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Magnetostatics

Magnetostatics is the regime of electromagnetism in which currents are steady (time-independent). The governing laws are Ampère’s law ∇×B=μ0J\nabla \times \mathbf{B} = \mu_0 \mathbf{J} and Gauss’s law for magnetism ∇⋅B=0\nabla \cdot \mathbf{B} = 0 (no magnetic monopoles).

Typical applications: permanent magnets, inductors, MRI magnets.

When it is relevant. Magnetostatics is the slow, static limit of Electrodynamics, the magnetic sibling of Electrostatics. It applies whenever the currents of interest are steady — or change so slowly that the fields they produce are essentially constant on the timescale of the problem. The criterion is the same as in electrostatics: the system’s size LL is small compared with the distance light travels in the relevant time, L≪c τL \ll c\,\tau, so the time-derivative terms in Maxwell’s equations vanish. Under that condition the magnetic field is produced only by steady currents (and by the steady magnetization of materials), and the whole problem is governed by two vector equations. This is the regime behind the field of a straight wire, the uniform field inside a solenoid, the force between parallel conductors, and the field of a permanent magnet.

This page treats vacuum magnetostatics first: the permeability is the vacuum value μ0\mu_0 throughout. How a material modifies the field — permeability, the auxiliary field H\mathbf{H}, and the constitutive relation B=μH\mathbf{B}=\mu\mathbf{H} — is developed in Section 5 and in Electromagnetic Media.

Authoritative sources

Part of Electromagnetism; the static limit of Electrodynamics.

1. Biot–Savart law

Definition. The magnetic field dBd\mathbf{B} at a point due to a small steady current element I dℓI\,d\boldsymbol{\ell} is

dB=μ04π I dℓ×r^r2d\mathbf{B} = \frac{\mu_0}{4\pi}\,\frac{I\,d\boldsymbol{\ell}\times\hat{\mathbf{r}}}{r^2}

where r^\hat{\mathbf{r}} points from the element to the field point and rr is the distance. The total field of a steady current is the vector integral (superposition) of these contributions. The field is always perpendicular to the current element and to the line joining it to the field point, and it circulates around the current according to the right-hand rule.

Details

The idea, step by step.

  1. A current element is the magnetic analogue of a point charge. Just as a point charge qq produces an electric field E∝q r^/r2\mathbf{E}\propto q\,\hat{\mathbf{r}}/r^2 (Coulomb’s law, Electrostatics), a tiny piece of wire carrying current II produces a magnetic field. The Biot–Savart law is the magnetic “Coulomb’s law”: it gives the field of the smallest source element, and everything else is built by superposition.

  2. The cross product sets the direction. The factor dℓ×r^d\boldsymbol{\ell}\times\hat{\mathbf{r}} means the field is perpendicular to both the current direction and the line to the field point. Curl the fingers of your right hand in the direction of the current; your thumb points along dℓd\boldsymbol{\ell} and your curled fingers show how B\mathbf{B} circles the wire. This is why the field of a straight wire is a set of concentric circles around it.

  3. Integrate to get the field of a real wire. For a long straight wire, every element contributes a field tangent to a circle of radius rr around the wire, and all contributions point the same way. Integrating around the whole wire gives the classic result

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

The field falls as 1/r1/r (not 1/r21/r^2 as in electrostatics) because a wire is an extended source, not a point.

↑ I B ∝ 1/r B circles the wire (right-hand rule); ↑ = current out of the page
  1. Authoritative source. See the Biot–Savart law and Griffiths, Introduction to Electrodynamics, Ch. 5.
Example problem

A long straight wire carries a steady current I=5.0 AI = 5.0\,\text{A}. Find the magnetic field at a distance r=0.10 mr = 0.10\,\text{m} from the wire.

Solution

Use the integrated Biot–Savart result for a long straight wire:

B=μ0I2πr=(4π×10−7)(5.0)2π(0.10)=2×10−7×5.00.10=1.0×10−5 T=10 μTB = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi\times10^{-7})(5.0)}{2\pi(0.10)} = \frac{2\times10^{-7}\times 5.0}{0.10} = 1.0\times10^{-5}\ \text{T} = 10\ \mu\text{T}

The field circles the wire; at the point in question it is tangent to the circle of radius rr (right-hand rule).

