Question

The Maxwell's equation \( \oint \vec{E} \cdot d\vec{l} \) is equal to

• A. 0
• B. \( -j\omega \int \vec{B} \cdot d\vec{l} \)
• C. \( -j\omega \int \vec{B} \cdot d\vec{s} \)
• D. \( j\omega \int \vec{B} \cdot d\vec{v} \)

Hint:

Maxwell's Equations (Integral Form):
Gauss's Law for E: \( \oint_S \vec{D} \cdot d\vec{s} = \int_V \rho_v dv \)
Gauss's Law for B: \( \oint_S \vec{B} \cdot d\vec{s} = 0 \)
Faraday's Law: \( \oint_C \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \iint_S \vec{B} \cdot d\vec{s} \)
Ampere-Maxwell Law: \( \oint_C \vec{H} \cdot d\vec{l} = \iint_S (\vec{J} + \frac{\partial \vec{D}}{\partial t}) \cdot d\vec{s} \)
For time-harmonic fields (\(e^{j\omega t}\)), \(\frac{\partial}{\partial t} \rightarrow j\omega\).

Answer 1

The Correct Option is C

The integral form of Faraday's Law of Induction (one of Maxwell's equations) is: \[ \oint_C \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \iint_S \vec{B} \cdot d\vec{s} \] where C is a closed contour and S is any surface bounded by C. This states that the electromotive force (EMF) induced in a closed loop C is equal to the negative rate of change of magnetic flux (\(\Phi_B = \iint_S \vec{B} \cdot d\vec{s}\)) through the surface S. If we consider time-harmonic fields (phasor form), where fields vary as \(e^{j\omega t}\), the time derivative \(\frac{d}{dt}\) can be replaced by \(j\omega\). So, \(-\frac{d}{dt}\) becomes \(-j\omega\). Therefore, in the frequency domain (phasor form), Faraday's Law becomes: \[ \oint_C \vec{E} \cdot d\vec{l} = -j\omega \iint_S \vec{B} \cdot d\vec{s} \] Or, using \( \int \vec{B} \cdot d\vec{s} \) for the surface integral of \(\vec{B}\) over S: \[ \oint_C \vec{E} \cdot d\vec{l} = -j\omega \int_S \vec{B} \cdot d\vec{s} \] This matches option (c). Let's look at the other options: (a) 0: This would be true for static electric fields (\(\nabla \times \vec{E} = 0\)), but not for time-varying fields. (b) \( -j\omega \int \vec{B} \cdot d\vec{l} \): This is a line integral of B, not related. (d) \( j\omega \int \vec{B} \cdot d\vec{v} \): This is a volume integral of B, not related. Therefore, option (c) correctly represents Faraday's Law in integral form for time-harmonic fields. \[ \boxed{-j\omega \int_S \vec{B} \cdot d\vec{s}} \]