PhysicsCBSE Class 12 · Electromagnetic Induction

Motional EMF Class 12: Rod Moving and Rotating in a Magnetic Field

Updated 15 October 2026 · Fact-checked

When a conductor moves through a magnetic field, the magnetic force on its free charges pushes them along it, creating a motional emf. A rod of length l moving at speed v perpendicular to B has ε = Blv; a rod rotating at ω about one end, perpendicular to B, has ε = ½Bωl².

Magnetic force qv×B pushes positive charges to end P, so the moving rod acts as a battery with emf ε = Blv and drives current round the loop.

Illustration drawn for QuestionBuddy.

Understand Motional EMF

Faraday's law says an emf appears when flux through a circuit changes. One way to change flux is to move part of the circuit through a field. The emf then has a very physical origin: the magnetic force on the charges inside the moving conductor.

Rod on rails (flux method). A straight rod PQ of length l slides with speed v along two parallel rails in a uniform field B perpendicular to the plane of the rails. The rails are joined at one end, so the rod and rails form a rectangle whose length x (the distance of the rod from the closed end) grows at the rate $\frac{d x}{d t} = v$ . The flux through the rectangle is $Φ = B l x$ , so $ε = \frac{d Φ}{d t} = B l \frac{d x}{d t} = B l v$ . With a total resistance R in the circuit, $I = \frac{B l v}{R}$ .

Rod on its own (force method). Even without rails, the free charges in the moving rod move with it at velocity $v$ , so each feels the Lorentz force $q v \times B$ , directed along the rod. Positive charges collect at one end and negative at the other until the electric field they create balances the magnetic force. The work done by the magnetic force per unit charge in moving it along the rod is $v B \times l$ , so $ε = B l v$ again. This explains why a moving conductor develops an emf: the magnetic force acts as the 'pump'. Use $v \times B$ (or Fleming's right-hand rule) to find which end is positive.

General direction. If $v$ is not perpendicular to $B$ , only the component of velocity perpendicular to B (and to the rod) counts: $ε = B l v sin θ$ . If the rod moves along its own length, or parallel to B, no emf is induced.

Rod rotating about one end. A rod of length l rotates with angular speed $ω$ about one end, in a plane perpendicular to a uniform field B. A small piece dr at distance r moves with speed $ω r$ and contributes $d ε = B (ω r) d r$ . Adding from 0 to l: $ε = \int_{0}^{l} B ω r d r = \frac{1}{2} B ω l^{2}$ . Equivalently, the rod sweeps area $\frac{1}{2} l^{2} θ$ , so the flux it cuts per second is $\frac{1}{2} B l^{2} ω$ . A wheel with many metal spokes gives the same emf between hub and rim, because the spokes are in parallel.

Loops entering or leaving a field. When a rectangular loop moves into a region of field, only its leading side cuts field lines: $ε = B l v$ while it enters. Once the loop is completely inside a uniform field, the flux stops changing and $ε = 0$ (even though both sides are moving). While it leaves, $ε = B l v$ again, in the opposite direction. The induced current always opposes the motion (Lenz's law), so a force is needed to keep the loop moving at a steady speed, and that work becomes heat in the loop.

Motional EMF: formulas and key facts

Motional emf of a straight rod: $ε = B l v$; l, v and B mutually perpendicular.
General form: $ε = B l v sin θ$; θ between v and B (v perpendicular to the rod).
Current in a circuit: $I = \frac{B l v}{R}$; R = total resistance of the closed circuit.
Rod rotating about one end: $ε = \frac{1}{2} B ω l^{2}$; Plane of rotation perpendicular to B; ω = 2πf.
Direction (force on charges): $F = q v \times B$; Positive charges collect at the end towards which $v \times B$ points.
Charge through a loop leaving a field: $q = \frac{ΔΦ}{R} = \frac{B A}{R}$; For one loop moving completely out of a uniform field.

How to solve Motional EMF questions

Use this method for rods, wheels and loops moving in a magnetic field.

1Identify the moving conductor, its length l and its velocity (or ω for rotation). Convert to SI units and rev/s to rad/s with ω = 2πf.
2Check the geometry: if v, l and B are mutually perpendicular use Blv; otherwise use the component of v perpendicular to B.
3For rotation about one end, use $\frac{1}{2} B ω l^{2}$ ; about the middle, the two halves give equal and opposite emfs between the ends.
4Find the polarity with $v \times B$ : positive charges move to the end in that direction, which becomes the positive terminal.
5If the conductor is part of a closed circuit, find $I = \frac{ε}{R}$ and its direction.
6For loops crossing a field boundary, split the motion into entering, fully inside and leaving; only the entering and leaving stages give an emf.
7Check with Lenz's law: the induced current should oppose the motion.

