Q.A straight wire carrying a current of $12\ \text{A}$ is bent into a semi-circular arc of radius $2.0\ \text{cm}$ as shown in Fig. 4.11(a). Consider the magnetic field $B$ at the centre of the arc.

The key idea is that the magnetic field at the centre of a current-carrying arc depends only on the arc's angle and radius, while straight segments along the radial line contribute zero field.

Reasoning:

1. Straight segments: For a straight wire, the magnetic field at a point on its extension is zero because the angle between the current element and the position vector is $0^\circ$ or $180^\circ$, making $\sin\theta = 0$. Here, both straight segments lie along radii to the centre, so their contribution is zero.

2. Semicircular arc: The field at the centre of a full circular loop is $B_{\text{loop}} = \frac{\mu_0 I}{2R}$. A semicircle subtends an angle $\pi$ radians (half of $2\pi$), so its contribution is half that of a full loop: $B_{\text{semi}} = \frac{1}{2} \cdot \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{4R}$.

3. Direction and comparison: The field direction is given by the right-hand rule -- perpendicular to the plane of the arc. For a semicircle, the field resembles that of a full loop in direction and in depending on $I/R$, but differs in magnitude (half) because only half the loop contributes.

4. Reversal of arc: If the arc bulges downward instead of upward (Fig. 4.11b), the current direction along the arc reverses relative to the centre. The magnitude remains $\frac{\mu_0 I}{4R}$, but the field direction reverses (into the page instead of out, or vice versa).

Why the Biot–Savart Law?

The Biot–Savart law tells us that a current element $\mathrm{d}\vec{l}$ produces a magnetic field $\mathrm{d}\vec{B}$ at a point given by

where $\hat{r}$ points from the element to the observation point. The key geometric insight: if the current element lies along the line joining it to the observation point, the cross product $\mathrm{d}\vec{l} \times \hat{r}$ is zero -- that element contributes nothing.

For the centre of a circular arc, every straight segment that points radially toward or away from the centre has $\mathrm{d}\vec{l}$ parallel (or antiparallel) to $\hat{r}$, so its contribution vanishes. Only the curved part, where $\mathrm{d}\vec{l}$ is perpendicular to $\hat{r}$, produces a field.

1. Straight segments: zero contribution

   In Fig. 4.11(a), the wire enters horizontally from the left, bends into a semicircle bulging upward, then continues horizontally to the right. At the centre of the semicircle, both straight segments lie along radii -- the left segment points directly toward the centre, the right segment points directly away from it. For any element on these straight parts, $\mathrm{d}\vec{l}$ is exactly along $\hat{r}$ (or opposite), so $\mathrm{d}\vec{l} \times \hat{r} = 0$. Hence the magnetic field from the straight segments is zero.

   Watch out

   A common mistake is to think the straight segments contribute like infinite wires. But the Biot–Savart law cares about the direction from element to point, not the overall wire shape. Here the centre lies exactly on the line of the straight segments -- the field from those segments is identically zero.

2. Semicircle vs. full circular loop

   For a full circular loop of radius $R$ carrying current $I$, the field at the centre is

directed perpendicular to the plane of the loop (right-hand rule).

For a semicircular arc, every current element $\mathrm{d}\vec{l}$ on the curved part is perpendicular to $\hat{r}$ (since $\hat{r}$ points radially inward to the centre), and the magnitude of $\mathrm{d}\vec{l} \times \hat{r}$ is just $\mathrm{d}l$. Integrating over the semicircle (half the circumference, $\pi R$) gives

This is exactly half the field of a full circular loop. The direction is the same as for the full loop -- perpendicular to the plane of the arc, following the right-hand rule.

Resemblance: Both produce a field perpendicular to the plane of the current, with magnitude proportional to $I/R$.

Difference: The semicircle gives exactly half the magnitude because only half the current elements contribute.

3. Flipping the arc -- Fig. 4.11(b)

   In Fig. 4.11(b), the wire is bent so the semicircle bulges downward instead of upward. The straight segments still lie along radii through the centre, so their contribution remains zero. The curved part still has the same radius and carries the same current -- the integration is identical. The magnitude of $B$ is unchanged: $B = \mu_0 I / (4R)$.

   What changes is the direction. Using the right-hand rule: for the upward bulge (a), current flows left-to-right along the arc, and the field points into the page. For the downward bulge (b), the current direction along the arc is reversed relative to the centre, so the field points out of the page. The magnitude is the same; only the sign flips.

   Tip

   Think of the semicircle as half a loop. Flipping the arc is like flipping the loop over -- the field reverses direction but keeps the same strength.

Method: Zero-Contribution Segments + Fraction-of-a-Loop for Biot-Savart Arcs

This method solves any "magnetic field at the centre of a bent/arc-shaped wire" problem by splitting the wire into pieces that contribute nothing and pieces that contribute a known fraction of the full-loop field.

Steps

Step 1: Check every straight segment for radial alignment

The Biot-Savart law gives $d\vec B \propto d\vec l \times \hat r$. Whenever a straight segment lies along the line joining it to the point where you want $B$ (it points directly toward or away from that point), $d\vec l$ is parallel or antiparallel to $\hat r$, so the cross product -- and hence the field from that segment -- is exactly zero. Scan the wire's geometry first and eliminate every such segment before doing any integration.

Step 2: Identify the curved portion and its subtended angle

For the remaining curved part (a full circle, a semicircle, a quarter-circle, etc.), every element $d\vec l$ is perpendicular to $\hat r$, so the whole element contributes. The field of a full circular loop of radius $R$ at its centre is the reference value:

An arc subtending angle $\phi$ (out of $2\pi$) contributes that same fraction of the full-loop field:

Step 3: Apply the direction via the right-hand rule, then track reversal

Curl the right-hand fingers in the direction of current flow around the arc; the thumb gives the field direction (into or out of the plane). If the arc's curvature is later reversed (bulging the other way) while the current path is unchanged, the magnitude from Steps 1-2 stays the same, but the current now circulates the opposite way around the same centre, so only the direction flips.

Applying to this problem: here $\phi=\pi$ (a semicircle), so $B_{\text{semi}} = \tfrac12\cdot\dfrac{\mu_0 I}{2R} = \dfrac{\mu_0 I}{4R}$, and the straight segments contribute zero by Step 1.