(1.) Here,
Cross- sectional area A=1.25×10−6m2
Drift velocity vd=2×10−4 m/s
Carrier Charge Density n=1.0\times 10^{29} \ \ \m3
Using the formula,
Current flowing through a copper wire
I=neAvd
=(1.0×1029)×(1.6×10−19)×(1.25×10−6)×(2×10−4)
=4 Amperes
Hence, Current flowing through a copper wire is 4 A
(2.) (a) The cross-sectional area of the wire is
A=πr2=π(0.321×10−3m)2=3.24×10−7 m2
and the resistivity of Nichrome is 1.5×10−6 Ω⋅m
So, the resistivity per unit length
lR=Aρ=3.24×10−7 m21.5×10−6 Ω⋅m=4.6 Ω/m
(b) Current in wire I=RΔV=4.6Ω10V=2.2A
(3.) R(150°C)=2.415Ω=R(0°C)(1+α(150−0)) ......(i)
R(20°C)=1.642 Ω=R(0°C)(1+α(20−0)) .....(ii)
Substitute first equation in second and solve for α
On Solving, we get
α=3.9×10−3 °C−1
and R(0°C)=1.5236 Ω
With α known, use any of the above equations to get
R(0°C)=1.5236Ω . Then use R(0oC)=ρ(0°C)L/A
1.5236Ω=0.25π(2×10−3m)2ρ(0°C)⋅300m⇒ρ(0°C)=1.596×10−8 Ωm
and ρ(20°C)=ρ(0°C)(1+α⋅20)=1.72×10−8 Ωm
Comments
Thank you very much for helping.