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A metal rod spins about one end in a magnetic field, and the question asks for the voltage between its ends. Students who know motional EMF, ε = Blv, often get stuck because different parts of the rod move at different speeds. This post derives ½BωL² two ways, then extends it to a circuit, a rod pivoted off-centre and a spinning disc.
What does NCERT cover?
The JEE (Main) 2026 syllabus lists Faraday's law and induced EMF in Unit 14, and the CBSE Physics curriculum for 2026-27 covers induced EMF in Class XII Chapter 6. The NCERT chapter Electromagnetic Induction works this case directly. Example 6.6 rotates a 1 m rod at 50 rev/s in a 1 T field and finds 157 V. Example 6.7 spins a wheel with 10 spokes and notes that the number of spokes does not matter. Check the syllabus for your exam year.
Where does ½BωL² come from?
Method 1: add up small pieces. A small piece of rod of length dr at distance r from the pivot moves at speed v = ωr, perpendicular to B. It contributes dε = Bv dr = Bωr dr. Add from r = 0 to r = L:
ε = ∫ Bωr dr = ½BωL².
You can read this as BLv with v equal to the average speed of the rod, ωL/2.
Method 2: area swept. In time dt the rod turns through ω dt and sweeps a thin sector of area ½L²ω dt. The rate of change of flux is B × ½L²ω, which again gives ½BωL². NCERT uses both methods in Example 6.6.
Worked example 1: a rod driving a current
A copper rod 0.50 m long rotates at 60 rad/s about one end in a field of 0.40 T along the axis. Its far end slides on a metal ring, and a 2.0 Ω resistor connects the ring to the pivot. Ignore the resistance of the rod and ring.
EMF: ε = ½ × 0.40 × 60 × 0.50² = 3.0 V.
Current: I = 3.0 / 2.0 = 1.5 A. Electrical power: εI = 4.5 W.
Magnetic force on the rod: F = ILB = 1.5 × 0.50 × 0.40 = 0.30 N. It is spread evenly along the rod, so it acts as if at the midpoint, giving a torque of 0.30 × 0.25 = 0.075 N m that opposes the rotation (Lenz's law).
Power needed to keep it turning: τω = 0.075 × 60 = 4.5 W.
The mechanical power in equals the electrical power out, which is a good check on any rotating-rod answer.
Worked example 2: a pivot that is not at the end
A rod 0.60 m long is pivoted 0.20 m from one end and rotates at 20 rad/s in a 0.50 T field along the axis. Find the EMF between its two ends.
Treat each arm separately, measured from the pivot:
Long arm (0.40 m): ½ × 0.50 × 20 × 0.40² = 0.80 V.
Short arm (0.20 m): ½ × 0.50 × 20 × 0.20² = 0.20 V.
Both tips are at the same kind of potential relative to the pivot, because the Lorentz force pushes charge outwards (or inwards) along both arms. So the EMF between the tips is the difference: 0.80 − 0.20 = 0.60 V. A rod pivoted at its midpoint gives zero between its ends.
Worked example 3: a spinning disc
A metal disc of radius 0.10 m spins at 3000 rev/min in a 0.80 T field along its axis. Think of it as many radial spokes connected in parallel, which is NCERT's argument in Example 6.7, so the EMF between centre and rim is ½BωR².
ω = 3000 × 2π / 60 ≈ 314 rad/s.
ε = ½ × 0.80 × 314 × 0.10² ≈ 1.26 V.
Which end is positive?
Apply F = qv × B to a positive charge in the rod. The end it is pushed towards is at the higher potential. For example, with B pointing out of the page and the rod turning anticlockwise, positive charge is pushed outwards, so the outer end is positive. Reverse either B or the rotation and the sign flips.
Which mistakes cost marks?
Dropping the ½. Using the tip speed ωL instead of the average speed doubles the answer.
Adding spokes in series. Spokes from axle to rim are in parallel, so the EMF stays ½BωR².
Leaving ω in rev/min or rev/s. Convert to rad/s first.
Using the wrong component of B. Only the component along the rotation axis counts, which is why NCERT's wheel uses the horizontal component of the earth's field.
Adding the two arms of an off-centre rod. Subtract them.
A practice routine
Day 1: Derive ½BωL² by both methods without notes.
Day 2: Redo NCERT Exercise 6.5 (1.0 m rod, 400 rad/s, 0.5 T) and check you get 100 V.
Day 3: Repeat worked example 1 with a 4.0 Ω resistor and confirm the two powers still match.
Day 4: Compare with a rod sliding on rails and a few Lenz's law problems.
Day 5: Do a mixed, timed set. Dunlosky and colleagues (2013) rated practice testing and spaced practice as the most useful study techniques.
If you cannot see why the tip and the pivot behave differently, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Tonight, work out the EMF for a rod pivoted at one quarter of its length and check that it is half the end-pivot value.
