PhysicsISC Class 12 · Electromagnetic Induction

Motional Emf: A Conductor Moving in a Magnetic Field

Updated 15 October 2026 · Fact-checked

A straight conductor of length l moving with velocity v perpendicular to a uniform field B has an induced emf $\varepsilon = Blv$ between its ends. In a closed circuit of resistance R it drives $I = \frac{Blv}{R}$; the field then exerts a retarding force $\frac{B^2l^2v}{R}$, and the mechanical power needed, $\frac{(Blv)^2}{R}$, equals the electrical power produced.

Understand Motional Emf

Faraday's laws. (1) Whenever the magnetic flux linked with a circuit changes, an emf is induced in it. (2) The induced emf equals the rate of change of flux linkage: $ε = - \frac{d ( N ϕ )}{d t}$ .
Lenz's law gives the sign: the induced current flows in the direction that opposes the change producing it. Lenz's law is a statement of energy conservation: if the induced current helped the change, energy would appear from nowhere.

Motional emf from flux change. A rod PQ of length l slides with speed v along two parallel rails joined at one end, in a uniform field B perpendicular to the plane of the rails. If the rod is a distance x from the closed end, the enclosed area is lx and the flux is $ϕ = B l x$ . As the rod moves, $\frac{d x}{d t} = v$ , so
$∣ ε ∣ = \frac{d ϕ}{d t} = B l \frac{d x}{d t} = B l v .$

Motional emf from the Lorentz force. Each free electron in the rod moves with the rod at velocity $v$ , so it feels $F = - e (v \times B)$ along the rod. Electrons drift to one end, leaving the other positive, until the electric field E they create balances the magnetic force: $e E = e v B$ . The p.d. between the ends is $E l = B l v$ : the same result. This works even with no closed circuit (an emf, but no current).

General case. If the rod, the velocity and the field are not mutually perpendicular, only the components that are perpendicular count: $ε = B l v sin θ$ for $v$ at θ to $B$ (with the rod perpendicular to both). A rod moving along its own length, or along the field, has no emf.

Direction. Use Fleming's right-hand rule (generator rule): stretch the thumb, first finger and second finger of the right hand mutually perpendicular; the First finger points along the Field, the thuMb along the Motion, and the seCond finger gives the induced Current. Or use Lenz's law: the current flows so as to oppose the change of flux. (The 2025 ISC examiners saw Fleming's left-hand and right-hand rules confused.)

Current, force and power. With total resistance R:
$I = \frac{B l v}{R} .$
The current in the rod sits in the field, so the field exerts a force $F = B I l = \frac{B ^{2} l ^{2} v}{R}$ opposite to the motion (Lenz's law again). To keep the rod moving steadily, an external force must supply power
$P = F v = \frac{B ^{2} l ^{2} v ^{2}}{R} = \frac{ε ^{2}}{R} = I^{2} R,$
which appears as heat in R. Mechanical energy becomes electrical energy: this is how a generator works.

Terminal speed. If a constant force F (or the rod's weight on vertical rails) pulls the rod, it speeds up until the retarding force balances it: $\frac{B ^{2} l ^{2} v _{t}}{R} = F$ , so $v_{t} = \frac{F R}{B ^{2} l ^{2}}$ .

Changing speed. If the position is given as x(t), find $v = \frac{d x}{d t}$ first; the emf $B l v$ then changes with time.

Motional Emf: formulas and key facts

Faraday's law: $ε = - \frac{d ( N ϕ )}{d t}$; Minus sign: Lenz's law.
Motional emf: $ε = B l v$; l, v, B mutually perpendicular.
Oblique velocity: $ε = B l v sin θ$; θ between v and B.
Current: $I = \frac{B l v}{R}$; Closed circuit.
Retarding force: $F = B I l = \frac{B ^{2} l ^{2} v}{R}$; Opposes the motion.
Power: $P = \frac{( B l v ) ^{2}}{R} = F v$; Mechanical → electrical → heat.
Terminal speed: $v_{t} = \frac{F R}{B ^{2} l ^{2}}$; Vertical rails: F = mg.

How to solve Motional Emf questions

Find the emf first; everything else follows from Ohm's law and $F = B I l$ .

