Question

Each cubic meter of the wire contains approximately 8.5x10^28 free electrons. The diameter of the wire is 2.5 mm and the length of wire within the magnetic field is 0.15 m. Deduce the speed of the electrons in the wire when the current is 0.20 A.

Answer 1

Every electron carries a charge $e=1.6\cdot 10^{-19}\,\text{C}$

then

$I=nSev\Rightarrow\\ v=\frac{I}{nqS} = \frac{I}{nq\pi r^2} = \\ =\frac{0.2}{8.5\cdot 10^{28}\cdot \pi\cdot (2.5\cdot 10^{-3})^2 \cdot 1.6\cdot 10^{-19}}=7.5\cdot 10^{-7} m/s$