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.

Motional emf in a moving fluid

Suppose an electromagnetic flow meter measures the speed of a conducting liquid in a pipe of diameter 5 cm. A magnet makes a uniform 0.1 T field across the pipe, and two electrodes on opposite sides of the pipe, at right angles to the field, measure the voltage. The liquid flows at 2 m/s.

(a) Treat the column of liquid between the electrodes as a conductor of length equal to the diameter. Find the emf. (b) What would the meter read if the flow speed halved? (c) Why must the liquid be conducting?

Try it first, then show the solutionSolution

$ε = B d v = 0.1 \times 0.05 \times 2 = 0.01$ V = 10 mV.
Halving v halves the emf: 5 mV. The reading is proportional to the flow speed.
The emf comes from the magnetic force on moving free charges; without free charges (an insulating liquid) there is nothing to separate.

Answer: (a) 10 mV; (b) 5 mV; (c) it needs free charges for the v × B force to act on.

Memory hook: Any conductor moving across B is a battery of emf Blv.

EMF of a rotating conductor

Imagine a 0.5 m metal blade spinning at 300 revolutions per minute about one end, in a plane perpendicular to a uniform 0.05 T field produced by a large lab magnet.

(a) Find the angular speed. (b) Find the emf between the centre and the tip. (c) How would the emf change if the blade were twice as long?

Try it first, then show the solutionSolution

$ω = \frac{300 \times 2 π}{60} \approx 31.4$ rad/s.
$ε = \frac{1}{2} B ω l^{2} = \frac{1}{2} \times 0.05 \times 31.4 \times 0.25 \approx 0.196$ V.
$ε \propto l^{2}$ , so doubling l makes it four times: about 0.79 V.

Answer: (a) about 31.4 rad/s; (b) about 0.20 V; (c) four times larger.

Memory hook: Rotating rod: half B omega l squared, since the tip moves fastest.

Tricks to remember Motional EMF

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

Blv, 'blue velvet'
Field × length × velocity, all at right angles.
Use it when: Recalling ε = Blv.
Half B omega L squared
A rotating rod is like a straight rod moving at its average speed ωl/2: B × l × ωl/2.
Use it when: Recalling ½Bωl².
Fleming's right hand for generators
Thumb motion, first finger field, second finger induced current.
Use it when: Direction of induced current.
Inside a uniform field, no emf
A loop fully inside a uniform field has constant flux: zero emf, however fast it moves.
Use it when: Graph questions for loops crossing a field.

Motional EMF: solved examples

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

Example 1: Rod on rails · 2 marks · Very short answerBoard level

A rod of length 0.5 m slides at 4 m/s on parallel rails in a uniform field of 0.3 T perpendicular to the plane of the rails. The circuit has a total resistance of 2 Ω. Find the emf and the current.

Try it first, then show the solutionSolution

$ε = B l v = 0.3 \times 0.5 \times 4 = 0.6$ V. (1)
$I = \frac{0.6}{2} = 0.3$ A. (1)

Answer: 0.6 V and 0.3 A.

Example 2: Derivation: motional emf · 3 marks · Short answerBoard level

A rod of length l moves with constant speed v on two parallel rails joined at one end, in a uniform magnetic field B perpendicular to the plane of the rails. Derive the motional emf (a) from the rate of change of flux and (b) from the Lorentz force on the charges in the rod.

Try it first, then show the solutionSolution

(a) If x is the distance of the rod from the closed end, the flux through the circuit is $Φ = B l x$ . Then $ε = - \frac{d Φ}{d t} = - B l \frac{d x}{d t} = - B l v$ ; the size is Blv. (1½)
(b) Each free charge q in the rod moves with velocity v and feels $F = q v B$ along the rod. The work done in moving it the length of the rod is $q v B l$ . (1)
Emf = work per unit charge = Blv, the same result. (½)

Answer: ε = Blv by both methods.

Example 3: Rod rotating about one end · 3 marks · Short answerHarder board level

A metal rod of length 0.8 m rotates at 10 revolutions per second about one end, in a plane perpendicular to a uniform field of 0.2 T. Derive the emf between its ends and evaluate it.

Try it first, then show the solutionSolution

A piece dr at distance r moves with speed ωr, giving $d ε = B ω r d r$ . (1)
$ε = \int_{0}^{l} B ω r d r = \frac{1}{2} B ω l^{2}$ . (1)
$ω = 2 π \times 10 = 20 π$ rad/s; $ε = \frac{1}{2} \times 0.2 \times 20 π \times 0.64 \approx 4.0$ V. (1)

Answer: ε = ½Bωl² ≈ 4.0 V.

