Electromotive Force and Kinetic Electromotive Force
Electromotive Force1
Let us denote the force that moves charges and generates current in a circuit as $\mathbf{f}$. This $\mathbf{f}$ can be divided into two types. One is the force of the circuit’s power source, $\mathbf{f}_{s}$, and the other is the electric force, $\mathbf{E}$, created by the charges accumulated in some part of the circuit. Here, the subscript $s$ stands for source. Therefore,
$$ \mathbf{f}=\mathbf{f}_{s}+\mathbf{E} $$
The force of the power source, $\mathbf{f}_{s}$, usually only affects a part of the circuit, and $\mathbf{E}$ transfers that effect throughout the entire circuit. The $\mathbf{f}$ that moves the charges in the circuit creating current is determined by integrating over the path of the circuit at that moment, and this is called the electromotive force, emf. (Although the electromotive force can be defined as the work done by the power source per unit charge, there is ambiguity. The calculated value of the work done on a unit charge and the electromotive force are the same, but the two principles are not the same and do not have the same meaning. This will be explained in detail in the section on motional electromotive force.)
$$ \mathcal{E} \equiv \oint \mathbf{f} \cdot d\mathbf{l} = \oint \mathbf{f}_{s}\cdot d\mathbf{l} + \oint \mathbf{E} \cdot d\mathbf{l} = \oint \mathbf{f}_{s}\cdot d\mathbf{l} $$
The value obtained by integrating the electric field, $\mathbf{E}$, over a closed path is $0$, so whether integrating over $\mathbf{f}$ or $\mathbf{f}_{s}$, the result is the same. Note that although it is called a force, it is actually not a force, but the integration of the force experienced by a unit charge over distance, and has the same units as volts. For example, in an ideal electromotive force source like a battery without internal resistance, since the net force on the charge is $0$,
$$ 0=\mathrm{f}_{s}+\mathbf{E}\ \implies \ \mathbf{E}=-\mathbf{f}_{s} $$
Thus, the potential is
$$ V=-\int_{a}^b \mathbf{E} \cdot d \mathbf{l} = \int_{a}^b \mathbf{f}_{s} \cdot d \mathbf{l} =\oint \mathbf{f}_{s} \cdot d \mathbf{l} -= \mathcal{E} $$
It is important to note that what this equation signifies is that the magnitude of the electromotive force is the same as that of the potential, not that potential and electromotive force have the same meaning. It should be understood that a battery creates a potential difference of the same magnitude as the electromotive force.
Motional Electromotive Force2
The electromotive force generated in a conductor moving in a magnetic field is called motional electromotive force, motional emf. For instance, in the situation depicted above, the motional electromotive force is
$$ \begin{align*} \mathcal{E} &= \oint \mathbf{f}_{m} \cdot d\mathbf{l} \\ &= \int_{ab} \mathbf{f}_{m} \cdot d\mathbf{l} + \int_{bc} \mathbf{f}_{m} \cdot d\mathbf{l} + \int_{ad} \mathbf{f}_{m} \cdot d\mathbf{l} \\ &= \int_{ab} vBdl \\ &= vBh \end{align*} $$
For the sides $bc$ and $ab$, the magnetic force $\mathbf{f}_{m}$ and the direction of the circuit’s path are perpendicular, so no electromotive force is generated. Among the remaining parts, excluding side $ab$, the magnetic field is $0$, so the integral value is also $0$. Therefore, only the integral term for side $ab$ remains. At this point, it is important to note that the integration path is not the actual path over which the charge moves. Although the charge moves from the bottom to the top of side $ab$ and side $ab$ moves to the right, meaning the charge actually moves diagonally upwards to the right, this is not the path of integration. The integration path is the same as the path of the circuit at a single moment as the loop moves.
As mentioned earlier, the electromotive force is not defined as the work done by a magnetic force on a unit charge because magnetic forces do not perform work. If it were defined this way, the calculated value of the motional electromotive force would always be $0$. Clearly, although magnetic forces generate electromotive force, they do not perform any work. The entity supplying energy (work) is not the magnetic force but the force exerted by a person pulling the loop. When a person pulls the loop and generates electromotive force, current flows through the loop.
Considering side $ab$, the current flows perpendicular, and the circuit itself moves horizontally. Therefore, the direction of the charge’s motion has both horizontal and vertical components. Let’s denote the horizontal speed as $\mathbf{v}$ and the vertical speed as $\mathbf{u}$.
According to the right-hand rule, the direction of the magnetic force, $\mathbf{f}_{m}$, is as shown in the figure above. Since the horizontal component of the magnetic force is $uB$, the magnitude of the force that a person must pull is also $\mathbf{f}_{pull}=uB$ of the same magnitude. The actual direction in which the charge moves is the same as the direction $\mathbf{u}+\mathbf{v}=\mathbf{w}$, and the distance moved is $\dfrac{h}{\cos \theta}$. Therefore, the work done by a person pulling the loop per unit charge is as follows:
$$ \begin{align*} \int \mathbf{f}_{pull}\cdot d\mathbf{l} &= (uB)\dfrac{h}{\cos \theta}\cos\left( \frac{\pi}{2}-\theta \right) \\ &= \dfrac{u}{\cos \theta} Bh \sin \theta \\ &= w\sin\theta Bh \\ &= vBh \\ &= \mathcal{E} \end{align*} $$
In reality, the work done on a unit charge is the same as the electromotive force. However, the relevant force and integration path are entirely different.
Relationship Between Motional Electromotive Force and Magnetic Flux
The electromotive force generated by a moving loop can be expressed in terms of the magnetic flux passing through the loop. Let’s consider the magnetic flux through the loop as $\Phi$.
$$ \Phi = \int \mathbf{B} \cdot d\mathbf{a} $$
Taking the rectangular loop shown at the top of the article as an example, its value is
$$ \Phi = Bhx $$
As shown in the figure, if the loop is pulled to the right, the area of the loop decreases, and the flux also decreases (this is why the sign is $-$).
$$ \dfrac{d \Phi}{dt}=Bh\dfrac{dx}{dt}=-Bhv $$
This is the opposite sign but of the same magnitude as the electromotive force calculated above. Therefore, the electromotive force generated in the loop can be represented by the rate of change of flux.
$$ \mathcal{E}=-\dfrac{d\Phi}{dt} $$
This equation is called the flux rule for motional electromotive force, and while a rectangular loop is used as an example, it generally applies to loops of any shape.