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A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduction electron density in silver is 5.8 × 1028 electrons/m3 and e = 1.60 × 10-19 c. what is the drift velocity of these electrons?

2 months ago

Solution 1

Guest #4326

2 months ago

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

The current I is any motion of charge from one region to another, so this is given by:

$I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$

Finally, for the drift velocity magnitude vd, we find:

$v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

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A light source emits a beam of photons, each of which has a momentum of 2.7 × 10-29 kg·m/s. (a) what is the frequency of the photons? (b) to what region of the electromagnetic spectrum do the photons belong?

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The frequency of the photons is equal to 1.22 ×10¹³ Hz and lies in the infrared region of the electromagnetic spectrum.

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A gravitational field vector points toward the earth; an electric field vector points toward an electron. why do electric field vectors point away from protons? 1. protons have much larger mass. 2. unlike electrons, protons will produce an electric field of their own. 3. protons have more net charge than electrons. 4. protons are positively charged. 5. none of these

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So the intensity is inversely proportional to the square of the distance from the source

$\rm I\approx \dfrac{1}{r^2}$

Intensity level (LI) can be formulated

$\rm LI=10\:log\dfrac{I}{I_o}$

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For the level of intensity of several sound sources as many as n pieces can be formulated:

LIn = LI1 + 10 log n

The intensity level of 1 baby is

$\rm LI=10\:log\dfrac{8.10^{-8}}{10^{-12}}$

LI = 10 log 8.10⁴

LI = 49

The intensity level of 5 babies :

LI5 = LI + 10 log n

LI5 = 49 + 10 log 5

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