An infinitely long wire, located on the z-axis, carries a current I along the +z-direction and produ

Q: Why did we use cylindrical coordinates instead of Cartesian?

The magnetic field of a straight wire has a natural cylindrical symmetry. In Cartesian coordinates, the dot product $\vec{B} \cdot d\vec{l}$ becomes a messy rational function of $x$ and $y$. In cylindrical coordinates, it simplifies to just a constant times $d\theta$, making the integration trivial!

Q: Does the straight line path matter? What if it was a curve?

That's the beauty of this result! As long as the path doesn't enclose the origin (the wire), the integral $\int \vec{B} \cdot d\vec{l}$ depends *only* on the starting and ending angles. Any path between $i$ and $f$ will give the exact same result.

Q: Why did we measure the angles from the y-axis?

You can measure the angles from any reference line, like the positive x-axis. We used the y-axis simply because the geometry of the points $(-\sqrt{3}a, a)$ and $(a, a)$ makes the triangles with the y-axis very easy to visualize ($30^\circ-60^\circ-90^\circ$ and $45^\circ-45^\circ-90^\circ$). The total angle difference $\Delta\theta$ will be the same regardless of your reference line.

Q: What if the path crossed the negative y-axis instead of the positive y-axis?

If the path went 'the long way around' the wire, the change in angle $\Delta\theta$ would be different (it would be $2\pi - 7\pi/12$). The line integral would have a different value. This is because the magnetic field is not a conservative field; the integral over a closed loop enclosing the current is not zero (Ampere's Law!).

Q: Why did the $dr$ and $dz$ terms disappear in the dot product?

The magnetic field $\vec{B}$ is purely in the azimuthal ($\hat{\theta}$) direction. The dot product of $\hat{\theta}$ with the radial direction $\hat{r}$ and the vertical direction $\hat{k}$ is zero because they are mutually perpendicular unit vectors.

Question

An infinitely long wire, located on the z-axis, carries a current $I$ along the $+z$-direction and produces the magnetic field $\vec{B}$. The magnitude of the line integral $\int \vec{B} \cdot d\vec{l}$ along a straight line from the point $(-\sqrt{3}a, a, 0)$ to $(a, a, 0)$ is given by
[$\mu_0$ is the magnetic permeability of free space.]

Options:

(A) $7\mu_0 I / 24$

(B) $7\mu_0 I / 12$

(C) $\mu_0 I / 8$

(D) $\mu_0 I / 6$

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine you are looking down at the x-y plane from above. Right at the origin, piercing through the screen towards you, is an infinitely long wire.

This wire carries a steady current $I$ along the positive z-axis.

Our mission is to calculate the line integral of the magnetic field, $\int \vec{B} \cdot d\vec{l}$, along a straight path connecting two specific points in this plane: an initial point $i(-\sqrt{3}a, a, 0)$ and a final point $f(a, a, 0)$.

### The Magnetic Field and the Path

First, let's recall what the magnetic field of an infinitely long straight wire looks like.

According to the right-hand thumb rule, the magnetic field lines form concentric circles around the wire.

Mathematically, we can express this field in cylindrical coordinates as:
$$ \vec{B} = \frac{\mu_0 I}{2\pi r} \hat{	heta} $$

Now, consider a tiny displacement vector $d\vec{l}$ along our path. In cylindrical coordinates, any general displacement can be written as:
$$ d\vec{l} = dr \hat{r} + r d	heta \hat{	heta} + dz \hat{k} $$

### The Magic of the Dot Product

Here is where the physics gets truly elegant. We need to evaluate the dot product $\vec{B} \cdot d\vec{l}$.

