Proofs for Maxwell's Equations and Foundational Laws
Gauss's Law is defined by the flux of the electric field $\mathbf{E}$ through a closed surface $\mathcal{S}$ being proportional to the enclosed charge $Q_{\text{enc}}$.
The integral form is:
We start with the integral form (1.1). The enclosed charge $Q_{\text{enc}}$ can be expressed as an integral of the volume charge density $\rho$ over the enclosed volume $\mathcal{V}$:
$$\oint_{\mathcal{S}} \mathbf{E} \cdot d\mathbf{A} = \frac{1}{\epsilon_0} \int_{\mathcal{V}} \rho \, d\tau$$Now, we apply the Divergence Theorem (or Gauss's Theorem) to the surface integral on the left side, which converts a surface integral into a volume integral over the volume $\mathcal{V}$ enclosed by $\mathcal{S}$:
$$\oint_{\mathcal{S}} \mathbf{E} \cdot d\mathbf{A} = \int_{\mathcal{V}} (\nabla \cdot \mathbf{E}) \, d\tau$$Equating the two expressions for the volume integral:
$$\int_{\mathcal{V}} (\nabla \cdot \mathbf{E}) \, d\tau = \int_{\mathcal{V}} \frac{\rho}{\epsilon_0} \, d\tau$$Since this equality must hold for *any* arbitrary volume $\mathcal{V}$, the integrands must be equal:
$$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0} \tag{1.2}$$This is the first of Maxwell's equations.
The absence of magnetic monopoles means that magnetic field lines ($\mathbf{B}$) always form closed loops, implying that the net magnetic flux out of any closed surface $\mathcal{S}$ is zero.
The integral form is:
We apply the Divergence Theorem directly to the integral form (2.1), converting the surface integral of $\mathbf{B}$ into a volume integral of its divergence over the enclosed volume $\mathcal{V}$:
$$\oint_{\mathcal{S}} \mathbf{B} \cdot d\mathbf{A} = \int_{\mathcal{V}} (\nabla \cdot \mathbf{B}) \, d\tau = 0$$Since the integral of $(\nabla \cdot \mathbf{B})$ must be zero for *any* arbitrary volume $\mathcal{V}$, the integrand itself must be zero everywhere:
$$\nabla \cdot \mathbf{B} = 0 \tag{2.2}$$This is the second of Maxwell's equations, stating that the magnetic field is always solenoidal (divergence-free).
As a consequence, $\mathbf{B}$ can be written as the curl of a vector potential $\mathbf{A}$:
$$\mathbf{B} = \nabla \times \mathbf{A} \tag{2.3}$$The Lorentz Force Law defines the total electromagnetic force $\mathbf{F}$ on a point charge $q$ moving with velocity $\mathbf{v}$ through electromagnetic fields $\mathbf{E}$ and $\mathbf{B}$.
The total force is:
We consider a segment of current-carrying wire. The charge carriers (e.g., electrons) have charge $q$, average drift velocity $\mathbf{v}_d$, and density $n$ (charges per unit volume). The differential volume element $d\tau$ in the wire is $A \, d\mathbf{l}$, where $A$ is the cross-sectional area and $d\mathbf{l}$ points in the direction of the current.
The total charge in the volume element $d\tau$ is:
$$dQ = n q d\tau = n q A \, dl$$The magnetic force on this differential charge element $dQ$ is (from $\mathbf{F}_{\text{mag}} = q (\mathbf{v} \times \mathbf{B})$):
$$d\mathbf{F} = dQ (\mathbf{v}_d \times \mathbf{B}) = (n q A \, dl) (\mathbf{v}_d \times \mathbf{B})$$The current $I$ is related to the drift velocity by $I = n q A v_d$. By convention, we align the vector $d\mathbf{l}$ with the direction of the current, making $\mathbf{v}_d dl = v_d d\mathbf{l}$. Therefore:
$$d\mathbf{F} = (n q A v_d) (d\mathbf{l} \times \mathbf{B}) = I (d\mathbf{l} \times \mathbf{B}) \tag{3.3}$$This expression is used in the Biot-Savart Law derivation and for calculating forces on wires.
