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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

3) Faraday’s Law:

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

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)

 

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