Acopper 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 free 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.

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

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation:

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