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

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Quantum Physics

Kinematics Physics

Classical Mechanics

Solid State Physics

Atoms and Radioactivity

Magnetism

Study anywhere. Anytime. Across all devices.

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

Quantum Physics

Kinematics Physics

Classical Mechanics

Solid State Physics

Atoms and Radioactivity

Magnetism

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.