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For a metal that has an electrical conductivity of 7.1 x 107 (Ω-m)-1, do the following: (a) Calculate the resistance (in Ω) of a wire 2.6 mm in diameter and 6.7 m long. (b) Calculate the current (in A) if the potential drop across the ends of the wire is 0.060 V. (c) Calculate the current density (in A/m^2). (d) Compute the magnitude of the electric field (in V/m) across the ends of the wire.

User EelkeSpaak
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1 Answer

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Answer:

(a) 0.0178 Ω

(b) 3.4 A

(c) 6.4 x 10⁵ A/m²

(d) 9.01 x 10⁻³ V/m

Step-by-step explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

A = 5.3 x 10⁻⁶ m²

L = length of the wire = 6.7 m

Resistance of the wire is given as


R=(L)/(A\sigma )


R=(6.7)/((5.3*10^(-6))(7.1*10^(7)) )

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as


i = (V)/(R)


i = (0.060)/(0.0178)

i = 3.4 A

(c)

Current density is given as


J = (i)/(A)


J = (3.4)/(5.3*10^(-6))

J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as


E = (J)/(\sigma )


E = (6.4 * 10^(5))/( 7.1 * 10^(7))

E = 9.01 x 10⁻³ V/m

User Bartosss
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