1Check that l, v and B are perpendicular; otherwise take the perpendicular components.
2If x(t) is given, differentiate to get v at the required time.
3Emf: $ε = B l v$ ; current: $I = \frac{ε}{R}$ .
4Force needed for steady motion: $F = B I l$ ; power: $F v$ (check it equals $I^{2} R$ ).
5Direction: Fleming's right-hand rule or Lenz's law; state which end is at higher potential.
6For terminal speed, set the driving force equal to $\frac{B ^{2} l ^{2} v}{R}$ .

Motional Emf in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Electromagnetic flow meter

Suppose a hospital flow meter places a blood vessel of diameter 5 mm in a field of 0.2 T, with electrodes on opposite sides of the vessel, and blood (a conductor) flows at 0.3 m s⁻¹ across the field.

What emf appears between the electrodes?

Try it first, then show the solutionSolution

The moving blood acts like a conductor of length d = 5 mm moving at v across B.
$ε = B d v = 0.2 \times 0.005 \times 0.3 = 3 \times 1 0^{- 4}$ V = 0.3 mV.

Answer: 0.3 mV, proportional to the flow speed.

Memory hook: Any conductor moving across a field: B × length × speed.

Emf across an aircraft wing

Suppose an aircraft with a wingspan of 40 m flies horizontally due north at 250 m s⁻¹ where the vertical component of the Earth's field is $4 \times 1 0^{- 5}$ T, pointing downwards.

Find the emf between the wingtips and say which tip is positive. Could it light a bulb connected by a wire across the wings?

Try it first, then show the solutionSolution

$ε = B l v = 4 \times 1 0^{- 5} \times 40 \times 250 = 0.4$ V.
Force on positive charges along $v \times B$ = north × down = west: the west wingtip is positive.
A wire joining the tips moves with the plane and has the same emf, so no current flows round the loop.

Answer: 0.4 V, west tip positive; no, it cannot drive a current through wires carried on the plane.

Memory hook: Emf needs relative motion between the circuit and the flux.

Tricks to remember Motional Emf

Short memory aids. Say each one aloud once, then use it on a question the same day.

BLV
"Blood Lost Very fast": ε = Blv.
Use it when: The formula.
Right hand generates
Right hand for generators (induced current); left hand for motors (force).
Use it when: Choosing Fleming's rule.
Lenz brakes
The induced current always produces a force that slows the motion.
Use it when: Direction of force.
Terminal speed
Driving force = $\frac{B ^{2} l ^{2} v}{R}$ .
Use it when: Rods on rails.

Motional Emf: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Direction of induced current · 1 mark · MCQBoard level

The direction of the current induced in a conductor moving in a magnetic field is given by:
(a) Fleming's left-hand rule
(b) Fleming's right-hand rule
(c) the right-hand thumb rule for a straight wire
(d) Coulomb's law

Try it first, then show the solutionSolution

Generator action: the right-hand rule.

Answer: (b).

Example 2: Emf and current · 2 marks · Short answerBoard level

A rod 0.8 m long moves at 5 m s⁻¹ perpendicular to a field of 0.25 T, on rails completing a circuit of resistance 4 Ω. Find the emf and the current.

Try it first, then show the solutionSolution

$ε = B l v = 0.25 \times 0.8 \times 5 = 1$ V. (1)
$I = \frac{1}{4} = 0.25$ A. (1)

Answer: 1 V; 0.25 A.

Example 3: Derivation and energy balance · 3 marks · Long answerHarder board level

A rod of length l slides at constant speed v on rails in a uniform field B perpendicular to the rails, the circuit having resistance R. Derive the emf from the rate of change of flux, and show that the power needed to keep the rod moving equals the power dissipated in R.

Try it first, then show the solutionSolution

Flux through the circuit $ϕ = B l x$ ; $∣ ε ∣ = \frac{d ϕ}{d t} = B l \frac{d x}{d t} = B l v$ . (1)
Current $I = \frac{B l v}{R}$ ; force on the rod $B I l = \frac{B ^{2} l ^{2} v}{R}$ , opposing the motion, so the same external force is needed. (1)
Mechanical power $F v = \frac{B ^{2} l ^{2} v ^{2}}{R}$ ; electrical power $I^{2} R = \frac{B ^{2} l ^{2} v ^{2}}{R}$ : equal. (1)

Answer: $ε = B l v$ ; both powers equal $\frac{B ^{2} l ^{2} v ^{2}}{R}$ .

