Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

5.3.5 Recap

Recap

Faraday's law, Eq. (5.3.7): $e = - d Φ/ d t = - \int_{S} (\partial B / \partial t) \cdot d S + \oint_{C} (v \times B) \cdot d l$ , transformer emf plus motional emf, with $d l$ and $d S$ by the right-hand rule. The minus sign is Lenz's law: the induced current opposes the change of flux, and this is energy conservation.
Transformer emf of a fixed coil of $N$ turns: $e = - N d Φ/ d t$ . The induced electric field has $\oint_{C} E \cdot d l \neq = 0$ and therefore no potential.
Motional emf of a bar of length $l$ moving at $v$ across $B$ : $e = B l v$ , the field brakes the bar with $F = i l \times B$ , and $P_{mech} = P_{elec}$ .