A wire of radius 0.15 mm is made of an unknown metal and carries a current of 8.4 mA.The drift velocity for electrons in the wire is measured to be 36 $\times$ 10^-6 m/s.What is the carrier density of the metal from which the wire is made? A) 1.3 $\times$ 10²⁹ m^-3 B) 4.2 $\times$ 10²⁸ m^-3 C) need more information to do this calculation. D) 4.2 $\times$ 10³⁰ m^-3

Question

Quizplus · Accepted Answer

The carrier density (n) can be calculated using the equation:

n = I / (Aqvd)

Where I is the current, A is the cross-sectional area of the wire, q is the charge on an electron, and vd is the drift velocity of electrons.

Cross-sectional area of the wire can be calculated using the formula:

A = πr2

Where r is the radius of the wire.

Substituting the given values, we get:

A = 3.14 x (0.15 x 10^-3)2
  = 7.07 x 10^-8 m2

n = (8.4 x 10^-3)/(7.07 x 10^-8 x 1.6 x 10^-19 x 36 x 10^-6)
  = 4.2 x 10^28 m^-3

Therefore, the answer is option B.

A Wire of Radius 0 ×\times× 10-6 M/s ×\times× 1029 M-3 B) 4

A Wire of Radius 0 $\times$ 10^-6 M/s $\times$ 10²⁹ M^-3
B) 4