Example 4: Position given as a function of time · 2 marks · Short answerHarder board level

A rod 0.5 m long moves on rails in a uniform 0.4 T field perpendicular to the rails. Its position is $x = 3 t^{2} + 2 t$ (x in m, t in s). Find the emf at t = 2 s.

Try it first, then show the solutionSolution

$v = \frac{d x}{d t} = 6 t + 2 = 14$ m s⁻¹ at t = 2 s. (1)
$ε = B l v = 0.4 \times 0.5 \times 14 = 2.8$ V. (1)

Answer: 2.8 V (and increasing with time).

Example 5: Terminal speed on vertical rails · 3 marks · Short answerHOTS / competency

A rod of mass 20 g and length 0.4 m slides without friction down a pair of vertical rails joined at the top by a 0.5 Ω resistor. A horizontal field of 0.5 T is perpendicular to the plane of the rails. Find the terminal speed. (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

At terminal speed, magnetic force = weight: $\frac{B ^{2} l ^{2} v _{t}}{R} = m g$ . (1)
$v_{t} = \frac{m g R}{B ^{2} l ^{2}} = \frac{0.02 \times 9.8 \times 0.5}{0.25 \times 0.16}$ . (1)
$v_{t} = \frac{0.098}{0.04} = 2.45$ m s⁻¹. (1)

Answer: 2.45 m s⁻¹.

Example 6: Pulled by a steady force · 4 marks · Case-basedHOTS / competency

Suppose a rod 0.5 m long lies across rails in a 0.6 T field perpendicular to the rails, the circuit resistance being 1.5 Ω. A constant force of 0.3 N pulls the rod.
(i) Explain why the rod reaches a steady speed. (1 mark)
(ii) Find that speed. (1 mark)
(iii) Find the current at that speed. (1 mark)
(iv) Show that the power supplied by the force equals the power dissipated. (1 mark)

Try it first, then show the solutionSolution

(i) As v rises, the opposing force $\frac{B ^{2} l ^{2} v}{R}$ grows until it equals 0.3 N; then the net force is zero.
(ii) $v_{t} = \frac{F R}{B ^{2} l ^{2}} = \frac{0.3 \times 1.5}{0.36 \times 0.25} = 5$ m s⁻¹.
(iii) $I = \frac{B l v}{R} = \frac{0.6 \times 0.5 \times 5}{1.5} = 1$ A.
(iv) $F v = 0.3 \times 5 = 1.5$ W; $I^{2} R = 1 \times 1.5 = 1.5$ W.

Answer: (ii) 5 m s⁻¹; (iii) 1 A; (iv) both 1.5 W.

Where students lose marks in Motional Emf

The slips examiners see most often, and what to write instead.

Mistake: Not finding v from x(t)
The ISC 2025 examiners noted candidates unclear about differentiating x to get the velocity.
Do this: Write $v = \frac{d x}{d t}$ as a separate line.
Usually costs: 1 mark.
Mistake: Left-hand rule for induced current
Also in 2025: Fleming's left-hand and right-hand rules were confused.
Wrong working
"Fleming's left-hand rule gives the induced current."
Right working
Right-hand (generator) rule, or Lenz's law.
Do this: Right hand = generator; left hand = motor.
Usually costs: 1 mark.
Mistake: Not knowing ε = Blv
2025 candidates did not know the motional emf formula.
Do this: Derive it from $ϕ = B l x$ once; it then sticks.
Usually costs: 1 to 2 marks.
Mistake: Force in the direction of motion
Students forget Lenz's law in force questions.
Do this: The magnetic force on the induced current always opposes the motion.
Usually costs: 1 mark.

Exam tips

Draw the rails, rod, field (⊗ or ⊙) and velocity, and mark the direction of the induced current.
State both laws when asked for directions: Fleming's right-hand rule for current, Lenz's law for the opposing force.
Check energy: Fv should equal I²R.
For changing speeds, give the emf at the stated instant and note that it varies.

