Chapter 27: Q. 3 (page 762)

$1.0 \times 10^{16}$ electrons flow through a cross section of a silver wire in $320 μ s$ with a drift speed of $8.0 \times 10^{- 4} m / s$ . What is the diameter of the wire?

Short Answer

Expert verified

The diameter of the wire is $0.93 m m$ .

Step by step solution

Given Information

Number of electrons $= 1.0 \times 10^{16}$

Time $= 320 μ s$

Drift speed $= 8.0 \times 10^{- 4} m / s$

Explanation

The electron density is

$i_{e} : = \frac{N}{t}$

The current density is known to be

$i_{e} = n_{e} A v_{d} .$

The area of a circle is,

$A = \frac{π D^{2}}{4} \Rightarrow D = \sqrt{\frac{4 A}{π}} .$

Write the area as

$A = \frac{i_{e}}{n_{e} v_{d}}$

Write the electron current and substitute,

$A = \frac{N}{n_{e} t v_{d}}$

Now, substituting this expression for the area at the one for the diameter,

$D = \sqrt{\frac{4 N}{π n_{e} t v_{d}}} .$

After substituting the values,

localid="1649322696741" $D = \sqrt{(\frac{4 \times 1 \times 10^{16} e}{π \times 5.8 \times 10^{28} e \times 3.2 \times 10^{- 4} μ s \times 8 \times 10^{- 4} m / s})}$

localid="1649322704585" $= 9.26 \times 10^{- 4} m .$

Final Answer

Hence, the diameter of the wire is $9.26 \times 10^{- 4} m .$

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$1.0 \times 10^{16}$ electrons flow through a cross section of a silver wire in $320 μ s$ with a drift speed of $8.0 \times 10^{- 4} m / s$ . What is the diameter of the wire?

Short Answer

Step by step solution

Given Information

Explanation

Final Answer

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1.0×1016electrons flow through a cross section of a silver wire in 320μs with a drift speed of 8.0×10-4m/s. What is the diameter of the wire?

Short Answer

Step by step solution

Given Information

Explanation

Final Answer

One App. One Place for Learning.

Most popular questions from this chapter

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Mechanics and Materials

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Study anywhere. Anytime. Across all devices.

$1.0 \times 10^{16}$ electrons flow through a cross section of a silver wire in $320 μ s$ with a drift speed of $8.0 \times 10^{- 4} m / s$ . What is the diameter of the wire?