2. Ampère’s law

Definition. The line integral of the magnetic field around any closed loop equals μ0\mu_0 times the net current threading the loop:

∮CB⋅dℓ=μ0 Ienc,∇×B=μ0J\oint_C \mathbf{B}\cdot d\boldsymbol{\ell} = \mu_0\, I_{\text{enc}},\qquad \nabla\times\mathbf{B} = \mu_0\mathbf{J}

Ampère’s law is the integral form of the curl of the field — the magnetic counterpart of Gauss’s law in Electrostatics. It is exact for any closed loop, but it is a calculation tool only when the current distribution has enough symmetry.

Details

The idea, step by step.

  1. The circulation counts the enclosed current. The line integral ∮B⋅dℓ\oint \mathbf{B}\cdot d\boldsymbol{\ell} sums the component of the field along a closed path. Ampère’s law says this circulation depends only on how much current passes through the surface bounded by the path — currents outside the loop contribute nothing to the net circulation. It is a “current counter,” exactly as Gauss’s law is a “charge counter.”

  2. Why the field of a straight wire satisfies it. Take a circular loop of radius rr around a long straight wire. By the right-hand rule B\mathbf{B} is everywhere tangent to the loop and has constant magnitude BB, so the integral is B(2πr)B(2\pi r). The enclosed current is II, so B(2πr)=μ0IB(2\pi r)=\mu_0 I, recovering B=μ0I/(2πr)B=\mu_0 I/(2\pi r) from Section 1 — but without doing the Biot–Savart integral.

  3. Symmetry turns it into a calculator. Choose an Ampèrian loop on which B\mathbf{B} is either constant and parallel to dℓd\boldsymbol{\ell} or perpendicular to it (so it contributes nothing):

    • Infinite straight wire (cylindrical symmetry): a coaxial circle gives B=μ0I/(2πr)B = \mu_0 I/(2\pi r).
    • Long solenoid (a coil of nn turns per unit length): a rectangular loop straddling the wall gives a uniform interior field B=μ0nIB = \mu_0 n I and (ideally) zero outside.
    • Toroid (a coil wound on a ring, NN total turns): a circular loop of radius rr inside the core gives B=μ0NI/(2πr)B = \mu_0 N I/(2\pi r), and zero outside.
↑ I Ampèrian loop C ∮ B·dℓ = μ₀ Iₜ₝₝ (B tangent to C, constant magnitude)
  1. Authoritative source. Ampère’s law is one of Maxwell’s equations; see Maxwell (1865) and Griffiths, Introduction to Electrodynamics, Ch. 5.
Example problem

A long solenoid has n=1000n = 1000 turns per metre and carries a current I=2.0 AI = 2.0\,\text{A}. Use Ampère’s law to find the magnetic field inside the solenoid.

Solution

Choose a rectangular Ampèrian loop with one side inside the solenoid (length ℓ\ell), parallel to the axis, and the other side far outside where B≈0\mathbf{B}\approx 0. Inside, B\mathbf{B} is uniform and parallel to the path, so the integral is BℓB\ell. The loop threads nℓn\ell turns, each carrying II, so Ienc=nℓII_{\text{enc}} = n\ell I:

B ℓ=μ0 nℓ I⇒B=μ0nI=(4π×10−7)(1000)(2.0)≈2.5×10−3 T=2.5 mTB\,\ell = \mu_0\, n\ell\, I \quad\Rightarrow\quad B = \mu_0 n I = (4\pi\times10^{-7})(1000)(2.0) \approx 2.5\times10^{-3}\ \text{T} = 2.5\ \text{mT}

The field is uniform and axial inside, and (for an ideal long solenoid) zero outside.

3. Magnetic vector potential

Definition. Because ∇⋅B=0\nabla\cdot\mathbf{B}=0, the magnetic field is the curl of a vector potential A\mathbf{A}:

B=∇×A,A(r)=μ04π∫I dℓ′r′\mathbf{B} = \nabla\times\mathbf{A},\qquad \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int \frac{I\,d\boldsymbol{\ell}'}{r'}

where r′r' is the distance from the source element to the field point. The vector potential is the magnetic analogue of the electrostatic scalar potential VV (with E=−∇V\mathbf{E}=-\nabla V): it is a bookkeeping quantity whose derivative gives the field.

Details

The idea, step by step.

  1. No monopoles force a vector potential. In electrostatics the field is conservative (∇×E=0\nabla\times\mathbf{E}=0), so it is the gradient of a scalar VV. In magnetostatics the field is solenoidal (∇⋅B=0\nabla\cdot\mathbf{B}=0, no magnetic monopoles), so it is the curl of a vector A\mathbf{A}. The two static regimes are exact mirrors: a scalar potential whose gradient gives E\mathbf{E}, and a vector potential whose curl gives B\mathbf{B}.

