Simulation · Physics · Class 12
Floating a rod on the magnetic force
From the lesson Force on a current-carrying conductor in Moving Charges and Magnetism. Change the values and watch what happens.
Floating a rod on the magnetic forcePhysics · Class 12
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.