Biot-Savart law — find magnetic field due to straight current-carrying wire

Question

Using Biot-Savart law, derive the expression for the magnetic field at a perpendicular distance $d$ from an infinitely long straight conductor carrying current $I$ .

(NCERT Class 12, Chapter 4)

Solution — Step by Step

The magnetic field $d\vec{B}$ due to a small current element $I \, d\vec{l}$ at a point P located at position vector $\vec{r}$ from the element is:

d\vec{B} = \frac{\mu_0}{4\pi} \frac{I \, d\vec{l} \times \hat{r}}{r^2}

This gives the magnitude:

dB = \frac{\mu_0 I \, dl \sin\theta}{4\pi r^2}

where $\theta$ is the angle between $d\vec{l}$ and $\hat{r}$ .

Let the wire lie along the y-axis. Point P is at perpendicular distance $d$ from the wire. Take a small element $dy$ at position $y$ from the foot of the perpendicular.

Then: $r = \sqrt{d^2 + y^2}$ , and $\sin\theta = \frac{d}{r} = \frac{d}{\sqrt{d^2 + y^2}}$ .

All $d\vec{B}$ contributions point in the same direction (by the right-hand rule), so we can add magnitudes directly.

B = \int_{-\infty}^{\infty} \frac{\mu_0 I}{4\pi} \frac{d \, dy}{(d^2 + y^2)^{3/2}}

Using the standard integral $\int_{-\infty}^{\infty} \frac{dy}{(d^2 + y^2)^{3/2}} = \frac{2}{d^2}$ :

B = \frac{\mu_0 I}{4\pi} \cdot d \cdot \frac{2}{d^2} = \frac{\mu_0 I}{2\pi d}

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

The field forms concentric circles around the wire (right-hand rule: curl fingers in the direction of $B$ , thumb along current).

Why This Works

The Biot-Savart law is the magnetic analogue of Coulomb’s law — it gives the field due to a small element of current. By integrating over the whole wire, we get the total field. The $1/d$ dependence (not $1/d^2$ ) arises because we’re summing contributions from an infinite line source.

The direction follows the right-hand rule: point the thumb along the current, and the curled fingers show the field direction. At point P, the field is perpendicular to the plane containing the wire and P.

Alternative Method — Using Ampere’s Circuital Law

For an infinitely long straight wire, take a circular Amperian loop of radius $d$ centred on the wire:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I

By symmetry, $B$ is constant along the loop and parallel to $d\vec{l}$ :

B \cdot 2\pi d = \mu_0 I

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

Ampere’s law is much faster for symmetric configurations (infinite wire, solenoid, toroid). Use Biot-Savart when symmetry is absent (finite wire, wire at a point not on the symmetry axis). JEE often asks for the field due to a finite wire segment — that requires Biot-Savart, not Ampere’s law.

Common Mistake

For a finite wire of length $L$ , the field is NOT $\mu_0 I/(2\pi d)$ . That formula applies only to an infinite wire. For a finite wire, the result involves the angles subtended at P by the two ends: $B = \frac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2)$ . Students who blindly apply the infinite wire formula to finite wire problems lose full marks.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Ampere’s Circuital Law

Common Mistake

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