  2. The potential is a sum of scalars, then differentiate. For a steady current the vector potential at a point is just a scalar sum over the current elements, A=(μ0/4π)∫I dℓ′/r′\mathbf{A} = (\mu_0/4\pi)\int I\,d\boldsymbol{\ell}'/r' — far easier to add than the vector Biot–Savart field. Once you have A\mathbf{A}, the field is its curl. This “sum, then differentiate” strategy parallels the electrostatic one and is the workhorse for more complicated geometries.

  3. Gauge freedom. A\mathbf{A} is not unique: adding the gradient of any scalar, A→A+∇χ\mathbf{A}\to\mathbf{A}+\nabla\chi, leaves B=∇×A\mathbf{B}=\nabla\times\mathbf{A} unchanged. Only B\mathbf{B} (and, in quantum mechanics, the line integral of A\mathbf{A}) is physically observable — the content of the Aharonov–Bohm effect.

scalar potential V vector potential A E = ∉V B = ∉×A gradient curl electrostatics ↔ magnetostatics: ∉V → E, ∉×A → B
  1. Authoritative source. See Griffiths, Introduction to Electrodynamics, Ch. 5 (vector potential), and Maxwell (1865).
Example problem

A long straight wire along the zz-axis carries current II. Show that a vector potential Az=−(μ0I/2π)ln⁡(r/r0)A_z = -(\mu_0 I/2\pi)\ln(r/r_0) reproduces the field of Section 1.

Solution

With A=Az(r) z^\mathbf{A} = A_z(r)\,\hat{\mathbf{z}} and Az=−(μ0I/2π)ln⁡(r/r0)A_z = -(\mu_0 I/2\pi)\ln(r/r_0), the curl in cylindrical coordinates has only a $`\phi`-component:

Bϕ=−∂Az∂r=−∂∂r[−μ0I2πln⁡rr0]=μ0I2π 1rB_\phi = -\frac{\partial A_z}{\partial r} = -\frac{\partial}{\partial r}\left[-\frac{\mu_0 I}{2\pi}\ln\frac{r}{r_0}\right] = \frac{\mu_0 I}{2\pi}\,\frac{1}{r}

This is exactly the field of the long straight wire. The arbitrary constant r0r_0 reflects the gauge freedom: shifting AzA_z by a constant (or by the gradient of a scalar) does not change B\mathbf{B}.

4. Magnetic force on currents

Definition. A charge qq moving with velocity v\mathbf{v} in a magnetic field feels the Lorentz force

F=q v×B\mathbf{F} = q\,\mathbf{v}\times\mathbf{B}

For a steady current II in a wire, the force on a segment is dF=I dℓ×Bd\mathbf{F} = I\,d\boldsymbol{\ell}\times\mathbf{B}. Two long parallel wires carrying currents I1I_1 and I2I_2 a distance dd apart exert a force per unit length

FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

on each other — attractive for currents in the same direction, repulsive for opposite currents.

Details

The idea, step by step.

  1. The magnetic force is always sideways. The cross product v×B\mathbf{v}\times\mathbf{B} means the force is perpendicular to both the velocity and the field, so it changes the direction of a moving charge but never its speed. A magnetic field does no work on a charge; it only bends the trajectory into a circle or helix. This is the origin of the circular field lines of Section 1 — a moving charge’s own field circles its path.

  2. A current in a field feels a force. The charges drifting through a wire each feel qv×Bq\mathbf{v}\times\mathbf{B}; summed over the wire this is dF=I dℓ×Bd\mathbf{F}=I\,d\boldsymbol{\ell}\times\mathbf{B}. This is the principle behind the electric motor (a current-carrying coil in a field experiences a torque) and the Hall effect.

  3. Parallel wires attract or repel. Wire 1 creates a circular field B1=μ0I1/(2πd)B_1 = \mu_0 I_1/(2\pi d) at the location of wire 2. The current in wire 2 then feels F/L=I2B1=μ0I1I2/(2πd)F/L = I_2 B_1 = \mu_0 I_1 I_2/(2\pi d). The right-hand rule shows the force pulls the wires together when the currents flow the same way and pushes them apart when they oppose. This force per unit length is in fact the historical definition of the ampere.

