# What is the relationship between magnetic field strength and current density?

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What is the relationship between magnetic field strength and current density?
1. ∇.H = J
2. ∇.J = H
3. ∇ × H = J
4. ∇ × J = H

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Correct Answer - Option 3 : ∇ × H = J

Maxwell Equations:

1) Modified Kirchhoff’s Current Law:

$\nabla .\vec J + \frac{{\partial \rho }}{{\partial t}} = 0$

J = Conduction Current density

2) Modified Ampere’s Law:

$\nabla \times \vec H = \vec J + \frac{{\partial \vec D}}{{\partial t}}$

Where $\frac{{\partial \vec D}}{{\partial t}}$ = Displacement current density

$\nabla .\vec E = - \frac{{\partial \vec B}}{{\partial t}}$

4) Gauss Law:

$\nabla .\vec D = \rho$

Maxwell's Equations for time-varying fields is as shown:

 Differential form Integral form Name $\nabla \times E = - \frac{{\partial B}}{{\partial t}}$ $\mathop \oint \nolimits_L^{} E.dl = - \frac{\partial }{{\partial t}}\mathop \smallint \nolimits_S^{} B.d S$ Faraday’s law of electromagnetic induction $\nabla \times H =J+ \frac{{\partial D}}{{\partial t}}$ $\mathop \oint \nolimits_L^{} H.dl = \mathop \smallint \nolimits_S^{} (J+\frac{{\partial D}}{{\partial t}}).dS$ Ampere’s circuital law ∇ . D = ρv $\mathop \oint \nolimits_S^{} D.dS = \mathop \smallint \nolimits_v^{} \rho_v.dV$ Gauss’ law ∇ . B = 0 $\mathop \oint \nolimits_S^{} B.dS = 0$ Gauss’ law of Magnetostatics (non-existence of magnetic monopole)