Emf of a rotating rod
"Emf of a rotating rod" is a free, interactive Class 12 Physics simulation for NEET and JEE, built on the NCERT section "Rotating rods and wheels" in Electromagnetic Induction. A rod or spoked wheel turning about one end in a field along the axis, with v = ωr growing along it and ε = ½BωR². JEE Main asked 29 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Rotating rods and wheels in Electromagnetic Induction. Change the values and watch what happens.
What it shows
NCERT §6.6
A rod or spoked wheel turning about one end in a field along the axis, with v = ωr growing along it and ε = ½BωR².
Controls
4 to try
- Sliderω 0–50 rev/s
- Sliderlength 0.1–1 m
- SliderB 0–1 T
- Sliderspokes 1–12
Try this
Reach 157 V with a 1 m rod in 1 T, then add spokes and show the emf does not change.
Common mistake
Using BωR², or adding the emfs of all spokes.
In JEE Main
42 shifts analysed
29 questions from Electromagnetic Induction across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §6.6
- A rod of length R turning at angular speed ω about one end, in a uniform field B along the axis, has different parts moving at different speeds v = ωr.
- Adding the emf B v dr from each small piece gives ε = ½BωR² between the pivot and the free end.
- The same result follows from Faraday's law: the rod sweeps a sector of area ½R²θ, and B times the rate of change of that area is ½BR²ω.
- Example 6.6: a 1 m rod turning at 50 rev/s in a 1 T field gives ε = ½ × 1.0 × 2π × 50 × 1² = 157 V between the centre and the rim.
- Example 6.7: a wheel with 10 spokes of 0.5 m, turning at 120 rev/min in the earth's horizontal field of 0.4 G, gives ε = ½ × 4π × 0.4 × 10⁻⁴ × 0.5² = 6.28 × 10⁻⁵ V between axle and rim.
- The number of spokes does not matter: the spokes are emfs in parallel, each giving the same value.
Questions
4 answers
- What does the "Emf of a rotating rod" simulation show?
- A rod or spoked wheel turning about one end in a field along the axis, with v = ωr growing along it and ε = ½BωR².
- What mistake does the "Emf of a rotating rod" simulation fix?
- A common mistake it clears up: using BωR², or adding the emfs of all spokes.
- Does JEE Main ask questions from Electromagnetic Induction?
- Yes. JEE Main asked 29 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Rotating rods and wheels" in Electromagnetic Induction (Class 12 Physics).
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