↑↑↑ ↑↑↑ attract same-direction currents attract; F/L = μ₀ I₁ I₂ / (2π d)
  1. Authoritative source. See Griffiths, Introduction to Electrodynamics, Ch. 5 (force on currents), and Maxwell (1865).
Example problem

Two long parallel wires are d=0.05 md = 0.05\,\text{m} apart and each carries a current of 4.0 A4.0\,\text{A} in the same direction. Find the force per unit length between them, and state whether it is attractive or repulsive.

Solution FL=μ0I1I22πd=(4π×10−7)(4.0)(4.0)2π(0.05)=2×10−7×160.05=6.4×10−5 N/m\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{(4\pi\times10^{-7})(4.0)(4.0)}{2\pi(0.05)} = \frac{2\times10^{-7}\times 16}{0.05} = 6.4\times10^{-5}\ \text{N/m}

Because the currents flow in the same direction, the force is attractive.

5. Magnetic materials: the H field and permeability

Definition. In a material the magnetic response is described by the auxiliary field H\mathbf{H}, related to B\mathbf{B} by the constitutive relation

B=μ H=μ0 μr H\mathbf{B} = \mu\,\mathbf{H} = \mu_0\,\mu_r\,\mathbf{H}

where μ=μ0μr\mu = \mu_0\mu_r is the permeability and μr\mu_r the relative permeability of the material. Ampère’s law is then written in terms of the free current: ∇×H=Jfree\nabla\times\mathbf{H} = \mathbf{J}_{\text{free}}. A permanent magnet is a material with a fixed magnetization M\mathbf{M} that produces a field even with no external current.

Details

The idea, step by step.

  1. Why a second field, H? In vacuum B=μ0H\mathbf{B}=\mu_0\mathbf{H} and the two are proportional, so either suffices. In a material, however, the total current that sources B\mathbf{B} has two parts: the free current you drive through a wire, and the bound current from the material’s microscopic magnetic dipoles aligning. The field H\mathbf{H} is defined so that Ampère’s law counts only the free current, ∇×H=Jfree\nabla\times\mathbf{H}=\mathbf{J}_{\text{free}}, leaving the material’s response in the relation B=μH\mathbf{B}=\mu\mathbf{H}. This is the magnetic twin of the D\mathbf{D}/E\mathbf{E} split in Electromagnetic Media.

  2. Permeability measures how easily a material “conducts” magnetic field. A vacuum has μr=1\mu_r=1. A diamagnet has μr\mu_r just below 1 (it weakly opposes the field); a paramagnet just above 1 (it weakly enhances it); a ferromagnet (iron, nickel, cobalt) has a large, field-dependent μr\mu_r that can reach thousands, which is why an iron core multiplies the field of a solenoid by a large factor.

  3. Permanent magnets are steady currents in disguise. A permanent magnet’s field comes from the alignment of atomic magnetic moments (electron spin and orbital motion). Macroscopically this is equivalent to a bound surface current flowing around the magnet, so the field of a bar magnet is exactly the field of a steady current loop — the dipole field of Section 6. Magnetostatics therefore describes a permanent magnet without any time dependence: the “current” is frozen into the material.

N S a permanent magnet ≅ a bound surface current loop (dipole field)
  1. Authoritative source. See Electromagnetic Media for the full constitutive treatment, and Griffiths, Introduction to Electrodynamics, Ch. 5 (magnetic materials).
Example problem

The solenoid of Section 2 (n=1000n=1000 turns/m, I=2.0 AI=2.0\,\text{A}) is filled with a core of relative permeability μr=200\mu_r = 200. Find the magnetic field inside.

Solution

Ampère’s law in terms of H\mathbf{H} depends only on the free current, so H=nIH = nI is unchanged by the core:

H=nI=(1000)(2.0)=2.0×103 A/mH = nI = (1000)(2.0) = 2.0\times10^{3}\ \text{A/m}

The field B\mathbf{B} is then multiplied by the permeability:

B=μ0μrH=(4π×10−7)(200)(2.0×103)≈0.50 TB = \mu_0 \mu_r H = (4\pi\times10^{-7})(200)(2.0\times10^{3}) \approx 0.50\ \text{T}

The iron core boosts the field from 2.5 mT2.5\,\text{mT} (air core) to about 0.5 T0.5\,\text{T} — a factor of μr=200\mu_r = 200.

6. Magnetic dipole moment and dipole field

Definition. A small current loop of current II and area AA has a magnetic dipole moment

m=I A n^\mathbf{m} = I\,A\,\hat{\mathbf{n}}

directed normal to the loop by the right-hand rule. Far from the loop (r≫r \gg loop size), its field is that of a magnetic dipole; on the axis of the dipole,

B=μ04π 2mr3B = \frac{\mu_0}{4\pi}\,\frac{2m}{r^3}
Details

The idea, step by step.

