outer radii $b$ and $c$. The inner wire carries a current $i_0$ outwards, and the outer shell carries an equal current inwards.

We need to find the magnetic field in different regions. For this, we will use Ampere's Circuital Law. This is a favorite concept for JEE. The law states that the line integral of $\mathbf{B} \cdot d\mathbf{l}$ equals $\mu_0$ times the enclosed current.

### The Master Equation: Ampere's Law
First, let's consider the region where $x < a$. We draw an Amperian loop of radius $x$ inside the inner wire. Since the current is uniformly distributed across the cross-section, the current enclosed by this loop will be a fraction of the total current.

The current density is $J = \frac{i_0}{\pi a^2}$, so the enclosed current is $I_{enc} = \frac{i_0}{\pi a^2} \cdot \pi x^2 = i_0 \frac{x^2}{a^2}$.

Now, let's substitute this into Ampere's Law. 
$$B_1 (2\pi x) = \mu_0 \left( i_0 \frac{x^2}{a^2} \right)$$

From this, we get the value of $B_1$ as:
$$B_1 = \frac{\mu_0 i_0 x}{2\pi a^2}$$

### Moving to the Gap
Let's look at the second case, where $a < x < b$. We draw a new Amperian loop. This loop completely encloses the inner wire, but it doesn't enclose any part of the outer shell. So, in this case, the enclosed current is simply $i_0$.

Applying Ampere's Law, we get:
$$B_2 (2\pi x) = \mu_0 i_0$$

This gives us the value of $B_2$ as:
$$B_2 = \frac{\mu_0 i_0}{2\pi x}$$

### Final Calculation
Finally, we need to find the ratio of $B_1$ and $B_2$. Let's substitute the values and get the answer.

$$\frac{B_1}{B_2} = \frac{\frac{\mu_0 i_0 x}{2\pi a^2}}{\frac{\mu_0 i_0}{2\pi x}}$$

When we divide them, $\mu_0$, $i_0$, and $2\pi$ will cancel out, and we are left with:
**$$\frac{B_1}{B_2} = \frac{x^2}{a^2}$$**

Imagine what would happen if $x > c$? The total enclosed current would be zero, because the current coming out equals the current going in. So, the magnetic field outside would be zero! This is exactly why co-axial cables are used, to prevent external interference.

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Visualized Solution

Co-axial Cable Setup

Ampere's Circuital Law

Region 1: $x < a$

Magnetic Field $B_{1}$ ( $x < a$ )

Region 2: $a < x < b$

Magnetic Field $B_{2}$ ( $a < x < b$ )

Ratio of Magnetic Fields

Beyond the Cable ( $x > c$ )

The Sigma Insight: Ampere's Circuital Law

Analyzing the Setup Imagine the cross-section of a co-axial cable

The Master Equation

Moving to the Gap Let's look at the second case, where $a < x < b$

Final Calculation Finally, we need to find the ratio of $B_{1}$ and $B_{2}$

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Visualized Solution

Co-axial Cable Setup

Ampere's Circuital Law

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