What Is Motional EMF
Consider a straight conductor moving in a uniform, time-independent magnetic field. NCERT §6.6 frames the standard arrangement: a rectangular conductor PQRS lies in a plane perpendicular to a uniform field $B$, and the arm PQ of length $\ell$ is free to slide along the two parallel rails. When PQ is pushed with a constant velocity $v$, the area enclosed by the loop PQRS changes continuously, so the magnetic flux threading the loop changes too.
By Faraday's law, that changing flux drives an induced EMF around the circuit. Because the change here is produced by the motion of the conductor rather than by a varying field, the EMF carries a special name: the motional electromotive force. It demonstrates that an EMF can be produced simply by moving a conductor through a steady field — the second of the two routes Faraday's law allows.
The arm PQ slides left with speed $v$; the enclosed area shrinks, flux falls, and an induced current $I$ circulates. The field pushes back on the current-carrying rod with a retarding force $F = BI\ell$, so an external agent must keep supplying force to maintain constant $v$.
Derivation From Flux Change
Let the rod PQ be at a distance $x$ from the closed end (so $RQ = x$) and let its length be $RS = \ell$. The flux enclosed by the loop is the field times the enclosed area:
$$\Phi_B = B\,\ell\,x$$
Only $x$ changes with time, so applying Faraday's law gives
$$\varepsilon = -\frac{d\Phi_B}{dt} = -\frac{d}{dt}(B\ell x) = -B\ell\frac{dx}{dt} = B\ell v$$
where we used $dx/dt = -v$ for the inward motion. The magnitude of this induced EMF, $\boxed{\varepsilon = B\ell v}$, is the motional EMF. The result is exact for a uniform field with $B$, $\ell$ and $v$ mutually perpendicular; if $v$ makes an angle with the rod or the field, only the perpendicular components contribute.
| Quantity | Symbol | Relation | SI unit |
|---|---|---|---|
| Motional EMF | ε | ε = Bℓv | volt (V) |
| Induced current | I | I = Bℓv / R | ampere (A) |
| Retarding force on rod | F | F = BIℓ = B²ℓ²v / R | newton (N) |
| Power dissipated | P | P = ε²/R = B²ℓ²v² / R | watt (W) |
The Lorentz-Force View
The same EMF can be obtained without invoking flux at all, by looking directly at the free charges in the moving rod. As NCERT notes, this microscopic picture explains why the EMF appears. Take any free charge $q$ inside PQ. When the rod moves with speed $v$ through the field $B$, that charge shares the rod's velocity, so it experiences a Lorentz force of magnitude $qvB$ directed along the rod (toward Q in the standard figure).
Every free charge in the rod feels the same force, in the same direction, regardless of its position. The work done by this force in moving a charge the full length $\ell$ from P to Q is
$$W = qvB\ell$$
Since EMF is the work done per unit charge,
$$\varepsilon = \frac{W}{q} = B\ell v$$
identical to the flux result. This is the deeper reason a motional EMF develops across an isolated moving rod even before any circuit is completed: charges accumulate at the ends until the electrostatic field they set up balances the magnetic force, leaving a steady potential difference $B\ell v$ across the rod.
The force that keeps the rod moving, and which way it points
Many students assume the field drives the rod. It does the opposite. The induced current $I$ flows through the rod sitting in field $B$, so the rod feels a force $F = BI\ell$. By Lenz's law this force is retarding — it points opposite to $v$, fighting the motion that caused it. To keep the rod at constant velocity, the external agent must push with an equal and opposite force of the same magnitude.
Force needed at constant $v$: $F_{\text{applied}} = BI\ell = \dfrac{B^2\ell^2 v}{R}$, directed along $v$. Net force on the rod is zero, so KE does not change.
Force, Current and Power
Close the circuit through a resistance $R$. The motional EMF $\varepsilon = B\ell v$ drives an induced current
$$I = \frac{\varepsilon}{R} = \frac{B\ell v}{R}$$
This current-carrying rod, immersed in the field $B$, experiences a magnetic force $F = BI\ell$. Substituting the current,
$$F = BI\ell = \frac{B^2\ell^2 v}{R}$$
The energy bookkeeping is the heart of the topic. NCERT stresses that no perpetual-motion machine can result: the work the external agent does against the retarding force is dissipated as Joule heating by the induced current. At constant velocity the mechanical power supplied equals the electrical power dissipated.
At constant speed the rod's kinetic energy is unchanged, so every joule the agent supplies reappears as Joule heat. Equating mechanical and electrical power gives $Fv = I^2R = B^2\ell^2v^2/R$, which is consistent with $F = B^2\ell^2 v/R$.
