The maximum value of induced potential difference in a conductor in motion inside magnetic field due to the Lorentz force on the free charges is called **motional emf**.

When a conductor moves in a magnetic field with a velocity $v$, we can consider that the free electrons in it, are also moving with the same velocity in the same direction with the conductor. Hence, they experience Lorentz force which is given by

\[\overrightarrow{F}=-e(\overrightarrow{v}×\overrightarrow{B})\]

where $e$ be the charge of each electron and $B$ be the strength of the magnetic field. Hence the electrons gain a motion in the direction of this force and induces an emf which is known as motional emf. Thus, the main cause of motional emf is the Lorentz force experienced by the free electrons of the conductor.

## Induced emf in a straight conductor moving in a uniform magnetic field: Theoretical proof of Faraday’s laws of electromagnetic induction

Consider a U-shaped conductor $SMNT$ on which a straight conductor $PQ$ can be slided freely. The straight conductor $PQ$ of length $l$ is placed at right angles to a uniform magnetic field $B$. If the conductor is set into the motion toward the right with a velocity $v$ perpendicular to both to its own length and to the magnetic field, then the charged particles contained in the conductor also move with the velocity $v$ with the conductor.

Hence, the charged particles of charge $q$ within a conductor experience a force $F$ equal to $Bqv$ directed along the length of the conductor. \[F=Bqv\]

According to Fleming left hand rule, the direction of the force on positive charge is from $Q$ to $P$ while the force on the negative charge is from $P$ to $Q$. Thus, the end $P$ becomes positive and $Q$ becomes negative, causing induced emf across the moving rod.

When a charge moves from $Q$ to $P$ through a distance $l$ as the length of the conductor, the work of the force $F$ is given by \[W=Fl=Bqvl\]

Since emf is the work per unit charge so if $\varepsilon$ be the induced emf in the conductor, we have \[\varepsilon=\frac{W}{q}=\frac{Bqvl}{q}\] \[\therefore\varepsilon=Bvl\text{ __(1)}\]

This gives the induced emf developed in the conductor moving in the uniform magnetic field. Since it is due to the motion of the conductor through the magnetic field, it is called **motional emf**.

Now, the magnetic flux associated with circuit $PMNQ$ of area $A_1$ is given by \[\phi=BA_1\]

Let the sliding conductor come in the new position $P’Q’$ such that it covers a small distance $dx$. Then, the magnetic flux associated with the circuit $P’MNQ’$ of area $A_2$ is given by \[\phi_2=BA_2\]

The change in magnetic flux is given by, \[d\phi=\phi_2-\phi_1=B(A_2-A_1)\] \[d\phi=B×\text{Area of PQQ’P’}\] \[d\phi=B(dx.l)\]

Dividing by $dt$, we get \[\frac{d\phi}{dt}=B\frac{dx}{dt}l\] \[\frac{d\phi}{dt}=Bvl\text{ __(2)}\]

where $v=\frac{dx}{dt}$ is the velocity of the conductor.

From equations $\text{(1)}$ and $\text{(2)}$, \[\varepsilon=\frac{d\phi}{dt}\] According to Lenz’s law, induced emf opposes the change in flux so \[\varepsilon=-\frac{d\phi}{dt}\] For $N$ turns of coil, \[\varepsilon=-N\frac{d\phi}{dt}\] It shows that the induced emf is directly proportional to the rate of change of magnetic flux which is the mathematical form of Faraday’s laws of electromagnetic induction. Hence, the laws are proved.

## Induced emf in a coil rotating in a magnetic field: Motional emf

Consider a rectangular coil of $N$ turns, each coil having area $A$ and whose normal makes an angle $\theta$ with magnetic field $B$.

Magnetic flux $\phi$ through each turn of the coil is given by \[\phi=BA\cos\theta\text{ __(i)}\]

If the coil rotates with angular velocity $\omega$ and turns through an angle $\theta$ in time $t$ then

\[\omega=\frac{\theta}{t}\Rightarrow\theta=\omega t\]

Hence, equation $\text{(1)}$ becomes \[\phi=BA\cos\omega t\]

The rotation of the coil changes the magnetic flux linked with it and hence an induced emf is set up in the coil. According to Faraday’s laws of electromagnetic induction,

\[\varepsilon=-N\frac{d\phi}{dt}=-N\frac{d}{dt}(BA\cos\omega t)\] \[=-NBA\frac{d}{dt}(\cos\omega t)\] \[=-NBA×-\sin\omega t×\omega\] \[\therefore\varepsilon=NBA\omega\sin\omega t\text{ __(ii)}\]

The magnitude of induced emf will be maximum if $\sin\omega t=1$. Hence, maximum value of induced emf $(\varepsilon_0$ is given by \[\varepsilon_0=NBA\omega\]

So, equation $\text{(ii)}$ becomes \[\varepsilon=\varepsilon_0\sin\omega t\]

Thus a coil rotating with a constant velocity in a uniform magnetic field produces sinusoidally alternating emf. The graph given below shows the waveform of induced emf.

This graph shows that both magnitude and direction of Induced emf varies sinusoidally with time. If a resistance $R$ is connected across the coil, then the resulting current will also be sinusoidal which is given by

\[I=\frac{\varepsilon}{R}=\frac{\varepsilon_0\sin\omega t}{R}=I_0\sin\omega t\]

where $I_0=\frac{\varepsilon_0}{R}$ is the maximum value of induced current.

Such a rotating coil in a uniform magnetic field is the basic operating principle of an ac generator.

**Previous:** Lenz’s Law

**Next:** Self Induction