A long straight wire of radius a carries a steady current I. The current is uniformly distributed ac

Q: Why isn't the enclosed current inside just $I/2$ since the radius is half?

This is a very common misconception! Current is distributed over the cross-sectional *area*, not just along the radius. Since area is proportional to the square of the radius ($A = \pi r^2$), halving the radius reduces the area to one-quarter ($1/2^2 = 1/4$). Therefore, the enclosed current is $I/4$.

Q: Does the magnetic field keep increasing inside the wire?

Yes, it does! Inside a solid wire with uniform current density, the magnetic field is directly proportional to the distance from the center ($B \propto r$). It starts at zero at the exact center and reaches its maximum value right at the surface of the wire ($r = a$).

Q: What happens to the magnetic field outside the wire?

Once you step outside the wire, the enclosed current stops increasing (it's capped at the total current $I$). However, the circumference of your Amperian loop keeps growing. Because of this, the magnetic field outside decreases inversely with distance ($B \propto 1/r$).

Q: Why do we use a circular Amperian loop?

We choose a circular loop because it perfectly matches the cylindrical symmetry of the wire. On a circular path centered on the wire, the magnetic field $\mathbf{B}$ has a constant magnitude and is always parallel to the path element $d\mathbf{l}$. This turns the complex dot product integral into a simple multiplication: $B \times 2\pi r$.

Q: What if the wire was a hollow pipe with current only on the surface?

If it were a hollow pipe (a thin cylindrical shell), any Amperian loop drawn *inside* the pipe would enclose absolutely zero current. According to Ampere's Law, this means the magnetic field everywhere inside the hollow region would be exactly zero!

Question

A long straight wire of radius $a$ carries a steady current $I$. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at $\frac{a}{2}$ and $2a$ is

Options:

(A) 1/4

(B) 4

(C) 1

(D) 1/2

Sigma Percentile · Accepted Answer

Imagine you are standing inside a massive, infinitely long cylindrical pipe. But instead of water, this pipe is carrying a steady, uniform flow of electric current, $I$. This is the classic setup for one of the most elegant applications of **Ampere's Circuital Law**.

In this problem, we are asked to compare the magnetic field at two very specific locations: one deep inside the wire at a distance of $a/2$ from the central axis, and another far outside the wire at a distance of $2a$. At first glance, you might think the magnetic field behaves completely differently in these two regions. And you would be right! But the beauty of physics lies in how these different behaviors can sometimes lead to surprising symmetries.

Let's embark on this journey by breaking down the problem into two distinct phases: exploring the inner depths of the wire, and then stepping outside to see the bigger picture.

### The Master Tool: Ampere's Circuital Law

Before we dive in, we need the right tool. When dealing with highly symmetric current distributions—like an infinitely long, straight cylinder—Biot-Savart's Law can be a nightmare of integration. Instead, we turn to its far more elegant cousin: **Ampere's Circuital Law**.

Ampere's Law states that the line integral of the magnetic field $\mathbf{B}$ around any closed loop is directly proportional to the net current enclosed by that loop:

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

Because of the cylindrical symmetry, if we choose a circular path (an Amperian loop) centered on the wire's axis, the magnetic field $\mathbf{B}$ will be constant in magnitude everywhere on that loop and perfectly parallel to the path element $d\mathbf{l}$. This simplifies our terrifying integral into a simple multiplication: $B 	imes (2\pi r) = \mu_0 I_{	ext{enc}}$.

### Phase 1: Journey Inside the Wire ($r = a/2$)

Let's shrink down and travel inside the wire to a distance $r = a/2$ from the center. We draw our imaginary circular Amperian loop here. The critical question is: **How much current is actually passing through this specific loop?**

It's not the total current $I$. Since the current is uniformly distributed across the entire cross-section of radius $a$, we need to find the **current density**, $J$. Think of current density as the "crowdedness" of the current.

$$J = \frac{	ext{Total Current}}{	ext{Total Area}} = \frac{I}{\pi a^2}$$

Now, to find the enclosed current $I_{	ext{enc}}$ for our small loop of radius $a/2$, we multiply this uniform density by the area of our specific loop:

$$I_{	ext{enc}} = J 	imes 	ext{Area of loop} = \left(\frac{I}{\pi a^2}ight) 	imes \pi \left(\frac{a}{2}ight)^2$$

$$I_{	ext{enc}} = \frac{I}{\pi a^2} 	imes \frac{\pi a^2}{4} = \frac{I}{4}$$

Fascinating! At half the radius, we only enclose one-quarter of the total current. This is a classic trap where students make silly mistakes—remember, area scales with the *square* of the radius.

Now, we plug this enclosed current back into Ampere's Law:

$$B_1 	imes 2\pi\left(\frac{a}{2}ight) = \mu_0 \left(\frac{I}{4}ight)$$

$$B_1 	imes \pi a = \frac{\mu_0 I}{4}$$

$$B_1 = \frac{\mu_0 I}{4\pi a}$$

This is the magnetic field strength at our inner point.

### Phase 2: Stepping Outside ($r = 2a$)

Now, let's teleport outside the wire to a distance $r = 2a$. We draw a massive Amperian loop that completely encircles the wire.

What is the enclosed current now? This is the easy part. Since our loop is larger than the wire itself, it captures **all** of the current flowing through the wire.

$$I_{	ext{enc}} = I$$

We apply Ampere's Law once more, this time with our new radius $2a$:

$$B_2 	imes 2\pi(2a) = \mu_0 I$$

$$B_2 	imes 4\pi a = \mu_0 I$$

$$B_2 = \frac{\mu_0 I}{4\pi a}$$

### The Grand Finale: The Ratio

Look at the two results we just derived.

Inside at $r = a/2$: $B_1 = \frac{\mu_0 I}{4\pi a}$

Outside at $r = 2a$: $B_2 = \frac{\mu_0 I}{4\pi a}$

They are exactly identical! The magnetic field strength halfway to the surface is precisely the same as the magnetic field strength at twice the radius outside.

Therefore, the ratio is simply:

$$\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{4\pi a}}{\frac{\mu_0 I}{4\pi a}} = 1$$

This problem is a beautiful demonstration of how the magnetic field behaves around a solid conductor. Inside the wire, the field grows linearly from zero at the center to a maximum at the surface. Outside the wire, it decays inversely with distance ($1/r$). The points $a/2$ and $2a$ just happen to be the perfect geometric counterparts where the rising inner field and the falling outer field perfectly match.

Animated Solution for Physics - Magnetic Effects of Current: A long straight wire of radius $a$ carries a steady current $I$ . The current is uniformly distributed across its cross-section. The ratio of the magnetic field at $\frac{a}{2}$ and $2 a$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Ampere's Circuital Law

Field Inside the Wire

Calculating $B_{1}$

Field Outside the Wire

Calculating $B_{2}$

Final Ratio

The Sigma Insight: Ampere's Circuital Law

The Master Tool

Phase 1

Phase 2

The Grand Finale

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