A 70.0-cm length of a cylindrical silver wire with a radius of 0.175 mm is extended horizontally between two leads. The potential at the left end of the wire is 3.15 V, and the potential at the right end is zero. The resistivity of silver is 1.586 ✕ 10−8 Ω · m.
a. What are the magnitude and direction of the electric field in the wire?
b. What is the resistance of the wire?
c. What are the magnitude and direction of the current in the wire?
d. What is the current density in the wire?

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onyebuchinnaji onyebuchinnaji · Answer 1 · 2020-03-04T08:34:32+01:00

Answer:

(a) the magnitude and direction of the electric field in the wire is 4.5 N/C towards the left end of the wire.

(b) the resistance of the wire is 0.1154 Ω

(c) the magnitude and direction of the current in the wire is 27.3 A towards the left end of the wire

(d) the current density in the wire is 4.053 x 10⁸ A/m³

Explanation:

Given;

Length of cylinder = 70cm = 0.7m

radius of cylinder = 1.75 x 10⁻⁴ m

potential V = 3.15 V

resistivity = 1.586 ✕ 10⁻⁸ Ω · m

Part (a) the magnitude and direction of the electric field in the wire

V = E x d

E = V/d

E = 3.15/0.7 = 4.5 N/C towards the left end of the wire.

Part (b) the resistance of the wire

[tex]R = \frac{\rho L}{A} \\\\R =\frac{\rho L}{\pi r^2} =R = \frac{1.586X10^{-8} X 0.7}{\pi (1.75X10^{-4})^2} = 0.1154 ohms[/tex]

R = 0.1154 Ω

Part (c) the magnitude and direction of the current in the wire

V = IR

I =V/R

I = 3.15/0.1154

I = 27.3 A towards the left end of the wire

Part (d) the current density in the wire

current density,J = current /volume

volume = πr²h = π x (1.75 × 10⁻⁴)² x 0.7 = 6.7357 x 10⁻⁸ m³

[tex]J = \frac{27.3}{6.7357 X10^{-8}} = 4.053 X10^8 \frac{A}{m^3}[/tex]

J = 4.053 x 10⁸ A/m³