Practice questions on Motional Emf

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A straight rod moves through a uniform magnetic field parallel to the field. The emf induced across its ends is:
(a) Blv
(b) maximum
(c) zero
(d) $\frac{B l v}{2}$
Hint
$v \times B = 0$ .
Answer
(c).
Q2 · 2 marks · Short answerBoard level
A rod of length 2 m moves perpendicular to a uniform field $B = 0.5$ T, its position being $x = 4 t$ (x in metres, t in seconds). Find the motional emf and name the rule for the direction of the induced current. (A variation on ISC 2025.)
Hint
$v = \frac{d x}{d t}$ .
Answer
$v = 4$ m s⁻¹, $ε = B l v = 0.5 \times 2 \times 4 = 4$ V; Fleming's right-hand rule.
Q3 · 1 mark · Very short answerHarder board level
A 1 m rod moves at 10 m s⁻¹ in a 0.2 T field, its velocity making 30° with the field (rod perpendicular to both). Find the emf.
Hint
Use $sin θ$ .
Answer
$0.2 \times 1 \times 10 \times sin 30° = 1$ V.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A conducting rod moving across a magnetic field with no closed circuit has an emf but no steady current.
Reason (R): The magnetic force moves electrons to one end until the electric field produced balances it.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
eE = evB.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
A rod 0.5 m long is moved at a steady 2 m s⁻¹ on rails in a 0.6 T field; the circuit resistance is 3 Ω. Find the emf, current, force needed and power.
Hint
$F = B I l$ .
Answer
0.6 V; 0.2 A; 0.06 N; 0.12 W.
Q6 · 3 marks · Short answerHOTS / competency
A square loop of side 10 cm and resistance 0.2 Ω is pulled at a steady 2 m s⁻¹ out of a region of uniform 0.5 T field perpendicular to the loop. Find the current while it leaves, the time taken, and the heat produced; check against the work done.
Hint
Only one side cuts field lines.
Answer
ε = 0.1 V, I = 0.5 A for 0.05 s; heat = $0.25 \times 0.2 \times 0.05 = 2.5 \times 1 0^{- 3}$ J; work = $B I l \times 0.1 = 0.025 \times 0.1 = 2.5 \times 1 0^{- 3}$ J.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a metal rod of length 1.2 m slides at 3 m s⁻¹ on rails joined by a 6 Ω resistor, in a field of 0.5 T perpendicular to the rails.
(i) Find the emf. (1 mark)
(ii) Find the current and the force on the rod. (1 mark)
(iii) In which direction does the force act relative to the motion, and which law explains it? (1 mark)
(iv) If the speed is doubled, by what factor does the power needed change? (1 mark)
Hint
$P \propto v^{2}$ .
Answer
(i) 1.8 V. (ii) 0.3 A; 0.18 N. (iii) Opposite to the motion; Lenz's law. (iv) Four times.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State Faraday's second law.Show answer
The induced emf equals the rate of change of magnetic flux linkage.
2.State Lenz's law.Show answer
The induced current flows in a direction that opposes the change producing it.
3.Motional emf for B = 0.5 T, l = 0.2 m, v = 10 m s⁻¹?Show answer
1 V.
4.Force needed to keep the rod moving steadily?Show answer
$\frac{B ^{2} l ^{2} v}{R}$ .
5.Which rule gives the direction of induced current in a moving conductor?Show answer
Fleming's right-hand rule.

Motional Emf: frequently asked questions

Where does the energy of the induced current come from?

From the work done by the external agent against the retarding magnetic force. Lenz's law ensures this force always opposes the motion.

Is motional emf the same as Faraday's law?

It is one case of it. The flux through the circuit changes because the area changes, and \frac{d\phi}{dt} gives Blv; the Lorentz-force argument gives the same result.

Why does a magnet fall slowly through a copper tube?

The moving magnet induces eddy currents in the tube that, by Lenz's law, produce a field opposing its motion, so it falls with a small terminal speed.

What does the ISC 2027 syllabus include here?

Faraday's laws, Lenz's law and energy conservation, motional emf ε = Blv and power P = \frac{(Blv)^2}{R}, and eddy currents qualitatively.

How was this tested recently?

In 2025 part of a 5-mark question asked for the motional emf of a rod whose position varied with time, and the law giving the direction of the induced current.