Question

Copper has 8.0 x 10²⁸ electrons per cubic metre carrying a current and lying at right angle to a magnetic field of strength 5 x 10^-3 T. experiences a force of 8.0 x 10^-2 N. Calculate the drift velocity of free electrons in the wire.

Answer 1

n = 8 x 10²⁸ m^-3 ; l = lm 

A= 8 x 10²⁸ m^-2 ; e = 1.6 x 10^-9C

Total charge contained in the wire 

q = volume of wire x ne = Alne = 8 x 10^-6 x 1 x 8 x10²⁸ x 1.6 x 10^-19 C

= 102.4 x 10³ C

If v_d is the drift speed of electrons, then

F = qv_d B sin 90° = qv_d B

v_d = 1.56 x 10 ms^-1