Question

Consider a silver wire (σ = 6.3 ✕ 107 Ω⁻¹ · m⁻¹) with a cross-sectional area of 1 mm2 (similar to your connecting wires) and carrying 0.3 amperes of current, which is about what you get in a circuit with a round bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.

Answer 1

Answer:

E=4.76 N/C

Explanation:

According to macroscopic ohm's law:

$J=\sigma*E\\where:\\J=Current\_density\\\sigma=Conductivity\\E=Electric\_field$

$J=\frac{I}{A_o}\\J=\frac{0.3A}{1*10^{-6}m^2}\\\\J=0.3*10^6A/m$

rearranging the formula:

$E=\frac{J}{\sigma}\\\\E=\frac{0.3*10^6A/m^2}{6.3*10^{7}}\\\\E=4.76*10^{-3}N/C$

Answer 2

The Electric Field Strength is calculated to be; E = 0.00476 N/C

What is the strength of the Electric Field?

We are given;

Conductivity; σ = 6.3 × 10⁷ Ω⁻¹.m⁻¹

Current; I = 0.3 A

Area; A = 1 mm² = 10⁻⁶ m²

Formula for current density is;

J = I/A

Thus;

J = 0.3/10⁻⁶

J = 300000 A/m

Now, to get the electric field strength we will use the formula;

E = J/σ

E = 300000/(6.3 × 10⁷)

E = 0.00476 N/C

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