Question

For electric and magnetic fields, \(\vec{E}\) and \(\vec{B}\), due to a charge density \(\rho(\vec{r},t)\) and a current density \(\vec{J}(\vec{r},t)\), which of the following relations is/are always correct?

• A. \(\nabla \times \vec{E} = 0\)
• B. \(\nabla \cdot \vec{B} = 0\)
• C. \(\nabla \cdot \vec{J} + \dfrac{\partial \rho}{\partial t} = 0\)
• D. \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\)

Hint:

The only equation not always valid is \(\nabla \times \vec{E} = 0\), which holds only in static conditions.

Answer 1

The Correct Option is B, D

Analyzing Maxwell's equations and electromagnetic relations:

(A) $\vec{\nabla} \times \vec{E} = 0$

From Faraday's law: $$\vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$

This equals zero only if $\vec{B}$ is time-independent (static magnetic field). For time-varying fields, $\vec{\nabla} \times \vec{E} \neq 0$.

NOT always correct 

(B) $\vec{\nabla} \cdot \vec{B} = 0$

This is Gauss's law for magnetism, stating that there are no magnetic monopoles. This is one of Maxwell's fundamental equations and is always true.

$$\vec{\nabla} \cdot \vec{B} = 0$$

ALWAYS correct 

(C) $\vec{\nabla} \cdot \vec{J} - \frac{\partial \rho}{\partial t} = 0$

This is the continuity equation (charge conservation): $$\frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot \vec{J} = 0$$

Rearranging: $$\vec{\nabla} \cdot \vec{J} - \frac{\partial \rho}{\partial t} = 0$$

This represents conservation of charge and is always true.

Wait, let me verify the sign. The continuity equation is: $$\frac{\partial \rho}{\partial t} + \vec{\nabla} \cdot \vec{J} = 0$$

So: $\vec{\nabla} \cdot \vec{J} = -\frac{\partial \rho}{\partial t}$

Therefore: $\vec{\nabla} \cdot \vec{J} - \frac{\partial \rho}{\partial t} = -\frac{\partial \rho}{\partial t} - \frac{\partial \rho}{\partial t} = -2\frac{\partial \rho}{\partial t} \neq 0$ in general.

Actually, the correct form is $\vec{\nabla} \cdot \vec{J} + \frac{\partial \rho}{\partial t} = 0$.

NOT always correct as written 

(D) $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$

This is the Lorentz force law, which gives the force on a charged particle moving in electromagnetic fields. This is a fundamental law and is always true.

ALWAYS correct 

Answer: (B) and (D) are always correct