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

5.3.2 The electromotive force

The quantity on the left of Eq. (5.3.3) is what a voltmeter connected into the loop measures. Recalling the definition of the magnetic flux, $Φ = \int_{S} B \cdot d S [Wb]$ , we define the electromotive force (emf) of the circuit as

emf = e (t) = - \frac{d Φ}{d t} [V], (5.3.4)

and find, from Eq. (5.3.3), the two equivalent forms

e (t) = \oint_{C} E \cdot d l + \oint_{C} (v \times B) \cdot d l, (5.3.5)

e (t) = - \int_{S} \frac{\partial B}{\partial t} \cdot d S + \oint_{C} (v \times B) \cdot d l . (5.3.6)

The first form says what the emf is: the work done on a unit charge carried once around the circuit, by the electric field and by the magnetic force $q v \times B$ on a charge that moves with the conductor. The second form says where it comes from: from the change of $B$ in time and from the motion of the circuit.

Faraday's law of induction

For a closed circuit $C$ , moving with velocity $v$ in a field $B$ that may vary in time, the induced electromotive force is

e = - \frac{d Φ}{d t} Φ = transformer emf - \int_{S} \frac{\partial B}{\partial t} \cdot d S + motional emf \oint_{C} (v \times B) \cdot d l, = \int_{S} B \cdot d S, (5.3.7)

where $d l$ and $d S$ are related by the right-hand rule. For a coil of $N$ close-wound turns the flux links the circuit $N$ times and $e = - N d Φ/ d t$ .

In plain words

Change the magnetic flux through a loop, by any means, and a voltage appears around it equal to the rate of change of the flux. The minus sign says that the voltage drives a current which opposes the change.