Lesson 2 of 12 · 6 min
Force on a current-carrying conductor
NCERT §4.2.3
Kabir lays a copper rod, 20 cm long and 12 g in mass, across the gap of a magnet giving 0.3 T, and hangs it on two thin flexible leads. As he turns up the current, the rod starts to float. At what current?
The lesson in notes
In short
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.