Amit & Deepali Sinha Foundation

Faraday's Law Keeps B Divergence-Free

Irodov 3.357DifficultyUniqueness

Problem Irodov 3.357

Show that the two Maxwell equations ∇×E=−∂B/∂t\nabla\times\vect E = -\partial\vect B/\partial t and ∇⋅B=0\nabla\cdot\vect B = 0 are compatible: the first does not contradict the second.

Given
  • ∇×E=−∂B/∂t\nabla\times\vect E = -\partial\vect B/\partial t
  • ∇⋅B=0\nabla\cdot\vect B = 0
Find
show that the first equation does not contradict the second

Solution

Step 1The divergence of Faraday’s law

Faraday’s law in differential form is

∇×E=−∂B∂t(1) \nabla\times\vect E = -\frac{\partial\vect B}{\partial t} \tag{1}

Taking the divergence of both sides, the left side vanishes identically, and on the right the space and time derivatives may be interchanged:

∇⋅(∇×E)=0=−∇⋅∂B∂t=−∂∂t(∇⋅B)\nabla\cdot(\nabla\times\vect E) = 0 = -\nabla\cdot\frac{\partial\vect B}{\partial t} = -\frac{\partial}{\partial t}\left(\nabla\cdot\vect B\right)

So the first equation demands only this:

∂∂t(∇⋅B)=0\frac{\partial}{\partial t}\left(\nabla\cdot\vect B\right) = 0
(2)
Step 2Why that is compatible

By (2) the divergence of B\vect B at each point keeps whatever value it has at any one moment:

∇⋅B=const in time\nabla\cdot\vect B = \text{const in time}

The second equation fixes that constant at zero. Faraday’s law never forces ∇⋅B\nabla\cdot\vect B to change, so it never contradicts ∇⋅B=0\nabla\cdot\vect B = 0: if the condition holds once, (1) keeps it holding at all later times.

Step 3The same in integral form

Let a contour CC cut a closed surface into two caps S1S_1 and S2S_2, each with its outward normal. Stokes’s theorem turns the flux of ∇×E\nabla\times\vect E through each cap into the circulation of E\vect E round CC, taken in the sense its own normal fixes by the right-hand rule.

A closed surface cut by the contour C into the caps S_1 and S_2, with outward normals n_1 and n_2. The arrow on C is the sense n_1 fixes; n_2 fixes the opposite one.

A closed surface cut by the contour CC into the caps S1S_1 and S2S_2, with outward normals n1\vect n_1 and n2\vect n_2. The arrow on CC is the sense n1\vect n_1 fixes; n2\vect n_2 fixes the opposite one.

The two outward normals give CC opposite senses, so the two circulations cancel. With (1) on each cap:

−∮S1+S2∂B∂t⋅dS=∮CE⋅dl−∮CE⋅dl=0-\oint_{S_1+S_2}\frac{\partial\vect B}{\partial t}\cdot d\vect S = \oint_{C}\vect E\cdot d\vect l - \oint_{C}\vect E\cdot d\vect l = 0

So the total flux of B\vect B out of any closed surface cannot change in time, the integral statement of (2); Gauss’s law for B\vect B sets that flux to zero.

Check

A concrete case: a uniform field B=B(t) k\vect B = B(t)\,\vect k that grows in time, with the eddy field E=−(B˙/2) k×r\vect E = -(\dot B/2)\,\vect k\times\vect r. Its curl is −B˙ k-\dot B\,\vect k, as (1) requires, and ∇⋅B=0\nabla\cdot\vect B = 0 at every instant: a changing B\vect B and a divergence-free B\vect B coexist without strain.

Remark

Faraday’s law does not imply ∇⋅B=0\nabla\cdot\vect B = 0; it implies that ∇⋅B\nabla\cdot\vect B cannot change. The source equation therefore acts as an initial condition that the curl equation then preserves. The pair in 3.356 has the same structure: the divergence of the circulation law for H\vect H, with Gauss’s law, is charge conservation.