Problem 4.1.1: Infinitely long thin wire (Biot–Savart)

Problem 4.1.1

Infinitely long thin wire (Biot–Savart)

Suppose that there is a constant current $I$ flowing along an infinitely long thin wire located on the $z$ -axis. Write the density of current and find the explicit expression of the magnetic field at a normal distance $ρ$ from the wire.

Try it yourself first.

Strategy

Biot–Savart's law, summed along the wire. The key observation is that every element of the wire contributes in the same direction $a_{ϕ}$ at P, so the vector integral becomes a scalar one. The integral is the same one that gave the field of the infinite line charge in Problem 3.1.1.

1. Symmetry

The wire is unchanged by translations along $z$ and rotations about $z$ , so the field cannot depend on $z$ or $ϕ$ . We therefore take the field point at P $(ρ, ϕ, z)$ in cylindrical coordinates and expect $B = B (ρ)$ .

2. Current density

The wire is thin and lies on the $z$ -axis, so the current is confined to $x^{'} = y^{'} = 0$ :

J = I δ (x^{'}) δ (y^{'}) a_{z} [A/ m^{2}],

and the volume element of Biot–Savart's law collapses to a line element, $J d v^{'} = I d z^{'} a_{z}$ .

3. Geometry: the points and R

The source point is Q $(0, 0, z^{'})$ . From Q to P (Figure 4.1.18a),

R R a_{z} \times R = ρ a_{ρ} + (z - z^{'}) a_{z}, = ρ^{2} + (z - z^{'})^{2}, = ρ (a_{z} \times a_{ρ}) = ρ a_{ϕ},

because $a_{z} \times a_{z} = 0$ and $a_{z} \times a_{ρ} = a_{ϕ}$ (Section 2.3). Every element of the wire therefore contributes in the same direction $a_{ϕ}$ , whatever $z^{'}$ : Biot–Savart's law leaves one component only.

4. Integrate

B = \frac{μ _{0}}{4 π} \int_{- \infty}^{\infty} \frac{I d z ^{'} a _{z} \times R}{R ^{3}} = \frac{μ _{0} I ρ}{4 π} a_{ϕ} \int_{- \infty}^{\infty} \frac{d z ^{'}}{[ ρ ^{2} + ( z - z ^{'} ) ^{2} ] ^{3/2}} .

With $u = z^{'} - z$ the integral is

\int_{- \infty}^{\infty} \frac{d u}{( ρ ^{2} + u ^{2} ) ^{3/2}} = [\frac{u}{ρ ^{2} ρ ^{2} + u ^{2}}]_{- \infty}^{\infty} = \frac{1}{ρ ^{2}} [1 - (- 1)] = \frac{2}{ρ ^{2}},

so that

B = \frac{μ _{0} I ρ}{4 π} \cdot \frac{2}{ρ ^{2}} a_{ϕ},

Answer

B H = \frac{μ _{0} I}{2 π ρ} a_{ϕ} [T], = \frac{I}{2 π ρ} a_{ϕ} [A/m]

(a) One element and its d\vec B (side view), and the lines of \vec B (top view). — (a) One element and its $d B$ (side view), and the lines of $B$ (top view).

(b) B versus \rho for I=1 A. Problem 4.1.1. (a) Every element I\,dz'\,\vec a_z gives at P a contribution along \vec a_\phi, since \vec a_z\times\vec R=\rho\,\vec a_\phi for every z', so the contributions simply add. Seen down the axis, the lines of \vec B are circles round the wire. (b) The 1/\rho law: 20 μT at 1 cm and 2 μT at 10 cm for 1 A. — **Figure 4.1.18.** (b) $B$ versus $ρ$ for $I = 1$ A. Problem 4.1.1. (a) Every element $I d z^{'} a_{z}$ gives at P a contribution along $a_{ϕ}$ , since $a_{z} \times R = ρ a_{ϕ}$ for every $z^{'}$ , so the contributions simply add. Seen down the axis, the lines of $B$ are circles round the wire. (b) The $1/ ρ$ law: $20$ μT at $1$ cm and $2$ μT at $10$ cm for $1$ A.

Check

Units: $μ_{0} I / ρ$ is (H/m)(A)/m $=$ (V s/A)(A)/m $^{2}$ $=$ V s/m $^{2}$ $=$ T. Direction: the right hand with the thumb along $+ z$ curls the fingers in the $+ a_{ϕ}$ sense (Figure 4.1.18a). A second route: on a circle of radius $ρ$ centred on the wire $H$ is tangential and of constant magnitude, hence Ampère's formula gives $\oint_{C} H \cdot d l = H_{ϕ} 2 π ρ = I$ , the same result in one line, which is Problem 4.1.4. The field falls off as $1/ ρ$ , not $1/ ρ^{2}$ , for the same reason as the field of a line charge (Problem 3.1.1): moving away one sees more of the wire at once. The result does not depend on $z$ , because the wire is infinite and every cross-section is equivalent. Numerically, $I = 1$ A gives $B = 20$ μT at $ρ = 1$ cm and $2$ μT at $10$ cm (Figure 4.1.18b).