Derive B = μ₀I/(2πr) for an infinite straight conductor.

Biot-Savart Law — Magnetic Field Due to Straight Wire

Question

A long straight conductor carries a steady current $I$ . Using the Biot-Savart Law, derive the expression for the magnetic field $B$ at a perpendicular distance $r$ from the wire.

This is the foundational result for magnetism: $B = \dfrac{\mu_0 I}{2\pi r}$

Solution — Step by Step

Place the wire along the Y-axis. We want the field at point P, which is at perpendicular distance $r$ from the wire (along the X-axis).

Consider a small current element $Id\vec{l}$ at position $y$ on the wire. The vector from this element to point P has magnitude $s = \sqrt{r^2 + y^2}$ .

The Biot-Savart Law gives the field from one small element:

d\vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{I\, dl \sin\theta}{s^2}

Here $\theta$ is the angle between $d\vec{l}$ and the line joining the element to P. By the right-hand rule, all $d\vec{B}$ contributions point in the same direction (out of the page, if current flows upward and P is to the right). That means we can simply add magnitudes.

This is where most derivations get messy — we switch to angle $\phi$ to make the integral clean.

Let $\phi$ be the angle that the line from element to P makes with the perpendicular. Then:

y = r\tan\phi, \quad dy = r\sec^2\phi\, d\phi, \quad s = \frac{r}{\cos\phi}

Also, $\sin\theta = \cos\phi$ (draw this — $\theta$ and $\phi$ are complementary).

Substituting:

dB = \frac{\mu_0 I}{4\pi} \cdot \frac{\cos\phi \cdot r\sec^2\phi\, d\phi}{r^2\sec^2\phi} = \frac{\mu_0 I}{4\pi r}\cos\phi\, d\phi

For an infinite wire, $\phi$ runs from $-\pi/2$ to $+\pi/2$ :

B = \int_{-\pi/2}^{+\pi/2} \frac{\mu_0 I}{4\pi r}\cos\phi\, d\phi = \frac{\mu_0 I}{4\pi r}\Big[\sin\phi\Big]_{-\pi/2}^{+\pi/2}

B = \frac{\mu_0 I}{4\pi r}\left[1 - (-1)\right] = \frac{\mu_0 I}{4\pi r} \times 2

\boxed{B = \frac{\mu_0 I}{2\pi r}}

Why This Works

The key insight is that a current element alone creates no measurable field — only the integrated effect of the entire wire produces the result we can measure. The Biot-Savart approach sums up infinitely many tiny contributions, each obeying the inverse-square-distance law, but the geometry means the net result falls off only as $1/r$ (not $1/r^2$ ). This is exactly analogous to how an infinite line charge gives $E \propto 1/r$ in electrostatics.

The variable substitution to angle $\phi$ is the technical heart of this derivation. Distance $s$ and element length $dy$ both change as we move along the wire, and expressing both in terms of a single angle makes the integral tractable. If you try integrating in terms of $y$ directly, you get $\int_{-\infty}^{\infty} \frac{dy}{(r^2 + y^2)^{3/2}}$ — it works, but it’s significantly harder.

The direction of $\vec{B}$ follows from the right-hand rule: wrap your right hand around the wire with the thumb pointing in the direction of current — your fingers curl in the direction of the field lines (concentric circles around the wire).

Alternative Method

For JEE problems, Ampere’s Law gives the same result in one line — use it whenever symmetry allows.

For an infinite straight wire, the field has cylindrical symmetry. Draw an Amperian loop — a circle of radius $r$ centred on the wire. By symmetry, $|\vec{B}|$ is constant everywhere on this loop and $\vec{B}$ is tangential to it.

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

B \cdot (2\pi r) = \mu_0 I

B = \frac{\mu_0 I}{2\pi r}

Same answer, ten times faster. Biot-Savart is needed when symmetry breaks down (finite wire, curved segments). For infinite straight wires and solenoids, always prefer Ampere’s Law in JEE.

Common Mistake

Applying this formula to a finite wire. The result $B = \mu_0 I / 2\pi r$ holds only for an infinitely long wire. For a finite wire of length $2L$ , the correct answer is $B = \frac{\mu_0 I}{4\pi r} \cdot \frac{2L}{\sqrt{r^2 + L^2}}$ . In JEE Main 2024, a variant asked for the field at the midpoint perpendicular of a finite wire — students who blindly applied the infinite-wire formula lost full marks. The limits of integration change from $\pm\pi/2$ to the actual angles subtended at point P.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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