Example 4: Rod moving at an angle to the field · 2 marks · Very short answerHarder board level

A rod 0.5 m long moves at 10 m/s, its velocity perpendicular to its length but at 60° to a uniform field of 0.4 T (the rod itself is perpendicular to the field). Find the emf. When would the emf be zero?

Try it first, then show the solutionSolution

Only the velocity component perpendicular to B matters: $ε = B l v sin 60° = 0.4 \times 0.5 \times 10 \times 0.866 \approx 1.73$ V. (1)
Zero if the rod moved parallel to B, or along its own length. (1)

Answer: About 1.73 V; zero when moving along B or along its own length.

Example 5: Loop leaving a field · 5 marks · Long answerHOTS / competency

A square loop of side 10 cm and resistance 0.5 Ω is pulled at a steady 2 cm/s out of a region of uniform field 0.5 T, the field being perpendicular to the loop.
(a) Find the emf and current while it leaves. (2)
(b) How long does the current last, and how much charge flows? Check the charge with q = ΔΦ/R. (2)
(c) Sketch how the emf varies with time from just before the loop starts leaving to just after it has left. (1)

Try it first, then show the solutionSolution

(a) Only the side still in the field (length 0.1 m) bounds the changing area: $ε = B l v = 0.5 \times 0.1 \times 0.02 = 1 0^{- 3}$ V. (1)
$I = \frac{1 0 ^{- 3}}{0.5} = 2 \times 1 0^{- 3}$ A. (1)
(b) Time to leave $= \frac{0.1}{0.02} = 5$ s; charge $= 2 \times 1 0^{- 3} \times 5 = 0.01$ C. (1)
Check: $q = \frac{B A}{R} = \frac{0.5 \times 0.01}{0.5} = 0.01$ C. (1)
(c) Zero while fully inside, a constant 1 mV during the 5 s of leaving, zero again once outside: a rectangular pulse. (1)

Answer: (a) 1 mV and 2 mA; (b) 5 s and 0.01 C (both ways); (c) a rectangular pulse of 1 mV lasting 5 s.

Example 6: Which end is positive? · 4 marks · Case-basedHarder board level

Read the passage and answer.

A rod 25 cm long lies along the y-axis and moves at 5 m/s along +x through a uniform field of 0.2 T directed into the page (−z). Its ends are joined through a 1 Ω resistor by sliding contacts on fixed rails.
(a) Find the emf. (1)
(b) Find the direction of the magnetic force on a positive charge in the rod. Which end of the rod is at higher potential? (1)
(c) Find the current and its direction in the rod. (1)
(d) What does Lenz's law predict about the force on the rod due to this current? (1)

Try it first, then show the solutionSolution

(a) $ε = B l v = 0.2 \times 0.25 \times 5 = 0.25$ V. (1)
(b) $v \times B = (5 \hat{i}) \times (- 0.2 \hat{k}) = + 1 \hat{j}$ : positive charges are pushed towards +y, so the upper (+y) end is at higher potential. (1)
(c) $I = \frac{0.25}{1} = 0.25$ A, flowing inside the rod from the lower end to the upper end (and round the circuit back through the resistor). (1)
(d) The current in the rod feels a force $I l \times B$ opposite to its motion (along −x), so a push is needed to keep it moving. (1)

Answer: (a) 0.25 V; (b) force along +y, upper end positive; (c) 0.25 A upward in the rod; (d) a force opposing the motion.

Where students lose marks in Motional EMF

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

Mistake: Writing Blv for a rotating rod.
Different points of the rod move at different speeds.
Wrong working
ε = Bl(ωl) = Bωl²
Right working
ε = ½Bωl² (integrate or use the average speed ωl/2)
Do this: For rotation, always integrate B ωr dr from 0 to l.
Usually costs: 1 to 2 marks.
Mistake: Using rev/s instead of rad/s for ω.
Forgetting ω = 2πf.
Do this: Convert frequency to angular speed before substituting.
Usually costs: The final mark.
Mistake: Saying a loop moving inside a uniform field has an emf.
Both sides move, so students add their emfs.
Do this: The emfs in the two sides cancel; the flux is constant, so ε = 0.
Usually costs: 1 mark in graph questions.
Mistake: Getting the polarity of the rod wrong.
Not working out v × B carefully.
Do this: Write v and B as unit vectors and take the cross product, or use Fleming's right-hand rule.
Usually costs: ½ to 1 mark.
Mistake: Writing a full energy-balance study for the sliding rod.
Older books include it.
Do this: The quantitative energy consideration is not in the 2026-27 syllabus; explain energy qualitatively with Lenz's law.
Usually costs: Time.

