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 × 10⁻⁶ m/s. What is the charge carrier density of the metal from which the wire is made? A) 1.3 × 10²⁹ m⁻³ B) 2.1 × 10²⁸ m⁻³ C) need more information to do this calculation. D) 4.2 × 10³⁰ m⁻³

Question

Quizplus · Accepted Answer

The charge carrier density $n$ can be calculated using the formula:
$$n = \frac{I}{Aq v_d}$$ 
where $I$ is the current, $A$ is the cross-sectional area of the wire, $q$ is the charge of an electron, and $v_d$ is the drift velocity of the electrons. 
The cross-sectional area of the wire is given by:
$$A = \pi r^2 = \pi (0.15 	imes 10^{-3})^2 = 7.07 	imes 10^{-8} \, 	ext{m}^2$$ 
The charge of an electron is $q = -1.6 	imes 10^{-19} \, 	ext{C}$. Substituting the given values, we get:
$$n = \frac{(8.4 	imes 10^{-3})}{(7.07 	imes 10^{-8})(-1.6 	imes 10^{-19})(36 	imes 10^{-6})} = 2.1 	imes 10^{28} \, 	ext{m}^{-3}$$ 
Therefore, the answer is (B) 2.1 × 10^28 m^-3.

A Wire of Radius 0