Chapter 2: Q46P (page 45)

Consider a silver wire with a cross-sectional area of $1 {mm}^{2}$ carrying $0.3 A$ of current. The conductivity of silver is $6.3 \times 10^{7} \frac{(A / m^{2})}{(V / m)}$ . Calculate the magnitude of the electric field required to drive this current through the wire.

Short Answer

Expert verified

$4.76 \times 10^{- 3} V / m$

Step by step solution

Given Data

$Area A = 1 {mm}^{2} Current I = 0.3 A Conductivity σ = 6.3 \times 10^{7} \frac{(A / m^{2})}{(V / m)}$

Concept

The force as per the test charge is known as the magnitude of electric field.

Calculate the magnitude of the electric field

First we have to find the resistivity,

$\begin{array}{rcl} σ & = & \frac{1}{ρ} (where, ρ is resistivity) \\ ρ & = & \frac{1}{σ} \\ = & \frac{1}{6.3 \times 10^{7}} \\ = & 1.58 \times 10^{- 8} Ωm \end{array}$

Electric Field,

localid="1662180919419" $\begin{matrix} E = ρ \frac{1}{A} (where E is electric field) \\ = 1.58 \times 10^{- 8} \times \frac{0.3 A}{1 \times 10^{- 6} m^{2}} \\ = 4.76 \times 10^{- 3} V / m \end{matrix}$

Hence, the magnitude of the electric field required to drive this current through the wire is $4.76 \times 10^{- 3} V / m$

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Consider a silver wire with a cross-sectional area of $1 {mm}^{2}$ carrying $0.3 A$ of current. The conductivity of silver is $6.3 \times 10^{7} \frac{(A / m^{2})}{(V / m)}$ . Calculate the magnitude of the electric field required to drive this current through the wire.

Short Answer

Step by step solution

Given Data

Concept

Calculate the magnitude of the electric field

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Consider a silver wire with a cross-sectional area of 1mm2carrying0.3Aof current. The conductivity of silver is6.3×107(A/m2)(V/m). Calculate the magnitude of the electric field required to drive this current through the wire.

Short Answer

Step by step solution

Given Data

Concept

Calculate the magnitude of the electric field

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Radiation

Fluids

Rotational Dynamics

Measurements

Conservation of Energy and Momentum

Engineering Physics

Study anywhere. Anytime. Across all devices.

Consider a silver wire with a cross-sectional area of $1 {mm}^{2}$ carrying $0.3 A$ of current. The conductivity of silver is $6.3 \times 10^{7} \frac{(A / m^{2})}{(V / m)}$ . Calculate the magnitude of the electric field required to drive this current through the wire.