Floating a rod on the magnetic force
"Floating a rod on the magnetic force" is a free, interactive Class 12 Physics simulation for NEET and JEE, built on the NCERT section "Force on a current-carrying conductor" in Moving Charges and Magnetism. A 200 g, 1.5 m rod carrying current in a horizontal field, lifted when IlB sin θ = mg. NEET has asked about this idea 2 times in 10 years of papers. JEE Main asked 56 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Force on a current-carrying conductor in Moving Charges and Magnetism. Change the values and watch what happens.
What it shows
NCERT §4.2.3
A 200 g, 1.5 m rod carrying current in a horizontal field, lifted when IlB sin θ = mg.
Controls
3 to try
- Slidercurrent 0–10 A
- SliderB 0–1 T
- Sliderangle between rod and field 0–90°
Try this
Float the rod with 2 A (B = 0.65 T), then turn it toward the field and show it falls.
Common mistake
Forgetting that a wire along the field feels no force.
In NEET
10 years of papers
Asked 2 times in 10 years of NEET.
See the past questions: Magnetic Force on Charges and Currents, Moving Charges and Magnetism (chapter)
In JEE Main
42 shifts analysed
56 questions from Moving Charges and Magnetism across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §4.2.3
- A straight rod of length l and cross-section A with n carriers per unit volume holds nlA carriers, each drifting at v_d. Adding their forces in a field B gives F = (nlA)q v_d × B.
- Since nqv_d A is the current I, the force becomes F = I l × B, where the vector l has the rod's length and points along the current. The current itself is a scalar; the direction sits on l.
- Size: F = IlB sin θ, with θ the angle between the wire and B. A wire along the field feels nothing; one at right angles feels the most, IlB.
- B in this formula is the external field, not the field of the wire itself. For a wire of any shape, add I dl × B over small straight pieces.
- Example 4.1: a 200 g, 1.5 m wire carrying 2 A floats in a horizontal field when IlB = mg, so B = (0.2 × 9.8)/(2 × 1.5) = 0.65 T. Only m/l matters, and the earth's field (about 4 × 10⁻⁵ T) is too small to count.
- Example 4.2: with B along +y and a particle moving along +x, v × B points along +z. A proton is pushed along +z and an electron along −z.
Questions
5 answers
- What does the "Floating a rod on the magnetic force" simulation show?
- A 200 g, 1.5 m rod carrying current in a horizontal field, lifted when IlB sin θ = mg.
- What mistake does the "Floating a rod on the magnetic force" simulation fix?
- A common mistake it clears up: forgetting that a wire along the field feels no force.
- Has NEET asked about this idea?
- Yes. NEET has asked about this idea 2 times in 10 years of papers.
- Does JEE Main ask questions from Moving Charges and Magnetism?
- Yes. JEE Main asked 56 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 "Force on a current-carrying conductor" in Moving Charges and Magnetism (Class 12 Physics).
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