A copper wire of resistivity - 1-6-10-8- ohms metre has a cross-sectional area of - 10- 10-4cm-2- Calculate the length of this wire required to make a - 10- ohm coil-

Given, R=10Ω A=10× 10^ 8m^2 ρ=1.6× 10^ 8Ω /m We know that, R=ρlA l=RAρ=10× 10× 10^ 81.6× 10^ 8 l=62.5 m The length of wire copper wi ...

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Standard XII

Physics

Ohm's Law in Scalar Form

A copper wire...

Question

A copper wire of resistivity $1.6 \times 10^{- 8}$ ohms metre has a cross-sectional area of $10 \times 10^{- 4} c m^{2}$ . Calculate the length of this wire required to make a $10$ ohm coil.

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Solution

Given,

$R = 10 Ω$

$A = 10 \times 10^{- 8} m^{2}$

$ρ = 1.6 \times 10^{- 8} Ω / m$

We know that,

$R = ρ \frac{l}{A}$

$l = \frac{R A}{ρ} = \frac{10 \times 10 \times 10^{- 8}}{1.6 \times 10^{- 8}}$

$l = 62.5 m$

The length of wire copper wire is $62.5 m$

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A copper wire of resistivity 1.6×10−8 ohms metre has a cross-sectional area of 10×10−4cm2. Calculate the length of this wire required to make a 10 ohm coil.

Given,R=10ΩA=10×10−8m2ρ=1.6×10−8Ω/mWe know that,R=ρlAl=RAρ=10×10×10−81.6×10−8l=62.5mThe length of wire copper wire is 62.5m

A copper wire of resistivity $1.6 \times 10^{- 8}$ ohms metre has a cross-sectional area of $10 \times 10^{- 4} c m^{2}$ . Calculate the length of this wire required to make a $10$ ohm coil.

Given,

$R = 10 Ω$

$A = 10 \times 10^{- 8} m^{2}$

$ρ = 1.6 \times 10^{- 8} Ω / m$

We know that,

$R = ρ \frac{l}{A}$

$l = \frac{R A}{ρ} = \frac{10 \times 10 \times 10^{- 8}}{1.6 \times 10^{- 8}}$

$l = 62.5 m$

The length of wire copper wire is $62.5 m$