The Biot-Savart law calculates the magnetic field $\mathbf{B}$ generated by a steady electric current $I$. It is the magnetic analogue of Coulomb's Law in electrostatics.
The magnetic field $d\mathbf{B}$ due to a current element $I d\mathbf{l}$ is:
Although the derivation of the Biot-Savart law from the fundamental principles of special relativity is complex, we can confirm its consistency using the vector potential $\mathbf{A}$ for steady currents:
$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int_{\mathcal{V}} \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d\tau'$$For a line current $I$, the volume integral reduces to a line integral, substituting $\mathbf{J} d\tau'$ with $I d\mathbf{l}'$:
$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0 I}{4\pi} \oint_{\mathcal{C}} \frac{d\mathbf{l}'}{|\mathbf{r}-\mathbf{r}'|}$$We know $\mathbf{B} = \nabla \times \mathbf{A}$. Applying the curl operation to the line integral form of $\mathbf{A}$ yields the Biot-Savart Law (4.2) after using standard vector identities (like $\nabla (1/r) = -\mathbf{\hat{r}}/r^2$):
$$\mathbf{B} = \nabla \times \mathbf{A} = \nabla \times \left( \frac{\mu_0 I}{4\pi} \oint_{\mathcal{C}} \frac{d\mathbf{l}'}{|\mathbf{r}-\mathbf{r}'|} \right) = \frac{\mu_0 I}{4\pi} \oint_{\mathcal{C}} \nabla \left(\frac{1}{|\mathbf{r}-\mathbf{r}'|}\right) \times d\mathbf{l}'$$Using the identity $\nabla (1/|\mathbf{r}-\mathbf{r}'|) = -\frac{\mathbf{\hat{r}}}{|\mathbf{r}-\mathbf{r}'|^2}$:
$$\mathbf{B} = \frac{\mu_0 I}{4\pi} \oint_{\mathcal{C}} \frac{d\mathbf{l}' \times \mathbf{\hat{r}}}{|\mathbf{r}-\mathbf{r}'|^2} \tag{4.2}$$Ampère's Law relates the circulation of the magnetic field $\mathbf{B}$ around a closed Amperian loop $\mathcal{C}$ to the total steady current $I_{\text{enc}}$ passing through the surface area defined by $\mathcal{C}$.
The integral form is:
We start with the integral form (5.1). The enclosed current $I_{\text{enc}}$ can be written as the flux of the current density $\mathbf{J}$ through the surface $\mathcal{S}$ bounded by $\mathcal{C}$:
$$\oint_{\mathcal{C}} \mathbf{B} \cdot d\mathbf{l} = \mu_0 \int_{\mathcal{S}} \mathbf{J} \cdot d\mathbf{A}$$Now, we apply Stokes' Theorem to the line integral on the left side, converting the line integral of $\mathbf{B}$ into a surface integral of its curl over the surface $\mathcal{S}$:
$$\oint_{\mathcal{C}} \mathbf{B} \cdot d\mathbf{l} = \int_{\mathcal{S}} (\nabla \times \mathbf{B}) \cdot d\mathbf{A}$$Equating the two expressions for the surface integral:
$$\int_{\mathcal{S}} (\nabla \times \mathbf{B}) \cdot d\mathbf{A} = \int_{\mathcal{S}} \mu_0 \mathbf{J} \cdot d\mathbf{A}$$Since this equality must hold for *any* arbitrary surface $\mathcal{S}$, the integrands must be equal:
$$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} \tag{5.2}$$This form is only valid for magnetostatics ($\nabla \cdot \mathbf{J}=0$).
Faraday's Law describes the phenomenon of electromagnetic induction: a changing magnetic flux ($\Phi_B$) through a surface induces a non-conservative electric field $\mathbf{E}$ that drives a current.
