Magnetic field

In physics, a magnetic field (commonly denoted by H) describes a magnetic field (a vector) at every point in space; it is a vector field. In non-relativistic physics, the space in question is the three-dimensional Euclidean space $\scriptstyle \mathbb {E} ^{3}$—the infinite world that we live in.

In general H is seen as an auxiliary field useful when a magnetizable medium is present. The magnetic flux density B is usually seen as the fundamental magnetic field, see the article about B for more details about magnetism.

The SI unit of magnetic field strength is ampere⋅turn/meter; a unit that is based on the magnetic field of a solenoid. In the Gaussian system of units |H| has the unit oersted, with one oersted being equivalent to (1000/4π)⋅A⋅turn/m.

Relation between H and B

The magnetic field H is closely related to the magnetic induction B (also a vector field). It is the vector B that enters the expression for magnetic force on moving charges (Lorentz force). Historically, the theory of magnetism developed from Coulomb's law, where H played a pivotal role and B was an auxiliary field, which explains its historic name "magnetic induction". At present the roles have swapped and some authors give B the name magnetic field (and do not give a name to H other than "auxiliary field").

In the general case, H is introduced in terms of B as:

${\begin{aligned}\mathbf {H} &={\frac {1}{\mu _{0}}}\mathbf {B} -\mathbf {M} \qquad &{\hbox{in SI units}}\\\mathbf {H} &=\mathbf {B} -4\pi \mathbf {M} \qquad \;\;&{\hbox{in Gaussian units}},\\\end{aligned}}$

with M(r, t) the magnetization of the medium.

For the most common case of linear materials, M is linear in H, and in SI units,

$\mathbf {B} =\mu _{0}(\mathbf {1} +{\boldsymbol {\chi }})\mathbf {H} ,$

where 1 is the 3×3 unit matrix, χ the magnetic susceptibility tensor of the magnetizable medium, and μ₀ the magnetic permeability of the vacuum (also known as magnetic constant). In Gaussian units the relation is

$\mathbf {B} =(\mathbf {1} +4\pi {\boldsymbol {\chi }})\mathbf {H} ,$

Many non-ferromagnetic materials are linear and isotropic; in the isotropic case the susceptibility tensor is equal to χ_m1, and H can easily be solved (in SI units)

$\mathbf {H} ={\frac {\mathbf {B} }{\mu _{0}(1+\chi _{m})}}\equiv {\frac {\mathbf {B} }{\mu _{0}\mu _{r}}},$

with the relative magnetic permeability μ_r = 1 + χ_m.

For example, at standard temperature and pressure (STP) air, a mixture of paramagnetic oxygen and diamagnetic nitrogen, is paramagnetic (i.e., has positive χ_m), the χ_m of air is 4⋅10⁻⁷. Argon at STP is diamagnetic with χ_m = −1⋅10⁻⁸. For most ferromagnetic materials χ_m depends on H, with a non-linear relation between H and B and is large (depending on the material) from, say, 50 to 10000 and strongly varying as a function of H.

The magnetic flux density B is a solenoidal (divergence-free, transverse) vector field because of one of Maxwell's equations

${\boldsymbol {\nabla }}\cdot \mathbf {B} =0\ ,$

This equation denies the existence of magnetic monopoles (magnetic charges) and hence also of magnetic currents.

The magnetic field H is not necessarily solenoidal, however, because it satisfies:

${\boldsymbol {\nabla }}\cdot \mathbf {H} =-{\boldsymbol {\nabla }}\cdot \mathbf {M} \ ,$

which need not be zero, although it will be zero in some common cases, for example, when B = μH.

Example

B-field lines near uniformly magnetized sphere; because B is solenoidal, field lines are closed.

To illustrate the roles of B and H a classic example is a sphere with uniform magnetization in an applied external uniform external magnetic flux density B. The figure shows the lines of the B-field when no external field is present. To this, a uniform external field is added along the horizontal direction in this example.

