physics2u
PhysicsMathsChemistryBiology
Tier
⌕ Search ⌘K
Derivation

Faraday's Law and the Induced Field

Statement

Starting from the experimental flux rule, that the electromotive force around a closed loop equals minus the rate of change of magnetic flux through it, we show that for a stationary loop this is equivalent to the local statement \( \nabla \times \vec{E} = -\dfrac{\partial \vec{B}}{\partial t} \). A time-varying magnetic field therefore sources an electric field whose curl is nonzero, so that field is non-conservative and cannot be written as the gradient of a scalar potential alone.

Why it matters

This is the differential form of Faraday's law, one of the four Maxwell equations. It is the precise sense in which electricity and magnetism are a single coupled field: change one and you generate the other. Without it there are no transformers, generators, inductors, or electromagnetic waves.

Conceptually it forces a revision of electrostatics. There we proved \( \nabla \times \vec{E} = 0 \), so \( \vec{E} = -\nabla V \). Once \( \partial \vec{B}/\partial t \neq 0 \) that curl no longer vanishes, the line integral of \( \vec{E} \) around a closed path is generally nonzero, and the notion of a unique scalar potential energy per charge breaks down.

Assumptions
The loop and the surface spanning it are held fixed in the lab frame.If the circuit moves or deforms, the total flux change also has a motional part \( \oint (\vec{v}\times\vec{B})\cdot d\vec{\ell} \). That piece comes from the Lorentz force on charges carried with the wire, not from \( \partial \vec{B}/\partial t \), and mixing the two obscures which physics the differential law actually captures.
The fields are single-valued and differentiable functions of position and time.Stokes' theorem and interchange of the time derivative with the surface integral both require continuity of the integrand and its derivatives. At a sharp material boundary one must instead use the integral form and matching conditions.
The flux rule holds for an arbitrary surface bounded by the loop.If flux depended on which capping surface we chose, \( \int (\nabla\times\vec{E})\cdot d\vec{A} \) would be ambiguous. Consistency requires \( \nabla\cdot(\partial\vec{B}/\partial t)=0 \), which is guaranteed here because \( \vec{B}=\nabla\times\vec{A} \) is automatically divergence-free.
Derivation
1
\[ \varepsilon = -\frac{d\Phi_B}{dt}, \qquad \Phi_B = \int_S \vec{B}\cdot d\vec{A} \]
Experimental flux rule for a fixed loop bounding surface \(S\); \(\varepsilon\) is the EMF and \(\Phi_B\) the magnetic flux. A
2
\[ \varepsilon = \oint_{\partial S} \vec{E}\cdot d\vec{\ell} \]
EMF is the work per unit charge done by the electric field around the loop; for a stationary loop only \(\vec{E}\) contributes (no motional \(\vec{v}\times\vec{B}\) term). A
3
\[ \oint_{\partial S} \vec{E}\cdot d\vec{\ell} = -\frac{d}{dt}\int_S \vec{B}\cdot d\vec{A} \]
Equate steps 1 and 2. This is the integral form we now localise. A
4
\[ \oint_{\partial S} \vec{E}\cdot d\vec{\ell} = \int_S (\nabla\times\vec{E})\cdot d\vec{A} \]
Stokes' theorem, legal because \(\vec{E}\) is differentiable on \(S\) (Assumption 2). B
5
\[ \frac{d}{dt}\int_S \vec{B}\cdot d\vec{A} = \int_S \frac{\partial \vec{B}}{\partial t}\cdot d\vec{A} \]
The surface \(S\) is time-independent (Assumption 1), so \(d/dt\) passes through the integral and acts only on the explicit time dependence of \(\vec{B}\), becoming a partial derivative. B
6
\[ \int_S (\nabla\times\vec{E})\cdot d\vec{A} = -\int_S \frac{\partial \vec{B}}{\partial t}\cdot d\vec{A} \]
Substitute steps 4 and 5 into step 3. Both sides are now surface integrals over the same \(S\). B
7
\[ \int_S \left(\nabla\times\vec{E} + \frac{\partial \vec{B}}{\partial t}\right)\cdot d\vec{A} = 0 \quad \forall\, S \]
Collect terms. The identity holds for every surface bounded by every loop; a continuous integrand whose flux vanishes through every surface must itself vanish pointwise. C
8
\[ \nabla\times\vec{E} = -\frac{\partial \vec{B}}{\partial t} \]
Local form. Consistency of "arbitrary \(S\)" needs \(\nabla\cdot(\partial\vec{B}/\partial t)=0\), automatic since \(\vec{B}=\nabla\times\vec{A}\) (assumed prior result). C
Result
\[ \nabla\times\vec{E} = -\frac{\partial \vec{B}}{\partial t} \]

Reading. The circulation density of the electric field at a point equals minus the rate at which the magnetic field there is changing. A magnetic field growing out of the page ( \(\partial\vec{B}/\partial t\) out of the page) drives \(\vec{E}\) to circulate clockwise, so the induced electric field opposes the change that created it, this is Lenz's law written locally. The field is genuinely non-conservative: \(\oint\vec{E}\cdot d\vec{\ell}\neq 0\).

Units check. \([\nabla\times\vec{E}] = \dfrac{\text{V/m}}{\text{m}} = \text{V}\,\text{m}^{-2}\). \([\partial\vec{B}/\partial t] = \dfrac{\text{T}}{\text{s}}\). Since \(1\,\text{T} = 1\,\dfrac{\text{V}\cdot\text{s}}{\text{m}^2}\), we get \(\dfrac{\text{T}}{\text{s}} = \dfrac{\text{V}\cdot\text{s}}{\text{m}^2\cdot\text{s}} = \text{V}\,\text{m}^{-2}\). Both sides match.

Limiting cases
  • Static fields \(\partial\vec{B}/\partial t\to 0\): recovers electrostatics \(\nabla\times\vec{E}=0\), so \(\vec{E}=-\nabla V\) (the assumed prior result).
  • Uniform \(\vec{B}(t)\) in a symmetric region: the induced \(\vec{E}\) is azimuthal, \(E_\phi = -\tfrac{r}{2}\,\dot B\) inside radius \(r\), independent of any charges.
  • Slowly varying (quasi-static) limit: the induced \(\vec{E}\) is small; treat \(\vec{B}(t)\) as adiabatic and neglect the displacement-current back-reaction.
  • High frequency: \(\partial\vec{B}/\partial t\) couples to the Ampère–Maxwell law, and the pair support propagating electromagnetic waves at speed \(c\).
Breaks when
  • Moving or deforming circuits. The full flux rule includes a motional EMF \(\oint(\vec{v}\times\vec{B})\cdot d\vec{\ell}\). The differential law \(\nabla\times\vec{E}=-\partial\vec{B}/\partial t\) captures only the transformer part; the flux rule can even fail entirely in pathological sliding-contact geometries where the "circuit" is not a material loop.
  • Discontinuous or singular fields. At a surface current sheet or material interface \(\vec{E}\) and \(\vec{B}\) are not differentiable, Stokes' theorem does not apply pointwise, and one must use the integral law with boundary matching conditions.
  • Non-inertial or relativistic frames without care. The split of the total field into "\(\vec{E}\) from \(\partial\vec{B}/\partial t\)" versus "motional" is frame-dependent; only the covariant field tensor \(F_{\mu\nu}\) statement is frame-invariant.
Failure modes
  • Sign amputation. Dropping the minus sign, which encodes Lenz's law and energy conservation; the induced field must oppose the flux change, not reinforce it.
  • Total-vs-partial confusion. Writing \(\nabla\times\vec{E}=-d\vec{B}/dt\). The local law uses a partial time derivative at fixed position; the total derivative belongs to the flux integral over a possibly moving surface.
  • Assuming \(\vec{E}=-\nabla V\) still holds. Once \(\nabla\times\vec{E}\neq 0\) the field needs both potentials: \(\vec{E}=-\nabla V - \partial\vec{A}/\partial t\).
  • Loop-orientation errors. Forgetting that \(d\vec{A}\) and the circulation sense of \(d\vec{\ell}\) are tied by the right-hand rule, which flips the sign of the computed EMF.
  • "Induced \(\vec{E}\) needs charges." The circulating field exists in vacuum where no charge is present; its source is \(\partial\vec{B}/\partial t\), not \(\rho\).
Discussion

The deepest lesson is that the electric field is no longer purely the gradient of a potential. Combining this law with \(\vec{B}=\nabla\times\vec{A}\) gives \(\nabla\times\vec{E}=-\partial(\nabla\times\vec{A})/\partial t=-\nabla\times(\partial\vec{A}/\partial t)\), so \(\nabla\times(\vec{E}+\partial\vec{A}/\partial t)=0\). The bracket is curl-free and may be written as \(-\nabla V\), giving \(\vec{E}=-\nabla V-\partial\vec{A}/\partial t\). Electrostatics is the special case \(\partial\vec{A}/\partial t=0\). This is the natural continuation of the two assumed prior results.

Physically the induced field is non-conservative: a charge carried once around a loop threaded by changing flux gains net energy \(q\varepsilon\), drawn from whatever agent changes \(\vec{B}\). This is exactly how a transformer transfers power between isolated windings and how a betatron accelerates electrons using no electrodes at all, only a rising magnetic field.

The minus sign is not a convention but a statement of energy conservation. If the induced current reinforced the flux change, the process would be self-amplifying and extract unlimited energy from nothing. Lenz's law, encoded in that sign, guarantees the induced effects oppose their cause, so driving the flux change always costs work.

At the covariant level this law and its magnetic partner \(\nabla\cdot\vec{B}=0\) together form the homogeneous Maxwell equations \(\partial_{[\mu}F_{\nu\lambda]}=0\), the Bianchi identity of the field tensor \(F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu\). They are therefore identities following from the existence of the four-potential \(A_\mu=(V/c,\vec{A})\), not independent dynamical constraints, whereas Gauss's and Ampère–Maxwell's laws are the true equations of motion sourced by charge and current.

Common misconceptions. "Faraday's law is about wires." No, the field law holds in empty space; the wire merely reveals the EMF as a current. "The induced \(\vec{E}\) points along \(\partial\vec{B}/\partial t\)." No, it circulates around it, perpendicular in the symmetric case. "A steady current's field induces an EMF." Only its rate of change does.

Worked examples
1
Solenoid with rising current. A long solenoid, \(n=1000\ \text{turns/m}\), carries \(I(t)\) rising at \(dI/dt=50\ \text{A/s}\). Find the induced electric field at radius \(r=2.0\ \text{cm}\) from the axis, inside the solenoid.
Symmetry: \(\vec{E}\) is azimuthal and depends only on \(r\); use the integral form on a circle of radius \(r\). A
2
\[ \oint \vec{E}\cdot d\vec{\ell} = E_\phi\,(2\pi r) = -\frac{d\Phi_B}{dt} = -\pi r^2 \frac{dB}{dt} \]
Uniform axial \(B=\mu_0 n I\) inside; flux through the circle is \(\pi r^2 B\). B
3
\[ E_\phi = -\frac{r}{2}\frac{dB}{dt} = -\frac{r}{2}\,\mu_0 n \frac{dI}{dt} \]
Solve for \(E_\phi\); substitute \(dB/dt=\mu_0 n\,dI/dt\). Symbols before numbers. B
4
\[ E_\phi = -\frac{(0.020)}{2}(4\pi\times10^{-7})(1000)(50) \approx -6.3\times10^{-4}\ \text{V/m} \]
Insert numbers in SI. A
\[ |E_\phi| \approx 6.3\times10^{-4}\ \text{V/m} \]

Reading. The field circulates so as to oppose the rising flux; magnitude \(0.63\ \text{mV/m}\) at \(2\ \text{cm}\). Units check. \(\text{m}\cdot(\text{T}\cdot\text{m}^{-1}\cdot\text{A}^{-1}\cdot\text{m})\cdot\text{A}\,\text{s}^{-1}\Rightarrow\text{T}\,\text{m}\,\text{s}^{-1}=\text{V/m}\).