The integral form (Motional and Transformer EMF) is:
We start with the integral form (6.1). Assuming the contour $\mathcal{C}$ is stationary ($\partial \mathcal{S} / \partial t = 0$), the time derivative can be brought inside the integral as a partial derivative:
$$\oint_{\mathcal{C}} \mathbf{E} \cdot d\mathbf{l} = - \int_{\mathcal{S}} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$$Now, we apply Stokes' Theorem to the line integral on the left side, converting it into a surface integral of the curl of $\mathbf{E}$ over the surface $\mathcal{S}$ bounded by $\mathcal{C}$:
$$\int_{\mathcal{S}} (\nabla \times \mathbf{E}) \cdot d\mathbf{A} = - \int_{\mathcal{S}} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$$Since this equality must hold for *any* arbitrary surface $\mathcal{S}$, the integrands must be equal:
$$\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t} \tag{6.2}$$This is the third of Maxwell's equations. It demonstrates that the electric field is no longer conservative ($\nabla \times \mathbf{E} \neq 0$) when the magnetic field is changing over time.
The Ampère-Maxwell Law corrects the magnetostatic Ampère's Law (5.2) by adding the displacement current term, $\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$, ensuring the conservation of charge ($\nabla \cdot \mathbf{J} = -\partial \rho / \partial t$).
In differential form, assuming linear, isotropic, homogeneous media (i.e., vacuum):
The existence of electromagnetic waves is derived by combining the two curl equations (7.3 and 7.4) in free space ($\rho=0, \mathbf{J}=0$).
Step 1: Start with Faraday's Law (7.3) and take the curl of both sides.
$$\nabla \times (\nabla \times \mathbf{E}) = \nabla \times \left(-\frac{\partial \mathbf{B}}{\partial t}\right)$$Step 2: Use the Vector Calculus Identity.
$$\nabla \times (\nabla \times \mathbf{V}) = \nabla (\nabla \cdot \mathbf{V}) - \nabla^2 \mathbf{V}$$Applying this to $\mathbf{E}$ and swapping the spatial ($\nabla \times$) and temporal ($\partial/\partial t$) derivatives on the right side:
$$\nabla (\nabla \cdot \mathbf{E}) - \nabla^2 \mathbf{E} = -\frac{\partial}{\partial t} (\nabla \times \mathbf{B})$$Step 3: Simplify using Maxwell I (7.1) and Substitute Maxwell IV (7.4).
In free space ($\rho=0$), Gauss's Law (7.1) becomes $\nabla \cdot \mathbf{E} = 0$. The first term on the left vanishes. Also, in free space ($\mathbf{J}=0$), Ampère-Maxwell Law (7.4) simplifies to $\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$.
$$(0) - \nabla^2 \mathbf{E} = -\frac{\partial}{\partial t} \left(\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\right)$$Step 4: Rearrange to the Standard Wave Equation Form.
$$-\nabla^2 \mathbf{E} = -\mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2}$$ $$\nabla^2 \mathbf{E} - \mu_0 \epsilon_0 \frac{\partial^2 \mathbf{E}}{\partial t^2} = 0 \tag{7.5}$$This is the standard 3D wave equation: $\nabla^2 \mathbf{E} - \frac{1}{v^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} = 0$. By comparison, the speed $v$ of the electromagnetic wave must be:
$$v = c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$$A similar derivation by starting with the curl of the Ampère-Maxwell Law yields the identical wave equation for the magnetic field $\mathbf{B}$.
Solutions are provided using a formula-first approach, showing the key mathematical steps.
Problem 1: Proton in Magnetic Field
Step 1: Calculate velocity $v$ from potential $V$ (Energy conservation).
$$\frac{1}{2} m v^2 = q V \implies v = \sqrt{\frac{2 q V}{m}}$$Step 2: Use Lorentz Force for circular motion ($F_{\text{mag}} = F_{\text{cent}}$).
$$q v B = \frac{m v^2}{R} \implies R = \frac{m v}{q B}$$Step 3: Substitute $v$ into the radius formula.
$$R = \frac{m}{q B} \sqrt{\frac{2 q V}{m}} = \frac{1}{B} \sqrt{\frac{2 m V}{q}}$$Result: $R \approx 0.057 \text{ m}$.