  1. A loop is the simplest magnetic dipole. A single current loop has no net “magnetic charge,” but its field is concentrated on one face and returns on the other — just like the north and south poles of a bar magnet. The strength and orientation of this dipole are captured by the single vector m=IAn^\mathbf{m}=IA\hat{\mathbf{n}}. More loops (a coil of NN turns) give m=NIAn^\mathbf{m}=NIA\hat{\mathbf{n}}.

  2. The far field falls as 1/r³. Close to the loop the field is complicated, but far away only the dipole term survives and it decays as 1/r31/r^3 — faster than the 1/r1/r of a wire or the 1/r21/r^2 of a point charge. On the axis the field is B=(μ0/4π)(2m/r3)B = (\mu_0/4\pi)(2m/r^3), pointing along m\mathbf{m}.

  3. Dipoles are everywhere. A bar magnet’s field is a dipole field (Section 5); the Earth’s magnetic field is, to first order, a dipole field from its core; and the magnetic moment of an electron is the quantum dipole that makes ferromagnetism possible. The dipole is the universal building block of magnetostatic fields.

N S dipole field: closed loops, B ∝ 1/r³ far away
  1. Authoritative source. See Griffiths, Introduction to Electrodynamics, Ch. 5 (magnetic dipole field), and Maxwell (1865).
Example problem

A circular loop of radius R=0.05 mR = 0.05\,\text{m} carries a current I=3.0 AI = 3.0\,\text{A}. Find its magnetic dipole moment and the field on its axis at r=0.20 mr = 0.20\,\text{m}.

Solution

The dipole moment is

m=IA=I πR2=(3.0) π(0.05)2≈2.4×10−2 A\cdotpm2m = I A = I\,\pi R^2 = (3.0)\,\pi(0.05)^2 \approx 2.4\times10^{-2}\ \text{A·m}^2

On the axis, far from the loop (r≫Rr \gg R), the dipole field is

B=μ04π 2mr3=(10−7) 2(2.4×10−2)(0.20)3=(10−7) 4.8×10−28.0×10−3≈6.0×10−7 TB = \frac{\mu_0}{4\pi}\,\frac{2m}{r^3} = (10^{-7})\,\frac{2(2.4\times10^{-2})}{(0.20)^3} = (10^{-7})\,\frac{4.8\times10^{-2}}{8.0\times10^{-3}} \approx 6.0\times10^{-7}\ \text{T}

The field points along the dipole moment (out of the “north” face of the loop).

Interactive: field of a current loop and solenoid

Trace the magnetic field of a single current loop or a solenoid (a stack of loops) using the Biot–Savart law. Use the sliders to change the current and the number of turns, and watch the field become uniform inside the solenoid and fringe out at the ends.

Part of Electromagnetism.

Quizz

Test yourself

Problem. A long solenoid of radius R=0.04 mR = 0.04\,\text{m} has n=2000n = 2000 turns per metre and carries a current I=1.5 AI = 1.5\,\text{A}.

  1. What is the magnetic field at the centre of the solenoid?
  2. What is the magnetic flux through a circular cross-section of the solenoid (radius RR)?
  3. If the current is doubled, by what factor does the field energy stored per unit length change? (The magnetic energy density is u=B2/(2μ0)u = B^2/(2\mu_0).)
Solution
  1. Inside a long solenoid the field is uniform and axial (Section 2):
B=μ0nI=(4π×10−7)(2000)(1.5)≈3.8×10−3 TB = \mu_0 n I = (4\pi\times10^{-7})(2000)(1.5) \approx 3.8\times10^{-3}\ \text{T}
  1. The flux through a cross-section of radius RR is Φ=B A=B πR2\Phi = B\,A = B\,\pi R^2:
Φ=(3.8×10−3) π(0.04)2≈(3.8×10−3)(5.0×10−3)≈1.9×10−5 Wb\Phi = (3.8\times10^{-3})\,\pi(0.04)^2 \approx (3.8\times10^{-3})(5.0\times10^{-3}) \approx 1.9\times10^{-5}\ \text{Wb}
  1. The energy density is u=B2/(2μ0)u = B^2/(2\mu_0), and B∝IB \propto I. Doubling the current doubles BB, so the energy density — and hence the stored energy per unit length — increases by a factor of 22=42^2 = 4.