KAHIBARO
Table of Contents
Moving conductors in magnetic fields
Motional EMF is the voltage produced when a conductor moves through a magnetic field. The key idea is that charges inside the conductor are also moving with the conductor, and magnetic forces act on those charges. This can separate positive and negative charges and create a potential difference between two ends of the conductor.
This effect is one of the simplest ways to generate electricity from motion. It is the basic idea behind electric generators, although full generators also involve changing magnetic flux and rotating loops, which belong more broadly to electromagnetic induction.
Magnetic force on charges in the conductor
Suppose a straight metal rod moves with velocity $\vec{v}$ through a magnetic field $\vec{B}$. A free charge $q$ inside the rod also moves with the rod, so it feels the magnetic force
$$
\vec{F}_B = q \, \vec{v} \times \vec{B}.
$$
This force pushes positive and negative charges in opposite directions along the rod. As charges build up at the ends, an electric field is created inside the rod. That electric field produces an electric force that opposes further charge separation.
Eventually, equilibrium is reached. At that point, the electric force and magnetic force balance:
$$
qE = qvB
$$
when $\vec{v}$ is perpendicular to $\vec{B}$ and the rod is oriented so the force acts along its length. So,
$$
E = vB.
$$
If the rod has length $\ell$, the potential difference between its ends is
$$
\mathcal{E} = E\ell = B \ell v.
$$
This voltage is called the motional EMF.
For a straight conductor of length $\ell$ moving at speed $v$ perpendicular to a magnetic field $B$, the motional EMF is
$$
\mathcal{E} = B \ell v.
$$
This is valid when the motion, field, and rod are mutually perpendicular in the correct way.
Direction of the induced voltage
The direction of charge separation is determined by the cross product $\vec{v} \times \vec{B}$. For positive charges, the magnetic force points in the direction of $\vec{v} \times \vec{B}$. Negative charges move in the opposite direction.
A useful way to determine which end becomes positive is to imagine the force on a positive test charge inside the rod. The end toward which positive charges are pushed becomes the higher potential end.
In the drawing, the rod moves to the right and the magnetic field points into the page. Then $\vec{v} \times \vec{B}$ points upward, so positive charges are pushed upward. The top end becomes positive.
Motional EMF from work per unit charge
Another way to understand motional EMF is by energy. The magnetic force separates charges, and external work must be done to keep the rod moving if current flows. The EMF is the work done per unit charge to move charge through the circuit.
If a rod moves through a field and sweeps out area, the magnetic environment of the circuit changes. This leads to an induced EMF. In the special case of a moving rod on rails, the motional EMF can be found directly from the rod's motion.
Sliding rod on rails
A classic example is a conducting rod sliding on two conducting rails in a uniform magnetic field. The rod, rails, and a connecting wire form a closed circuit. As the rod moves, the area of the loop changes, and an EMF is induced.
If the rod has length $\ell$ and moves with speed $v$ perpendicular to a magnetic field $B$, then
$$
\mathcal{E} = B \ell v.
$$
If the total circuit resistance is $R$, the induced current is
$$
I = \frac{\mathcal{E}}{R} = \frac{B\ell v}{R}.
$$
This setup is important because the moving rod not only develops a voltage, but also drives a current around the circuit.
General expression
The simple formula $\mathcal{E} = B\ell v$ is a special case. More generally, motional EMF in a moving conductor is related to the magnetic force per unit charge:
$$
\mathcal{E} = \int (\vec{v} \times \vec{B}) \cdot d\vec{\ell}.
$$
This says that the EMF depends on how the conductor moves through the field and on the orientation of the conductor.
If the velocity is not perpendicular to the magnetic field, only the perpendicular component contributes. Then the magnitude becomes
$$
\mathcal{E} = B \ell v \sin\theta,
$$
where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.
Only the component of motion perpendicular to the magnetic field produces motional EMF.
$$
\mathcal{E} = B\ell v \sin\theta
$$
If the conductor moves parallel to the field, then $\sin\theta = 0$ and no motional EMF is produced.