To begin, the fields introduced by the magnetized sphere alone are found. The magnetic field H satisfies:

$\mathrm {curl} \ \mathbf {H} =0\ ,$

which allows the introduction of a potential Φ:

$\mathbf {H} =-\ \mathrm {grad} \ \Phi \ .$

Taking the divergence, and using the zero divergence of B:

$\nabla ^{2}\Phi =\mathrm {div} \mathbf {M} \ ,$

which is Poisson's equation. For constant M, div M = 0. However, the solutions inside and outside the sphere must satisfy the boundary conditions on the sphere's surface that the normal component of B is continuous:

$\mathbf {{\hat {u}}_{12}\cdot } \left(\mathbf {B_{1}-B_{2}} \right)=\mathbf {{\hat {u}}_{12}\cdot } \mu _{0}\left(\mathbf {H_{1}+M_{1}-H_{2}-M_{2}} \right)=0\ ,$

while the normal component of H may be discontinuous if there exist surface "magnetic" charges, σ₁ and σ₂, that is:

$\mathbf {{\hat {u}}_{12}\cdot } \left(\mathbf {H_{1}-H_{2}} \right)=\sigma _{1}+\sigma _{2}\ ,$

where unit vector û₁₂ points into region 2 from region 1. Taking region 1 inside the sphere where M₁ = M and region 2 outside where M₂=0 and σ₂ = 0,

$\mathbf {{\hat {u}}_{12}\cdot } \mu _{0}\left(\mathbf {H_{1}+M-H_{2}} \right)=0\ ,$

or:

$\mathbf {{\hat {u}}_{12}\cdot } \left(\mathbf {H_{1}-H_{2}} \right)=-\mathbf {{\hat {u}}_{12}\cdot M} \ =\sigma _{1}.$

The solution to Poisson's equation with this surface charge is (see Poisson's equation):

${\begin{aligned}\Phi (\mathbf {r} )&=\iint _{S}{\frac {\sigma _{1}(\mathbf {r'} )}{4\pi |\mathbf {r} -\mathbf {r'} |}}d{S'}\\&=\int _{-\pi }^{\pi }\sin \theta 'd\theta '\int _{0}^{\pi }d\phi '{\frac {M\cos \theta '}{4\pi |\mathbf {r} -\mathbf {r'} |}}\\&={\frac {M}{3}}R^{2}\cos \theta {\frac {r_{<}}{r_{>}^{2}}}\ ,\end{aligned}}$

with R the radius of the sphere and r_< referring to the minimum of r and R, r_> to the maximum of r and R. The magnetic field is given inside the sphere by the gradient as:

$\mathbf {H} =-{\frac {M}{3}}\mathbf {{\hat {u}}_{z}} \ ,\ \mathbf {B} =\mu _{0}{\frac {2M}{3}}\mathbf {{\hat {u}}_{z}}$

with û_z a unit vector in the horizontal direction of magnetization. Evidently, the magnetic field H points oppositely to the magnetic flux B inside the sphere. Outside the sphere the field is that of a magnetic dipole:

$\mathbf {H} ={\frac {MR^{3}}{3r^{3}}}\left({\hat {\mathbf {r} }}\ 2\cos \theta +{\hat {\boldsymbol {\theta }}}\sin \theta \right)\ ,\ \mathbf {B} =\mu _{0}\mathbf {H} \ .$

In particular, along the horizontal z-axis, θ = 0, and the magnetic field outside the sphere along its axis is:

$\mathbf {H} ={\frac {2M}{3}}\mathbf {{\hat {u}}_{z}} \ ,\ \mathbf {B} =\mu _{0}{\frac {2M}{3}}\mathbf {{\hat {u}}_{z}} \ ,$

showing the continuity of the normal component of B.

If an external magnetic field is imposed, H_a = H_aû_z it can be added to the zero field solution above. Then, inside the sphere the magnetic field H is H_a − M/3. The geometric factor 1/3, which depends upon the shape of the object being a sphere, is called the "demagnetizing factor".