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Non conservative electric field and scalar potential V

  1. Nov 11, 2015 #1
    Hello forum.

    The electric field generated by a changing magnetic field is not conservative. A conservative field is a field with the following features:
    • the closed line integral is zero
    • the line integral from point A to point B is the same no matter the path followed to go from A to B
    • it is the (negative) gradient of a scalar potential field V
    The steady state E field inside DC current carrying wires should be an example of a non conservative field. Any time-varying E field should also be non conservative. That said, the line integral of a non conservative field is still set equal to a potential difference between two spatial points, as it is done in electric circuits...Why? If the field is not conservative, then we should not be able to use the concept of scalar potential...

    thanks,
    fog37
     
  2. jcsd
  3. Nov 11, 2015 #2
    Great observation! It's important to remember that Kirchoff's laws are only approximations that break down in certain circumstances. Let's derive Kirchoff's voltage law from Maxwell's equations.

    I presume you know the integral form of Faraday's law: $$\oint _C \mathbf{E} \cdot d\mathbf{\ell} = -\frac{\partial}{\partial t} \iint _S \mathbf{B} \cdot d\mathbf{S}$$
    Now, we can write the left side of that equation, if we lump all the circuit parameters together, as the sum of voltage drops around the loop (instead of continuous voltage drop, it's a sum of discrete voltage drops): $$\oint _C \mathbf{E} \cdot d\mathbf{\ell} = \sum V$$
    In other words, we can say $$\sum V = -\frac{\partial}{\partial t} \iint _S \mathbf{B} \cdot d\mathbf{S}$$
    and $$-\frac{\partial}{\partial t} \iint _S \mathbf{B} \cdot d\mathbf{S} = -\frac{\partial \Phi}{\partial t} = \frac{\partial}{\partial t} L_s i = -L_s \frac{\partial i}{\partial t}$$

    This last step, if you'll notice, is because by definition, the inductance ##L## is the ratio of the magnetic flux to the given current ##\Phi/I##. So we have
    $$\sum V = -L_s \frac{\partial i}{\partial t}$$
    Where ##L_s## is the stray inductance of the circuit (this is geometry-dependent). For small circuits and low frequencies, the right side of this equation is very small, so we can just say ##\sum V = \oint _C \mathbf{E} \cdot d\mathbf{\ell} = 0##. This is Kirchoff's voltage law, as you know, and this condition implies the field is conservative (note that it really isn't--it's just close).

    Reference: Balanis, Advanced Engineering Electromagnetics
     
    Last edited: Nov 11, 2015
  4. Nov 12, 2015 #3
    Thanks axmls.

    But in the DC case (steady state current), the magnetic field is time-invariant and the rate of change of the current is perfectly zero.

    So it seems that the E field inside the wires is conservative but the E field inside the battery is nonconservative and gives rise to a nonzero line integral.

    The total line integral is the sum of the line integral within the battery and the line integral along the close metal path formed by the wires and the result is nonzero....
     
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