12.3 Magnetic Fields of Current-Carrying Wires and the Biot-Savart Law

1. Example 1Calculator allowed

   Deriving the field at the center of an arc

   A wire bent into a circular arc of radius R spans an angle θ (in radians) and carries current I. Use the Biot–Savart law to find the field at the arc's center.

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   1. 1Step 1: Every piece $d\vec{\ell}$ of the arc is perpendicular to the line from it to the center, so the angle between $d\vec{\ell}$ and $\hat{r}$ is 90° and sin 90° = 1. Every piece is the same distance R away.
   2. 2Step 2: By the right-hand rule, every piece's $d\vec{B}$ points the same way (perpendicular to the plane of the arc), so the sizes simply add.
   3. 3Step 3: $B = \displaystyle\int \frac{\mu_0}{4\pi}\frac{I\,d\ell}{R^2} = \frac{\mu_0 I}{4\pi R^2}\int d\ell = \frac{\mu_0 I}{4\pi R^2}(R\theta) = \frac{\mu_0 I\theta}{4\pi R}$.
   4. 4Step 4: Check: a full loop has θ = 2π, giving $\dfrac{\mu_0 I}{2R}$, the known result.

   Answer: $B = \dfrac{\mu_0 I\theta}{4\pi R}$, perpendicular to the plane of the arc

2. Example 2Calculator allowed

   A semicircle with straight leads (classic trap)

   A wire carries 8.0 A. It runs straight toward a point P, bends around a semicircle of radius 5.0 cm centered on P, then runs straight away from P along the same line, on the other side of P. Find the magnetic field at P.

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   1. 1Step 1: The straight parts point directly toward or away from P, so for each piece $d\vec{\ell}$ is parallel to $\hat{r}$ and $d\vec{\ell} \times \hat{r} = 0$. They add nothing at P.
   2. 2Step 2: Only the semicircle counts, with θ = π: $B = \dfrac{\mu_0 I\pi}{4\pi R} = \dfrac{\mu_0 I}{4R}$.
   3. 3Step 3: $B = \dfrac{(4\pi \times 10^{-7})(8.0)}{4(0.050)} \approx 5.0 \times 10^{-5}\text{ T}$, perpendicular to the plane of the semicircle.
   4. 4Step 4: The trap is adding a long-wire field for each straight lead. That formula is for a point off to the side of a wire, not on the wire's own line.

   Answer: About 5.0 × 10⁻⁵ T, perpendicular to the plane of the semicircle

3. Example 3Calculator allowed

   Force on a wire in a uniform field

   A straight wire 0.40 m long carries 3.0 A in a uniform 0.25 T magnetic field. The wire makes a 30° angle with the field. Find the size of the magnetic force on it.

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   1. 1Step 1: $F = I\ell B\sin\theta = (3.0)(0.40)(0.25)\sin 30^\circ$.
   2. 2Step 2: $F = (0.30)(0.50) = 0.15\text{ N}$, directed perpendicular to both the wire and the field, along $\vec{\ell} \times \vec{B}$.

   Answer: 0.15 N

Common mistakes

• !Curling the left hand, or pointing the thumb along the electron flow. Use your right thumb along the conventional current.
• !Applying the long-wire formula to a short segment or to a point on the wire's own line.
• !Adding field magnitudes from different pieces without checking direction. Add vectors, or use symmetry to show the pieces line up.
• !Thinking parallel currents repel because like charges repel. Parallel currents in the same direction attract.

On the exam

• ✓Free-response questions often ask you to derive a field with the Biot–Savart law. Show the expression for dB, explain why sin θ or r is constant (or how they vary), and write the integral with clear limits.
• ✓Expect superposition: a shape made of arcs and straight pieces, where you add each part's field at one point, paying attention to whether each points into or out of the page.