Question

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Current Density and Drift Speed A group of charges, each with charge q, moves with velocity v^vector. The number of particles per unit volume is n. What is the current density J^vector of these charges, in magnitude and direction? Make sure that your answer has units of A/m^2. We want to calculate how long it takes an electron to get from a car battery to the starter motor after the ignition switch is turned. Assume that the current flowing is115 A, and that the electrons travel through copper wire with cross-sectional area 31.2 mm^2 and length 85.5 cm. What is the current density in the wire? The number density of the conduction electrons in copper is 8.49 times 10^28/m^3. Given this number density and the current density, what is the drift speed of the electrons? How long does it take for an electron starting at the battery to reach the starter motor? [Ans: 3.69 times 10^6 A/m^2, 2.71times 10^- 4 m/s, 52.5 min.]

Answer 1

(a) The magnitude of current density is given by

       $\left | \vec{J} \right |=\left | nq\vec{v} \right |$

The direction of $\vec{J}$ and $\vec{v}$ is same for positive charge carriers and opposite for negative charge carriers.

(b) Given:

    I=115A, A=31.2mm²,   L=85.5cm, n=8.49*10²⁸/m³

   The drift velocity of the electron is given by

    $v=\frac{I}{neA}$

$\Rightarrow v=\frac{115}{8.49*10^{28}*1.6*10^{-19}*31.2*10^{-6}}=2.71*10^{-4}m/s$

The current density in the wire is

   $J=nqv=8.49*10^{28}*1.6*10^{-19}*2.71*10^{-4}=3.69*10^{6}A/m^{2}$

The time taken is

$t=\frac{L}{v}=\frac{85.5*10^{-2}}{2.71*10^{-4}}=3.15*10^{3}s=52.58min$