Problem 2: Force on a Rectangular Loop
Step 1: Magnetic field $B(r)$ from the long wire.
$$B(r) = \frac{\mu_0 I_1}{2\pi r}$$Step 2: Force on the inner ($F_{\text{in}}$) and outer ($F_{\text{out}}$) sides.
$$F_{\text{in}} = I_2 L B(r_1) = I_2 L \frac{\mu_0 I_1}{2\pi d}$$ $$F_{\text{out}} = I_2 L B(r_2) = I_2 L \frac{\mu_0 I_1}{2\pi (d+W)}$$Step 3: Calculate Net Force ($F_{\text{net}} = F_{\text{in}} - F_{\text{out}}$).
$$F_{\text{net}} = \frac{\mu_0 I_1 I_2 L}{2\pi} \left( \frac{1}{d} - \frac{1}{d+W} \right) = \frac{\mu_0 I_1 I_2 L W}{2\pi d (d+W)}$$Result: $F_{\text{net}} \approx 1.0 \times 10^{-4} \text{ N}$ (Attractive, towards $I_1$).
Problem 3: Undeflected Electron Motion
Step 1: Set Total Force to Zero (Undeflected motion).
$$\mathbf{F} = q (\mathbf{E} + \mathbf{v} \times \mathbf{B}) = 0 \implies \mathbf{E} = -(\mathbf{v} \times \mathbf{B})$$Step 2: Substitute vectors and calculate $\mathbf{v} \times \mathbf{B}$.
$$\mathbf{v} \times \mathbf{B} = (v \mathbf{\hat{i}}) \times (B_z \mathbf{\hat{k}}) = v B_z (\mathbf{\hat{i}} \times \mathbf{\hat{k}}) = -v B_z \mathbf{\hat{j}}$$Step 3: Equate $\mathbf{E} = -(\mathbf{v} \times \mathbf{B})$ and solve for $B_z$.
$$E_x \mathbf{\hat{i}} + E_y \mathbf{\hat{j}} = v B_z \mathbf{\hat{j}}$$ $$B_z = \frac{E_y}{v}$$Result: $B_z = 5.0 \times 10^{-5} \text{ T}$. $\mathbf{B} = 5.0 \times 10^{-5} \mathbf{\hat{k}} \text{ T}$.
Problem 4: External Force on a Moving Loop (Motional EMF)
Step 1: Calculate induced EMF ($\mathcal{E}$) and current ($I$).
$$\mathcal{E} = B L v$$ $$I = \frac{\mathcal{E}}{R} = \frac{B L v}{R}$$Step 2: Calculate magnetic braking force ($F_{\text{mag}}$).
$$F_{\text{mag}} = I L B = \left(\frac{B L v}{R}\right) L B = \frac{B^2 L^2 v}{R}$$Step 3: External force ($F_{\text{ext}}$) must balance $F_{\text{mag}}$.
$$F_{\text{ext}} = \frac{B^2 L^2 v}{R}$$Problem 5: Magnetic Field of a Solenoid
Step 1: Apply Ampère's Law ($\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}$).
$$B l_{\text{loop}} = \mu_0 (n l_{\text{loop}}) I$$Step 2: Solve for $B$, where $n=N/l$ is turns per unit length.
$$B = \mu_0 n I = \mu_0 \frac{N}{l} I$$Result: $B \approx 3.02 \times 10^{-3} \text{ T}$.
Problem 6: Rotating Uniformly Charged Sphere
Step 1: Current Density $\mathbf{J}$.
$$\mathbf{J} = \rho \mathbf{v} = \rho (\mathbf{\omega} \times \mathbf{r}) = \rho \omega r \sin\theta \, \mathbf{\hat{\phi}}$$Step 2: Magnetic Moment $\mathbf{m}$ (for $r>R$).
$$\mathbf{m} = \frac{1}{2} \int \mathbf{r} \times \mathbf{J} d\tau = \frac{4\pi \rho \omega R^5}{15} \mathbf{\hat{z}}$$Step 3: Vector Potential $\mathbf{A}(r>R)$.
$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \frac{\mathbf{m} \times \mathbf{\hat{r}}}{r^2} = \frac{\mu_0 \rho \omega R^5}{15} \frac{\sin\theta}{r^2} \mathbf{\hat{\phi}}$$Step 4: Magnetic Field $\mathbf{B}(r
Problem 7: Proof of $\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$ from Biot-Savart
Step 1: Start from the Vector Potential $\mathbf{A}$.
