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Virty [35]
3 years ago
13

A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the

density of free electrons in the metal?Express your answer numerically in m−3 to two significant figures.
Physics
2 answers:
podryga [215]3 years ago
3 0

Answer:

6.9\times 10^{28}m^{-3}

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=rr=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m

1mm=10^{-3} m

Current=I=8 A

Drift velocity=v_d=5.4\times 10^{-5} m/s

We have to find the density of free electrons in the metal

We know that

Density of electron=n=\frac{I}{v_deA}

Using the formula

Density of free electrons=\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}

By using Area of wire=\pi r^2

\pi=3.14\\e=1.6\times 10^{-19} C

Density of free electrons=6.9\times 10^{28}m^{-3}

Furkat [3]3 years ago
3 0

Explanation:

We will calculate the density as follows.

               n = \frac{i}{v_{d}eA}

As the metallic wire is in the shape of a circle. So, its area will be calculated as follows.

                 Area = \pi \times r^{2}

                          = 3.14 \times (2.06 \times 10^{-3})^{2}

                v_{d} = 5.40 \times 10^{-5}

                        i = 8 A

Hence, we will calculate the density of free electrons as follows.

              n = \frac{i}{v_{d}eA}

                 = \frac{8}{5.40 \times 10^{-5} \times 3.14 \times (2.06 \times 10^{-3})^{2} \times 1.60 \times 10^{-19}}

                  = 6.950 \times 10^{28} m^{-3}

Thus, we can conclude that the density of free electrons in the metal is 6.950 \times 10^{28} m^{-3}.

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