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.
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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 :
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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 :
Important Questions from Maxwell's Equations
∇ × H = J is differential form of
Maxwell's divergence equation for the magnetic field is given by _______.
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?
Maxwell's third equation is derived from _______.
Which law is represented by the given expression?
\(\int B.dl = \mu_oi_c+\mu_0\epsilon_0 \frac{d \Phi_E}{dt}\)