Because the magnetic field $\vec{B}$ points purely in the azimuthal direction ($\hat{	heta}$), the dot product with the radial ($\hat{r}$) and vertical ($\hat{k}$) components of $d\vec{l}$ is exactly zero!

$$ \vec{B} \cdot d\vec{l} = \left( \frac{\mu_0 I}{2\pi r} \hat{	heta} ight) \cdot (dr \hat{r} + r d	heta \hat{	heta} + dz \hat{k}) $$

$$ \vec{B} \cdot d\vec{l} = \frac{\mu_0 I}{2\pi r} (r d	heta) $$

Notice what happens next. The radial distance $r$ cancels out completely!

$$ \vec{B} \cdot d\vec{l} = \frac{\mu_0 I}{2\pi} d	heta $$

### Path Independence

This cancellation is a profound result. It tells us that the tiny contribution to the line integral depends **only** on the change in the angle $d	heta$, and not on how far away we are from the wire ($r$).

Therefore, the total line integral from point $i$ to point $f$ is simply the integral of $d	heta$:

$$ \int_i^f \vec{B} \cdot d\vec{l} = \int_{	heta_i}^{	heta_f} \frac{\mu_0 I}{2\pi} d	heta = \frac{\mu_0 I}{2\pi} (	heta_f - 	heta_i) = \frac{\mu_0 I}{2\pi} \Delta	heta $$

This means the integral is **path-independent** for any path that doesn't loop around the wire. It only depends on the total angle $\Delta	heta$ subtended by the start and end points at the origin.

### Calculating the Angles

Now, we just need to find this total angle $\Delta	heta$. Let's look at the coordinates of our points and measure their angles from the positive y-axis for simplicity.

For the initial point $i(-\sqrt{3}a, a, 0)$, the x-coordinate is negative and the y-coordinate is positive. The angle it makes with the y-axis is:
$$ 	heta_1 = 	an^{-1}\left(\frac{|x|}{y}ight) = 	an^{-1}\left(\frac{\sqrt{3}a}{a}ight) = 	an^{-1}(\sqrt{3}) = \frac{\pi}{3} $$

For the final point $f(a, a, 0)$, both coordinates are positive. The angle it makes with the y-axis is:
$$ 	heta_2 = 	an^{-1}\left(\frac{x}{y}ight) = 	an^{-1}\left(\frac{a}{a}ight) = 	an^{-1}(1) = \frac{\pi}{4} $$

### The Final Calculation

The total angular sweep from point $i$ to point $f$ is the sum of these two angles:

$$ \Delta	heta = 	heta_1 + 	heta_2 = \frac{\pi}{3} + \frac{\pi}{4} $$

To add these fractions, we find a common denominator, which is 12:

$$ \Delta	heta = \frac{4\pi}{12} + \frac{3\pi}{12} = \frac{7\pi}{12} $$

Finally, we substitute this total angle back into our simplified integral expression:

$$ \left| \int_i^f \vec{B} \cdot d\vec{l} ight| = \frac{\mu_0 I}{2\pi} 	imes \frac{7\pi}{12} = \frac{7\mu_0 I}{24} $$

This is our final answer, matching option (A).

### The Ampere's Law Connection

Before we finish, let's appreciate the deeper symmetry here. Ampere's Law states that the closed loop integral $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{	ext{enclosed}}$.

If we had formed a closed loop by traveling from $i$ to $f$ along the straight line, then radially inward to the origin, and finally radially outward back to $i$, the integral along the radial segments would be zero.

Thus, the integral along our path is exactly the same as the integral along a circular arc subtending the same angle. It represents a fraction of the full Ampere's loop, specifically $\frac{\Delta	heta}{2\pi}$ of the total $\mu_0 I$.

This beautiful connection between geometry and electromagnetism is what makes physics so incredibly satisfying!

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Magnetic Field of a Long Wire}

\text{Evaluating } \vec{B} \cdot d\vec{l}

\text{The Line Integral}

\text{Calculating } \Delta\theta

\text{Final Calculation}

\text{Ampere's Law Connection}

The Sigma Insight: Ampere's Circuital Law

Visualizing the Setup

The Magnetic Field and the Path

The Magic of the Dot Product

Path Independence

Calculating the Angles

The Final Calculation

The Ampere's Law Connection

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Visualizing the Setup} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Magnetic Field of a Long Wire} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Evaluating } \vec{B} \cdot d\vec{l} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{The Line Integral} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Calculating } \Delta\theta .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Final Calculation} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Ampere's Law Connection} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }