A copper wire of resistivity 2.6 × 10-3 Ωm, has a cross sectional area of 30 × 10-4 cm3. Calculate the length of this wire required to make a 10 Ω coil.
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Answers
Answered by
8
Answer:
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Explanation:
Given,
R=10Ω
A=10×10
−8
m
2
ρ=1.6×10
−8
Ω/m
We know that,
R=ρ
A
l
l=
ρ
RA
=
1.6×10
−8
10×10×10
−8
l=62.5m
The length of wire copper wire is 62.5m
Answered by
20
Answer:
1.15 x 10^-3 m
Explanation:
R=ρL/A
R=resistance of the coil
ρ=resistivity of the wire
L=length of the wire
A=cross sectional area of the wire
given,
ρ=2.6 x 10^-3 Ωm , A= 30 x 10^-4cm^2=30 x 10^-8 m^2,R=10Ω
10=2.6 x 10^-3 x L/30 x 10^-8
L= 10 x 30x 10^-8/2.6 x 10^-3=1.15 x 10^-3 m
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