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The starter motor of a car engine draws a current of 170 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.930 s until the car engine starts How much charge passes through the starter motor?

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lestat345

Answer:

The charge that passes through the starter motor is [tex]\Delta Q=158.1 C[/tex].

Explanation:

Known Data

  • Avogadro's Number [tex]N_{A}=6.02x10^{23}[/tex]
  • Current, [tex]I=170A=170\frac{C}{s}[/tex]
  • Charge in an electron, [tex]q=1.60x10^{-19}C[/tex]
  • Time, [tex]\Delta t=0.930s[/tex]
  • Diameter, [tex]d=4.60mm=0.0046m[/tex]
  • Transversal Area, [tex]A=(\frac{d}{2})^{2} \pi=(\frac{0.0046m}{2})^{2} \pi=1.66x10^{-5} m^{2}[/tex]
  • Volume, [tex]V=Length*A=(1.2m)(1.66x10^{-5} m^{2})=1.99x10^{-5} m^{3}[/tex]

First Step: Find the number of the electrons per unit of volume in the wire

We use the formula [tex]n=\frac{N_{A}}{V}= \frac{6.02x10^{23} electrons}{1.99x10^{-5} m^{3}} =3.02x10^{28}el/ m^{3}[/tex].

Second Step: Find the drag velocity

We can use the following formula [tex]v_{d}=\frac{I}{nqA}=\frac{170C/s}{(3.02x10^{28}m^{-3})(1.60x10^{-19}C)(1.66x10^{-5} m^{2})}  =2.11x10^{-3} m/s[/tex]

Finally, we use the formula [tex]\Delta Q=(nAv_{d}\Delta t)q=(3.02x10^{28} m^{-3})(1.66x10^{-5} m^{2})(2.11x10^{-3} m/s)(0.930s)(1.60x10^{-19}C)=158.1 C[/tex].

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