# Answer to Question #73947 in Electromagnetism for Ayrill

Question #73947

2. A wire 100cm long and 2mm in diameter has a resistivity of 4.8x10

-8Ωm.

(i) Calculate the resistance of the wire?

(ii) A second wire of the same material has the same weight as the 100m length, but twice its diametric. Evaluate its resistance.

-8Ωm.

(i) Calculate the resistance of the wire?

(ii) A second wire of the same material has the same weight as the 100m length, but twice its diametric. Evaluate its resistance.

Expert's answer

The electrical resistance R is defined as:

R=ρ l/A

where

R is the electrical resistance of a uniform specimen of the material (measured in ohms, Ω)

ρ is the electrical resistivity (Ωm)

l is the length of the piece of material (measured in metres, m)

A is the cross-sectional area of the specimen (measured in square metres, m2).

(i)

A_1=(πd^2)/4=π×(2×〖10〗^(-3) m)^2/4=3.14×〖10〗^(-6) m^2

R_1=(4.8×〖10〗^(-8) Ωm) (1 m)/(3.14×〖10〗^(-6) m^2 )=0.015 Ω

(ii)

l_2=100 m;

A_2=(πd_2^2)/4=π×(4×〖10〗^(-3) m)^2/4=12.57×〖10〗^(-6) m^2

R_2=(4.8×〖10〗^(-8) Ωm) 100/(12.57×〖10〗^(-6) m^2 )=0.38 Ω

Answer: (i) 0.015 Ω; (ii) 0.38 Ω.

R=ρ l/A

where

R is the electrical resistance of a uniform specimen of the material (measured in ohms, Ω)

ρ is the electrical resistivity (Ωm)

l is the length of the piece of material (measured in metres, m)

A is the cross-sectional area of the specimen (measured in square metres, m2).

(i)

A_1=(πd^2)/4=π×(2×〖10〗^(-3) m)^2/4=3.14×〖10〗^(-6) m^2

R_1=(4.8×〖10〗^(-8) Ωm) (1 m)/(3.14×〖10〗^(-6) m^2 )=0.015 Ω

(ii)

l_2=100 m;

A_2=(πd_2^2)/4=π×(4×〖10〗^(-3) m)^2/4=12.57×〖10〗^(-6) m^2

R_2=(4.8×〖10〗^(-8) Ωm) 100/(12.57×〖10〗^(-6) m^2 )=0.38 Ω

Answer: (i) 0.015 Ω; (ii) 0.38 Ω.

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