$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d\tau'$$Step 2: Calculate $\nabla \times \mathbf{B} = \nabla \times (\nabla \times \mathbf{A})$.
$$\nabla \times \mathbf{B} = -\nabla^2 \mathbf{A} = -\nabla^2 \left( \frac{\mu_0}{4\pi} \int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d\tau' \right)$$Step 3: Use the Laplacian Identity.
$$\nabla \times \mathbf{B} = \frac{\mu_0}{4\pi} \int \mathbf{J}(\mathbf{r}') \left( -\nabla^2 \frac{1}{|\mathbf{r}-\mathbf{r}'|} \right) d\tau'$$Step 4: Substitute $-\nabla^2 (1/|\mathbf{r}-\mathbf{r}'|) = 4\pi \delta^3(\mathbf{r}-\mathbf{r}')$.
$$\nabla \times \mathbf{B} = \frac{\mu_0}{4\pi} \int \mathbf{J}(\mathbf{r}') \left( 4\pi \delta^3(\mathbf{r}-\mathbf{r}') \right) d\tau'$$Step 5: Apply the Sifting Property.
$$\nabla \times \mathbf{B} = \mu_0 \mathbf{J}(\mathbf{r})$$Problem 8: Magnetic Field in a Coaxial Cable
Use Ampère's Law: $B(r) 2\pi r = \mu_0 I_{\text{enc}}$
Case 1: $r < a$ (Inside inner conductor)
$$I_{\text{enc}} = I \frac{r^2}{a^2} \implies B(r) = \frac{\mu_0 I r}{2\pi a^2}$$Case 2: $a < r < b$ (Between conductors)
$$I_{\text{enc}} = I \implies B(r) = \frac{\mu_0 I}{2\pi r}$$Case 3: $b < r < c$ (Inside outer conductor)
$$I_{\text{enc}} = I - I \frac{r^2 - b^2}{c^2 - b^2}$$ $$B(r) = \frac{\mu_0 I}{2\pi r} \frac{c^2 - r^2}{c^2 - b^2}$$Case 4: $r > c$ (Outside cable)
$$I_{\text{enc}} = I + (-I) = 0 \implies B(r) = 0$$Problem 9: Magnetic Field of a Rotating Charged Ring
Step 1: Calculate differential current $dI$ and differential field $dB$.
$$dI = \frac{dQ}{T} = \frac{\lambda_0 R \omega}{2\pi} \cos(\phi) d\phi$$ $$dB = \frac{\mu_0 dI}{2 R} = \frac{\mu_0 \lambda_0 \omega}{4\pi} \cos(\phi) d\phi$$Step 2: Integrate the $x$-component $dB_x = dB \cos\phi$.
$$B_x = \int_0^{2\pi} \left( \frac{\mu_0 \lambda_0 \omega}{4\pi} \cos(\phi) \right) \cos(\phi) d\phi = \frac{\mu_0 \lambda_0 \omega}{4\pi} \int_0^{2\pi} \cos^2(\phi) d\phi$$Step 3: Evaluate integral $\int_0^{2\pi} \cos^2(\phi) d\phi = \pi$.
$$B_x = \frac{\mu_0 \lambda_0 \omega}{4\pi} (\pi)$$Result: $\mathbf{B}$ at the center is $B_x = \frac{\mu_0 \lambda_0 \omega}{4}$.
Problem 10: Circuit with Motional EMF and Inductance
Step 1: Apply Kirchhoff's Loop Rule ($\mathcal{E}_{\text{mot}} - IR - L_{\text{ind}} \frac{dI}{dt} = 0$).
$$\mathcal{E}_{\text{mot}} = B L v$$Step 2: Differential Equation for $I(t)$.
$$\frac{dI}{dt} + \frac{R}{L_{\text{ind}}} I = \frac{B L v}{L_{\text{ind}}}$$Step 3: Solve the ODE with initial condition $I(0)=0$.
The solution is $I(t) = I_{\text{steady}} (1 - e^{-t/\tau})$, where $I_{\text{steady}} = BLv/R$ and $\tau = L_{\text{ind}}/R$.Result: The current as a function of time is:
$$I(t) = \frac{B L v}{R} \left( 1 - e^{-(R/L_{\text{ind}})t} \right)$$