What is the negative sign in Faraday's Law/EMF with respect to?

  • Thread starter etotheipi
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  • #1
A common definition seems to be that emf is an electrical action produced from a non-electrical source. So, for instance, a voltage might develop across a resistor due to a gradient of electric charge across the resistor, however this isn't an emf since the source is electrostatic in nature.

As an example, the voltage across an inductor might be reported as ##L\frac{di}{dt}## or ##-L\frac{di}{dt}## depending on our choice of sign convention, however the emf is only ever written as ##\mathcal{E} = -L\frac{di}{dt}##. This is because there is a minus sign in Faraday's law. I understand the operational definition of this negative sign (Lenz's law), however don't understand why it needs to be there from a mathematical perspective. Negative with respect to what? In essence, I'm confused as to why there is only ever one correct sign for the emf. Thank you!
 

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  • #2
alan123hk
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I believe the reason for this negative sign is to satisfy Faraday's equation.

## ∇×\vec E = -\frac {φ \vec B} {φt} ##

Note that all the E, curl E and B are vectors, and the direction of the curl is the axis of rotation, as determined by the right-hand rule , therefore, in order to depict the correct direction of the induced circular E field, the direction of curl E must be opposite to the direction of B, so there must be a negative sign.

In other words, if we can redefine the curl operation to follow the left-hand rule instead of the right-hand rule, then it seems that the minus sign should be removed. 🤔
 
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  • #3
Ah okay, so the negative sign in Faraday's Law is due to the conventional choice of a right-handed coordinate system, and then we can just say

$$\nabla \times \vec{E} = -\frac{\partial B}{\partial t}$$ $$\int_{S} \nabla \times \vec{E} \cdot d\vec{A} = -\frac{d}{dt} \int_{S} \vec{B} \cdot d\vec{A}$$ $$\oint \vec{E} \cdot d\vec{s} = -\frac{d}{dt} \int_{S} \vec{B} \cdot d\vec{A}$$ which is just $$\mathcal{E} = - \frac{d\Phi}{dt}$$
 

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