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slava [35]
4 years ago
12

A copper wire of 1-m length and 1-mm radius has a resistance of 5.5 × 10 − 3 [Ω] and carries a current of 4 [A]. The number of f

ree electrons per unit volume in copper is 8.43 × 10 28 [m − 3 ]. Find (a) the current density, (b) the conductivity of the copper, (c) the power dissipated in the wire, (d) the (average) drift velocity of free electrons, and (e) the mobility of free electrons.
Physics
1 answer:
SOVA2 [1]4 years ago
8 0

Answer: a)1.27 *10^6 A/m^2; b) 57.9*10^6 Ω*m; c) 94.16*10^-6 m/s;

d) 4.29*10^-3 m^2/V*s

Explanation: In order to explain this problem we have to take into account the following expressions:

The current density is equal:

J=I/A  where A is the area of the wire.

J=4A/(π*1*10^-6m^2)=1.27 *10^6 A/m^2

The resistence of the wire is given by:

R=L/(σ*A) where σ and L are the conductivity and the length of the wire, respectively.

Then we have:

σ= L/(R*A)=1/(5.5*10^-3*π*1*10^-6)=57.9 *10^6 Ω*m

The drift velocity of free electrons is given by:

Vd=i/(n*A*e) where n is the numer of electrons per volume. e is the electron charge.

then we have:

Vd=J/(n*e)= 1.27 *10^6/(8.43*10^28*1.6*10^-19)=94.16 *10^-6 m/s

Finally, the mobility of free electrons is equal to:

μ=Vd/E  ;  E=J/σ

μ=E*σ/J= 94.16*10^-6*57.9 *10^6/1.27 *10^6=4.29*10^-3 m^2/V*s

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laila [671]

Answer:

2 seconds

Explanation:

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For maxima and minima, put the value of dh / dt is equal to zero. we get

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Answer:

Acceleration of the proton will be equal to 4.79\times 10^{14}m/sec^2

Explanation:

We have given electric field E=5\times 10^6N/C

Mass of proton is equal to m=1.67\times 10^{-27}kg

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So acceleration of the proton will be equal to 4.79\times 10^{14}m/sec^2

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