The power dissipated, equivalently written through the EMF, is
$$P = I^2 R = \frac{\varepsilon^2}{R} = \frac{B^2\ell^2 v^2}{R}$$
If the agent stops pushing, only the retarding force acts; the rod decelerates and its kinetic energy is converted to heat until it halts. This is the same Lenz's-law mechanism that makes a magnet falling near a conductor slow down.
The retarding-force direction here is a direct consequence of Lenz's law — revise it to never get the sign wrong.
Motional EMF of a Rotating Rod
NEET frequently extends the idea to a rod rotating in a field rather than sliding. Consider a rod of length $R$ rotating about one fixed end with angular speed $\omega$, in a uniform field $B$ parallel to the rotation axis. Different points along the rod move at different speeds: a point at distance $r$ from the pivot has speed $v = \omega r$.
Treat an element of length $dr$ at distance $r$. Its motional EMF is $d\varepsilon = Bv\,dr = B\omega r\,dr$. Integrating from the pivot to the tip,
$$\varepsilon = \int_0^{R} B\omega r\,dr = \tfrac{1}{2}B\omega R^2$$
This $\tfrac{1}{2}B\omega R^2$ is the EMF between the centre and the rim. NCERT's Example 6.6 applies exactly this result to a $1\,\text{m}$ rod spinning at $50$ rev/s in a $1\,\text{T}$ field, obtaining $157\,\text{V}$; Example 6.7 uses the spoked-wheel version to get $6.28\times10^{-5}\,\text{V}$ for spokes in the earth's field. The number of spokes is irrelevant because their EMFs are in parallel.
Rotating rod uses ½BωR², not BℓV
For a sliding rod every point moves at the same $v$, so $\varepsilon = B\ell v$. For a rotating rod the speed varies along its length, so you must integrate — the answer is $\tfrac{1}{2}B\omega R^2$, with the factor of $\tfrac{1}{2}$. Plugging the tip speed $v = \omega R$ into $B\ell v$ would double the correct value.
Sliding rod: $\varepsilon = B\ell v$. Rotating rod (about one end): $\varepsilon = \tfrac{1}{2}B\omega R^2 = \tfrac{1}{2}BvR$ with $v = \omega R$ the tip speed.
Worked Examples
A rod of length $0.5\,\text{m}$ slides on frictionless rails at $4\,\text{m s}^{-1}$ in a uniform field $0.2\,\text{T}$ perpendicular to its plane. The circuit resistance is $2\,\Omega$. Find the EMF, current, retarding force and power dissipated.
EMF: $\varepsilon = B\ell v = 0.2 \times 0.5 \times 4 = 0.4\,\text{V}$.
Current: $I = \varepsilon/R = 0.4/2 = 0.2\,\text{A}$.
Retarding force: $F = BI\ell = 0.2 \times 0.2 \times 0.5 = 0.02\,\text{N}$ (an equal applied force keeps $v$ constant).
Power: $P = B^2\ell^2v^2/R = (0.2)^2(0.5)^2(4)^2/2 = 0.08\,\text{W}$, which equals $Fv = 0.02 \times 4 = 0.08\,\text{W}$.
A metallic rod of length $1\,\text{m}$ is rotated at $50$ rev/s about one end, the other end touching a ring of radius $1\,\text{m}$, in a uniform field $1\,\text{T}$ parallel to the axis. Find the EMF between centre and ring.
Angular speed $\omega = 2\pi \times 50 = 100\pi\ \text{rad s}^{-1}$. Using $\varepsilon = \tfrac{1}{2}B\omega R^2 = \tfrac{1}{2}(1)(100\pi)(1)^2 \approx 157\,\text{V}$, matching NCERT's value.
Motional EMF in one screen
- A conductor moving through a field changes the enclosed flux, inducing a motional EMF; for a sliding rod $\varepsilon = B\ell v$.
- Flux route: $\Phi_B = B\ell x \Rightarrow \varepsilon = -d\Phi_B/dt = B\ell v$.
- Lorentz route: each charge feels $qvB$, work $= qvB\ell$, so $\varepsilon = W/q = B\ell v$. EMF exists across an isolated rod; a closed loop is needed only for current.
- Closed circuit: $I = B\ell v/R$, retarding force $F = BI\ell = B^2\ell^2v/R$ (opposes $v$), power $P = B^2\ell^2v^2/R$.
- Energy: at constant $v$, agent's mechanical power $Fv$ equals Joule heating $I^2R$ — no perpetual motion.
- Rotating rod about one end: $\varepsilon = \tfrac{1}{2}B\omega R^2$ (mind the factor of $\tfrac{1}{2}$).