Question

the electron density of copper is 8 * 10²⁸ electrons per m³.a copper wire of length 1m area of cross section 8mm² is carrying current. it is placed at right angle to a magnetic field of 5 * 10^-3T experience a force of 8 * 10^-2N . calculate drift velocity of free electreon in the wire

Answer 1

Here,

n = 8 ×10²⁸ electrons/m³

B = 5 × 10^{-3 T}

F = 8×10^{-2 N}

A = 8×10^-6 m²

L = 1 m

θ = π/2 rad

We know force experienced by a current carrying wire kept in a magnetic field is given by

F = BiLsinθ

=> 8×10^-2 = 5 × 10^-3×i×1×sin(π/2)

=> i = 16

Again we know

I = neAv_d

=> 16 = 8 ×10²⁸ ×1.6×10^-19×8×10^-6 ×v_d

=> v_d = 1.6×10^-4 ms^-1 (approx.)