Current, force, and mechanical work
When the conductor is part of a closed circuit, the induced EMF drives a current. That current-carrying rod in a magnetic field then experiences a magnetic force. This force opposes the motion in many practical situations.
For a rod carrying current $I$ in a field $B$, the magnetic force magnitude is
$$
F = I\ell B
$$
when the rod is perpendicular to the field.
This opposing force means that an external agent must keep pulling or pushing the rod to maintain constant speed. Mechanical energy is converted into electrical energy.
The electrical power delivered is
$$
P_{\text{elec}} = \mathcal{E} I.
$$
Using $\mathcal{E} = B\ell v$ and $F = I\ell B$, we get
$$
P_{\text{mech}} = Fv = I\ell B \, v = \mathcal{E} I.
$$
So the mechanical power supplied equals the electrical power produced, neglecting losses other than those already included in the circuit.
In motional EMF systems, mechanical work is converted into electrical energy.
For ideal conditions,
$$
Fv = \mathcal{E} I.
$$
Relation to changing flux
Motional EMF is closely connected to magnetic flux change. In a sliding rod circuit, the rod changes the area of the loop, so the magnetic flux changes with time. This gives the same result as the direct force-on-charges picture.
If the loop area is $A = \ell x$, where $x$ is the rod position, then
$$
\Phi_B = BA = B\ell x.
$$
Differentiating with respect to time gives
$$
\left| \frac{d\Phi_B}{dt} \right| = B\ell \frac{dx}{dt} = B\ell v.
$$
So the induced EMF magnitude is
$$
\mathcal{E} = B\ell v.
$$
This shows that motional EMF is one practical way that electromagnetic induction appears.
Common cases
The table below summarizes several simple situations.
| Situation | Motional EMF |
|---|---|
| Rod moves perpendicular to rod and field | $\mathcal{E} = B\ell v$ |
| Rod moves at angle $\theta$ to field | $\mathcal{E} = B\ell v \sin\theta$ |
| Motion parallel to field | $\mathcal{E} = 0$ |
| Open circuit | Voltage appears, but no sustained current |
| Closed circuit | Voltage appears and current can flow |
A simple example
Consider a rod of length $\ell = 0.50 \, \text{m}$ moving at $v = 4.0 \, \text{m/s}$ through a uniform magnetic field $B = 0.30 \, \text{T}$. Assume the rod moves perpendicular to the field.
Then
$$
\mathcal{E} = B\ell v = (0.30)(0.50)(4.0) = 0.60 \, \text{V}.
$$
So the potential difference between the ends of the rod is $0.60 \, \text{V}$.
If the circuit resistance were $2.0 \, \Omega$, then the current would be
$$
I = \frac{0.60}{2.0} = 0.30 \, \text{A}.
$$
Physical meaning
Motional EMF is not produced because the magnetic field directly does work on charges. Magnetic force is always perpendicular to the charge's instantaneous velocity, so magnetic force alone does no work on a single charge. Instead, the magnetic force rearranges charges, creating an electric field inside the conductor. In a complete circuit, an external force maintaining the motion provides the energy.
This is an important conceptual point. The voltage appears because motion through the magnetic field causes charge separation.
A magnetic field can create charge separation in a moving conductor, but the energy delivered to the circuit comes from the mechanical work done to keep the conductor moving.
Summary formula set
For beginners, the most useful formulas for motional EMF are these:
$$
\vec{F}_B = q \, \vec{v} \times \vec{B}
$$
$$
\mathcal{E} = B\ell v
$$
$$
\mathcal{E} = B\ell v \sin\theta
$$
$$
I = \frac{\mathcal{E}}{R}
$$
$$
P = \mathcal{E} I = Fv
$$
These formulas describe how motion in a magnetic field can produce a voltage, a current, and energy conversion from mechanical to electrical form.