Physics · Ch 6 — Electromagnetic Induction

Motional Electromotive Force

6.6

Motional Electromotive Force

What is Motional EMF?

When a conductor moves through a uniform, time-independent magnetic field, an emf is induced across it. This is called motional electromotive force. It arises because the motion changes the magnetic flux enclosed by the circuit, or equivalently, because the moving charges inside the conductor experience a Lorentz force.

Derivation from Faraday's Law (Flux Change)

Consider a rectangular loop $PQRS$ where arm $PQ$ (length $l$ ) slides to the left with constant speed $v$ . The loop is in a uniform magnetic field $B$ perpendicular to its plane.

Let $RQ = x$ and $RS = l$ .
Area of loop: $A = lx$ .
Magnetic flux through the loop:

\Phi_B = B \cdot A = Blx

As $x$ decreases with time, the flux changes. The induced emf is given by Faraday's law:

\varepsilon = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(Blx) = -Bl \frac{dx}{dt}

Since the rod moves left, $dx/dt = -v$ (negative because $x$ decreases). Substituting:

\varepsilon = -Bl(-v) = Blv

This is the motional emf:

\boxed{\varepsilon = Blv}

$\varepsilon$ = induced emf (V)
$B$ = magnetic field strength (T)
$l$ = length of the moving conductor (m)
$v$ = speed of the conductor perpendicular to $B$ (m/s)

Derivation from Lorentz Force (Charge Perspective)

Consider a charge $q$ inside the moving rod $PQ$ . The rod moves with speed $v$ in magnetic field $B$ . The Lorentz force on the charge is:

F = qvB

Direction: towards $Q$ (using right-hand rule). All charges experience the same force.

Work done to move the charge from $P$ to $Q$ (distance $l$ ):

W = F \cdot l = qvBl

Since emf is work done per unit charge:

$\varepsilon = \frac{W}{q} = Blv$ …

Figure 6.10The arm PQ is moved to the left side, thus decreasing the area of the rectangular loop. This movement induces a current I as shown.

Fig. 6.10 — The arm PQ is moved to the left side, thus decreasing the area of the rectangular loop. This movement induces a current I as shown.

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

The figure shows a rectangular conducting loop PQRS placed in a uniform magnetic field $\mathbf{B}$ directed into the page (represented by $\times$ symbols). The loop consists of:

A fixed left side SR of length $l$ (vertical double-headed arrow).
Top and bottom horizontal rails extending to the right, labelled SP and RQ.
A movable right side PQ — a thick sliding rod that can move left or right along the rails.
The rod PQ is shown with a bold velocity arrow $\mathbf{v}$ pointing left, indicating it is being pushed toward the fixed side SR, thereby decreasing the area of the loop.
The horizontal distance from the fixed side to the rod is labelled $x$ (double-headed arrow along the bottom rail).
A current $I$ is shown flowing in the loop (arrow labelled I along the top rail).

The physical idea is motional emf: when a conductor moves in a magnetic field, the free charges inside experience a Lorentz force, leading to an induced emf and current. Here, the rod PQ moves left, reducing the area $A = lx$ of the loop. The magnetic flux through the loop is

\Phi_B = B \cdot A = B l x

Since $x$ changes with time, the flux changes, inducing an emf. Using Faraday’s law:

\varepsilon = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(Blx) = -Bl\frac{dx}{dt}

The rod moves left with speed $v$ , so $\frac{dx}{dt} = -v$ (because $x$ decreases). Substituting gives the motional emf:

\boxed{\varepsilon = Blv}

Here:

$B$ = magnitude of uniform magnetic field (into page)
$l$ = length of the rod PQ (also the side SR)
$v$ = speed of the rod (magnitude of velocity) …

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