10.6 Induced EMF and Faraday's Law

10.6 Induced EMF and Faraday's Law

It has been observed experimentally that when a conductor moves across a magnetic field, an electromotive force (emf) is induced between its ends. The induced emf in the moving conductor is similar to that of a battery. That is, if the ends of the conductor are connected by a wire to form a closed circuit, a current will flow through it.

The emf induced by the motion of a conductor across a magnetic field is called motional emf.

Consider an experiment as shown in Fig. 10.13. A conducting rod of length $L$ is placed on two parallel metal rails separated by a distance $L$. A galvanometer is connected between the ends $c$ and $d$ of the rails. This forms a complete conducting loop. A uniform magnetic field $B$ is applied directed into the page. Initially, when the rod is stationary, galvanometer indicates no current in the loop. If the rod is pulled to the right with constant velocity $v$, the galvanometer indicates a current flowing through the loop. Obviously, the current is induced due to the motion of the conducting rod across the magnetic field. The moving rod is acting as a source of emf $\varepsilon = V_{b} - V_{a} = \Delta V$.

When the rod moves, a charge $q$ within the rod also moves with the same velocity $v$ in the magnetic field $B$ and experiences a force given by Eq. 10.13:

The magnitude of the force is:

Since angle $\theta$ between $v$ and $B$ is $90^{\circ}$, so $F = qvB$ $\ldots (10.16)$

Applying right hand rule, we see that the force $F$ acting on the charge $q$ is directed from point $a$ to point $b$ along the rod. As a result, charges migrate to the top end of the conductor. As more charges move, a concentration of charge builds up at the top end, while there is a deficiency of charges at the bottom end. This redistribution of charge creates an electrostatic field $E$ directed from $b$ to $a$. The electrostatic force on the charge is $F = qE$ directed from $b$ to $a$. The system quickly reaches an equilibrium state where these two forces on the charge are balanced. If $E$ is the electric field intensity in this state, then:

The motional emf $\varepsilon$ will be equal to the potential difference $\Delta V = V_{a} - V_{b}$, between the two ends of the moving conductor in this equilibrium state. The gradient of potential is given by $\Delta V/L$. As the electric intensity is given by the negative of the gradient, therefore:

or $\Delta V = -LE = -(LvB)$

The motional emf is: $\varepsilon = \Delta V = LvB$ $\ldots (10.19)$

This is the magnitude of motional emf. However, if the angle between $v$ and $B$ is $\theta$, then:

Due to the induced emf, positive charges flow along the path $abcd$; therefore, the induced current is anticlockwise in the diagram. As the current flows, the quantity of charge at the top decreases, which reduces the electric field intensity, while the magnetic force remains unchanged. This imbalance disturbs the equilibrium in favour of the magnetic force. Consequently, as the charges reach the end '$a$' of the conductor due to the current flow, they are carried back to the top end '$b$' by the unbalanced magnetic field, and the current continues to flow.