Q.An element $\Delta \vec{l} = \Delta x\,\hat{i}$ is placed at the origin and carries a large current $I = 10\ \text{A}$ (Fig. 4.8). What is the magnetic field on the $y$-axis at a distance of $0.5\ \text{m}$. $\Delta x = 1\ \text{cm}$.

Concept: Biot–Savart law for a current element.

For a short element, $d\vec B=\dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec l\times\hat r}{r^2}$. Here $d\vec l=\Delta x\,\hat i=(0.01\ \text{m})\hat i$, $I=10\ \text{A}$, and the point is on the $y$-axis at $r=0.5\ \text{m}$, so $\hat r=\hat j$ and the angle between $d\vec l$ and $\hat r$ is $90^\circ$. …

Why Biot–Savart (not the long-wire formula)

The segment $\Delta x=1\ \text{cm}$ is tiny compared with $r=0.5\ \text{m}$, so it acts as a point source of field and we use the Biot–Savart law directly, with its $1/r^2$ fall-off:

1. Identify the vectors

The element is at the origin pointing along $+x$: $d\vec l=(0.01\ \text{m})\hat i$. The field point is on the $y$-axis, so $\vec r=0.5\,\hat j\ \text{m}$ and $\hat r=\hat j$. The angle between $d\vec l$ and $\hat r$ is $90^\circ$.

2. Direction

so the field points along $+z$ — out of the page. The magnitude of the cross product is $\Delta x\sin90^\circ=\Delta x$.

3. Magnitude

Using $\dfrac{\mu_0}{4\pi}=10^{-7}\ \text{T·m/A}$ and $r=0.5\ \text{m}$: …

Method: Biot–Savart Law

The Biot–Savart law gives the magnetic field due to a small current element:

Steps

1. Identify the given quantities

• Current element: $d\vec{l} = \Delta x \,\hat{i} = 0.01 \,\hat{i}\ \text{m}$
• Current: $I = 10\ \text{A}$
• Point of interest: on the $y$-axis at $y = 0.5\ \text{m}$
• $\mu_0 = 4\pi \times 10^{-7}\ \text{T·m/A}$

2. Determine the position vector and unit vector

From the origin to the point $(0, 0.5, 0)$:

3. Compute the cross product $d\vec{l} \times \hat{r}$

4. Apply the Biot–Savart law …

Common Mistakes in Magnetic Field Due to a Current Element (Biot–Savart Law)

Mistake 1: Forgetting the Vector Nature of the Cross Product

The error: Students often plug magnitudes into the Biot–Savart law without considering the direction of $\Delta \vec{l} \times \hat{r}$.

Why it happens: The formula $dB = \frac{\mu_0}{4\pi} \frac{I \, dl \sin\theta}{r^2}$ is memorised as a scalar, but $\theta$ is the angle between $\Delta \vec{l}$ and $\hat{r}$ — not just any angle in the diagram.

How to avoid: Always write the full vector form first:

Then compute the cross product explicitly. Here, $\Delta \vec{l} = \Delta x \, \hat{i}$ and $\hat{r}$ points from the origin to the point on the $y$-axis, so $\hat{r} = \hat{j}$. Since $\hat{i} \times \hat{j} = \hat{k}$, the field is along $\hat{k}$ (out of the page).

Mistake 2: Using the Wrong Distance $r$

The error: Taking $r = 0.5\ \text{m}$ directly as the distance from the element to the field point.

Why it happens: The problem gives "distance on the $y$-axis = $0.5\ \text{m}$", which is the $y$-coordinate. But the current element is at the origin, so the distance from the element to the point $(0, 0.5)$ is indeed $0.5\ \text{m}$ — this is correct here. However, students often confuse this when the element is not at the origin.

How to avoid: Always identify the position vector from the current element to the field point. Here:

• Element at $(0,0,0)$
• Field point at $(0, 0.5, 0)$
• So $\vec{r} = 0.5\,\hat{j}$ and $r = 0.5\ \text{m}$ ✓

Mistake 3: Forgetting Unit Conversion for $\Delta x$

The error: Using $\Delta x = 1\ \text{cm}$ as $1$ in the formula without converting to metres.

Why it happens: The current is in amperes and distance in metres, but $\Delta x$ is given in cm. Students overlook the mismatch.

How to avoid: Convert all lengths to metres before plugging in:

The correct magnitude becomes:

Mistake 4: Confusing $\mu_0/4\pi$ with $\mu_0$

The error: Using $\mu_0 = 4\pi \times 10^{-7}$ directly without the $1/4\pi$ factor.

Why it happens: Many remember $\mu_0 = 4\pi \times 10^{-7}$, but the Biot–Savart law has $\mu_0/4\pi = 10^{-7}$.

How to avoid: Memorise the constant as:

This is the number you actually multiply by. Using $\mu_0$ alone gives an answer $4\pi$ times too large.

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