As per the Maxwell's equations and their applications {Let all symbols are used with their usual meaning} A. The tangential component of electric flux density, at a conducting surface, is a non zero quantity. B. Electric field intensity can be found as \(E = \nabla V\). C. For a time varying field the value of \(\oint \vec{E}\cdot \vec{dl}\) will be non-zero D. The value of current flowing in the wire will be equal to \(\oint \vec{H}\cdot \vec{dl}\) E. The magneto static field is conservative in nature. Choose the correct answer from the options given below :
C and D only
Test each statement against the relevant Maxwell equation or boundary condition.
A — "The tangential component of electric flux density at a conducting surface is non-zero." FALSE. The boundary condition at a perfect conductor is
\(E_{tan}=0, \qquad D_{normal}=\rho_s\)
Any tangential field would drive a surface current and be shorted out instantly by the free charges, so the field at a conductor is always perpendicular to the surface — which is also why a conductor surface is an equipotential.
B — "\(E=\nabla V\)." FALSE — the minus sign is missing. The correct relation is
\(\vec{E}=-\nabla V\)
The negative sign expresses that the field points from high to low potential, i.e. down the potential gradient. (And this potential form is valid only in electrostatics; with time-varying fields one needs \(\vec{E}=-\nabla V-\partial\vec{A}/\partial t\).)
C — "For a time-varying field \(\oint\vec{E}\cdot d\vec{l}\) is non-zero." TRUE. Faraday's law gives
\(\oint\vec{E}\cdot d\vec{l}=-\dfrac{d\Phi_B}{dt}\neq 0\)
so a time-varying electric field is non-conservative — it can drive current round a closed loop, which is how transformers and generators work.
D — "The current in the wire equals \(\oint\vec{H}\cdot d\vec{l}\)." TRUE. This is Ampère's circuital law:
\(\oint\vec{H}\cdot d\vec{l}=I_{enc}\)
(with Maxwell's displacement-current term added when fields vary rapidly). It is the law used to derive \(H=I/2\pi r\) around a long wire.
E — "The magnetostatic field is conservative." FALSE. Statement D itself proves it is not: a field whose closed line integral equals the enclosed current cannot be conservative, since a conservative field must give zero round every closed path. Equivalently \(\nabla\times\vec{H}=\vec{J}\neq 0\), so no single-valued scalar magnetic potential exists where current flows. (The electrostatic field is the conservative one.)
Hence, the correct answer is C and D only.
Similar Questions
For a steady magnetic fields, which of the following is true :
1. The tangential component of magnetic field is continuous across any boundary except the surface of perfect conductor.
2. The tangential component of magnetic flux density is continuous across any boundary.
3. The normal component of magnetic flux density is continuous across any boundary.
4. The normal component of electric field is continuous across the boundary.Which one of the following is correct ?
Which of the following are Maxwell equations ?
1. \(B=\mu H\)
2. \(E=\dfrac{D}{\epsilon}\)
3. \(E=\dfrac{J}{\sigma}\)4. \(E=\epsilon D\)
Select the correct answer :
An electromagnetic wave is propagating in free space. Identify the correct situation out of the following
The unit of \(\left(\sigma E + \frac{\partial D}{\partial t}\right)\) is
Which one of the fundamental equation was modified by Maxwell to form the basis of electro magnetic theory?
A medium has the value of displacement flux density
\(\overline{D}=20xy^{2}(z+1)\hat{a}_{x}+20x^{2}y(z+1)\hat{a}_{y}+10x^{2}y^{2}\hat{a}_{z}\ \text{Coulomb/m}^{2}\)
The volume charge density at a point P(0.3, 0.4, 0.5) is given by :
Consider the following statements regarding Maxwell's equations in differential form (Symbols have their usual meanings) :
(a) For free space \(\nabla\times\overline{H}=(\sigma+j\omega\epsilon)\overline{E}\)
(b) For free space \(\nabla\cdot\overline{D}=\rho\)
(c) For steady current \(\nabla\times\overline{H}=\overline{J}\)
(d) For static electric field \(\nabla\cdot\overline{D}=\rho\)Of these statements :
Match the following :
List - I List - II (a) \(\nabla\cdot\overline{D}\) (i) 0 (b) \(\nabla\cdot\overline{B}\) (ii) \(\overline{J}+\dfrac{\partial\overline{D}}{\partial t}\) (c) \(\nabla\times\overline{H}\) (iii) \(-\dfrac{\partial\overline{B}}{\partial t}\) (d) \(\nabla\times\overline{E}\) (iv) \(\rho_{v}\) Codes :
Match List – I with List – II and select the correct answer using codes given below :
List – I List – II a. \(\nabla\times\overline{H}\) i. \(\rho_{v}\) b. \(\nabla\cdot\overline{D}\) ii. \(-\dfrac{\partial B}{\partial t}\) c. \(\nabla\times\overline{E}\) iii. \(\overline{J}+\dfrac{\partial\overline{D}}{\partial t}\) d. \(\nabla\cdot\left(\nabla\times\overline{B}\right)\) iv. 0 Codes :
If Es is the field intensity vector identified as a phasor by its subscript ‘S’ and ko is the wave number, equation \(\nabla^{2}\mathbf{E}_S=-k^{2}\mathbf{E}_S\) is known as :
Important Questions from Maxwell's Equations
∇ × H = J is differential form of
"Time-varying magnetic field will always produce an electric field".
The given statement is true for:
Which of the following equations is based on Ampere's circuit law?
If flux density is represented by 'B' and magnetic field is represented by 'H' in a magnetic circuit, then what will be the energy density in the magnetic field?
Which of the following laws do not form a Maxwell’s equation?