1
EMF in a search coil. A flat coil of \(N=200\) turns, area \(A=25\ \text{cm}^2\), sits in a uniform field along its axis that decays as \(B(t)=B_0 e^{-t/\tau}\), with \(B_0=0.40\ \text{T}\), \(\tau=0.10\ \text{s}\). Find the EMF at \(t=0\).
Coil stationary, so only the transformer term contributes. A
2
\[ \varepsilon = -N\frac{d\Phi_B}{dt} = -N A\frac{dB}{dt} \]
\(N\) turns multiply the flux; \(A\) constant. B
3
\[ \frac{dB}{dt} = -\frac{B_0}{\tau}e^{-t/\tau} \;\Rightarrow\; \varepsilon = \frac{N A B_0}{\tau}e^{-t/\tau} \]
Differentiate the exponential; the two minus signs cancel. B
4
\[ \varepsilon(0) = \frac{(200)(25\times10^{-4})(0.40)}{0.10} = 2.0\ \text{V} \]
Evaluate at \(t=0\), \(e^0=1\). A
\[ \varepsilon(0) = 2.0\ \text{V} \]

Reading. A decaying field induces an EMF driving current to sustain the flux (Lenz). It too decays with time constant \(\tau\). Units check. \(\text{m}^2\cdot\text{T}\cdot\text{s}^{-1}=\text{Wb}\,\text{s}^{-1}=\text{V}\).

Problems
  1. A magnetic field points out of the page and increases at \(0.20\ \text{T/s}\) within a circular region of radius \(R=5.0\ \text{cm}\). Find the induced electric field magnitude at \(r=3.0\ \text{cm}\).
    Solution Inside the region, \(E_\phi=\tfrac{r}{2}\tfrac{dB}{dt}=\tfrac{0.030}{2}(0.20)=3.0\times10^{-3}\ \text{V/m}=3.0\ \text{mV/m}\), directed clockwise (opposing the increasing outward flux).
  2. For the same region, find \(E_\phi\) at \(r=8.0\ \text{cm}\) (outside \(R\)).
    Solution Outside, all flux \(\pi R^2\tfrac{dB}{dt}\) is enclosed: \(E_\phi(2\pi r)=\pi R^2\tfrac{dB}{dt}\Rightarrow E_\phi=\tfrac{R^2}{2r}\tfrac{dB}{dt}=\tfrac{(0.05)^2}{2(0.08)}(0.20)=3.1\times10^{-3}\ \text{V/m}\). It falls off as \(1/r\).
  3. A square loop of side \(0.10\ \text{m}\) and resistance \(2.0\ \Omega\) lies in a field normal to its plane, \(B(t)=0.5t\ \text{T}\) (t in s). Find the induced current.
    Solution \(\Phi=A B=0.01\cdot0.5t=5\times10^{-3}t\ \text{Wb}\); \(\varepsilon=-d\Phi/dt=-5\times10^{-3}\ \text{V}\); \(|I|=|\varepsilon|/R=2.5\times10^{-3}\ \text{A}=2.5\ \text{mA}\).
  4. Show that \(\nabla\times\vec{E}=-\partial\vec{B}/\partial t\) is consistent with \(\nabla\cdot\vec{B}=0\).
    Solution Take the divergence of both sides: \(\nabla\cdot(\nabla\times\vec{E})=0\) identically, so \(0=-\partial(\nabla\cdot\vec{B})/\partial t\), i.e. \(\nabla\cdot\vec{B}\) is time-independent. If it is zero at any instant it stays zero, consistent with the absence of magnetic monopoles.
  5. A field \(\vec{B}=B_0\cos(\omega t)\,\hat{z}\) fills all space. Find the induced \(\vec{E}\) consistent with \(\nabla\times\vec{E}=-\partial\vec{B}/\partial t\), assuming azimuthal symmetry about the \(z\)-axis.
    Solution \(\partial\vec{B}/\partial t=-B_0\omega\sin(\omega t)\hat{z}\). By symmetry \(\vec{E}=E_\phi(r,t)\hat{\phi}\); the integral law on a circle gives \(E_\phi\,2\pi r=-\pi r^2\,\partial B/\partial t\), so \(E_\phi=-\tfrac{r}{2}\partial B/\partial t=\tfrac{r}{2}B_0\omega\sin(\omega t)\). (This ideal grows with \(r\); at large \(r\) the neglected displacement current matters and the exact solution involves Bessel functions.)