Exam tips

Give both derivations of ε = Blv if asked to 'derive'; the flux method with a clear diagram is the one usually expected.
For a rotating rod, show the element dr, its speed ωr and the integral; the final ½Bωl² alone may not earn full marks.
Board papers have asked motional emf, including the rotating rod, in 2023 (55/1/1 Q7), 2024 (55/5/1 Q24(a)), 2025 (55/7/1 Q6) and 2026 (55/1/1 Q33(a)).
State the polarity and current direction whenever a figure is given.
In loop-and-boundary questions, divide the motion into stages and draw a quick emf-time sketch.

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 parallel to a uniform magnetic field. The emf induced across its ends is
(a) Blv (b) ½Blv (c) zero (d) 2Blv
Hint
v × B = 0.
Answer
(c) zero.
Q2 · 1 mark · Very short answerBoard level
A 1 m rod moves at 5 m/s perpendicular to a 0.2 T field, its length perpendicular to both. Find the emf.
Hint
ε = Blv.
Answer
$ε = 0.2 \times 1 \times 5 = 1$ V.
Q3 · 2 marks · Very short answerHarder board level
A rod of length 0.5 m rotates at 100 rad/s about one end in a plane perpendicular to a 0.4 T field. Find the emf between its ends.
Hint
½Bωl².
Answer
$ε = \frac{1}{2} \times 0.4 \times 100 \times 0.25 = 5$ V.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): No emf is induced in a rod moving along its own length in a magnetic field perpendicular to it.
Reason (R): The magnetic force on the free charges is then perpendicular to the rod and cannot drive them along it.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Direction of v × B.
Answer
(a).
Q5 · 2 marks · Very short answerHOTS / competency
A rod of length l rotates with angular speed ω about its midpoint in a plane perpendicular to a uniform field B. Find the emf (a) between the two ends and (b) between the centre and one end.
Hint
Each half acts like a rod rotating about one end.
Answer
(b) Each half gives $\frac{1}{2} B ω (\frac{l}{2})^{2} = \frac{B ω l ^{2}}{8}$ between centre and end. (a) Both ends are at the same potential relative to the centre, so the emf between the ends is zero.
Q6 · 3 marks · Short answerHOTS / competency
A square loop of side 0.2 m moves at a steady 0.1 m/s through a region 0.6 m wide containing a uniform field of 0.5 T perpendicular to the loop, entering at t = 0. Describe the emf from t = 0 to t = 8 s.
Hint
Track the leading and trailing edges.
Answer
0 to 2 s (entering): $ε = B l v = 0.5 \times 0.2 \times 0.1 = 0.01$ V. 2 to 6 s (fully inside): flux constant, ε = 0. 6 to 8 s (leaving): 0.01 V in the opposite direction. After 8 s: zero.
Q7 · 1 mark · Very short answerHarder board level
A metal rod moves to the right across a field directed out of the page, the rod lying vertically in the page. Which end of the rod becomes positive?
Hint
Find v × B for v to the right and B out of the page.
Answer
$\hat{i} \times \hat{k} = - \hat{j}$ : positive charges are pushed downwards, so the lower end becomes positive.

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.Motional emf of a straight rod?Show answer
ε = Blv.
2.EMF of a rod rotating about one end?Show answer
½Bωl².
3.EMF when a rod moves parallel to B?Show answer
Zero.
4.What provides the force that separates charges in a moving rod?Show answer
The magnetic Lorentz force q v × B.
5.EMF in a loop moving entirely inside a uniform field?Show answer
Zero.
6.Rule for the direction of induced current in a moving conductor?Show answer
Fleming's right-hand rule (or v × B for positive charges).

Motional EMF: frequently asked questions

Is motional emf the same as Faraday's law?

It is a special case. Moving a conductor changes the flux through the circuit it is part of, and dΦ/dt gives Blv. The force picture (q v × B) explains the same result in terms of forces on the charges.

Why is the emf of a rotating rod ½Bωl² and not Bωl²?

Points near the pivot move slowly and the tip moves fastest (ωl). The average speed along the rod is ωl/2, so the emf is Bl × ωl/2 = ½Bωl².

Does a wheel with many spokes give more emf than one spoke?

No. Each spoke has the same emf ½Bωl² between hub and rim, and the spokes are in parallel, so the emf between hub and rim is the same; only the internal resistance is smaller.

Why does a sliding rod slow down when the circuit is closed?

The induced current in the rod feels a magnetic force opposite to its motion (Lenz's law). Without a pushing force the rod slows, its kinetic energy turning into heat in the circuit.

Is the energy consideration for motional emf in the syllabus?

The quantitative study of energy in the moving-rod circuit is not in the 2026-27 syllabus. Faraday's laws, induced emf and current, Lenz's law and self and mutual induction are.