Sure, let's break down each part of the question:

(a) To find the density of mobile charges in a piece of copper, we need to find the number of free electrons per unit volume. We know that each copper atom contributes one free electron. The number of atoms per unit volume can be found by dividing the density of copper by the atomic mass and then multiplying by Avogadro's number.

So, the number of free electrons per unit volume (n) is given by:

n = δ * NA / M
n = 9 g/cm³ * 6 × 10²³ mol⁻¹ / 64 g/mol
n = 8.4375 × 10²² cm⁻³

The density of mobile charges (ρ) is then given by multiplying the number of free electrons per unit volume by the charge of an electron:

ρ = n * e
ρ = 8.4375 × 10²² cm⁻³ * 1.6 × 10⁻¹⁹ C
ρ = 1.35 × 10⁴ C/cm³

(b) The average electron velocity (v) in a copper wire can be found using the formula for current (I), which is the charge (Q) divided by the time (t). In this case, the charge is the number of free electrons times the charge of an electron, and the time is the length of the wire divided by the velocity.

So, v = Q / (I * t)
v = (n * e) / (I * (L / v))
v = √(I / (n * e * π * (d/2)²))
v = √(1 A / (8.4375 × 10²² cm⁻³ * 1.6 × 10⁻¹⁹ C * π * (0.1 cm)²))
v = 6.93 × 10⁻⁵ cm/s

(c) The force of attraction between two such wires can be found using Ampere's law, which states that the magnetic field (B) around a current-carrying wire is given by μ₀I / (2πr), where μ₀ is the permeability of free space. The force (F) is then given by B * I * L, where L is the length of the wire.

So, F = B * I * L
F = (μ₀I / (2πr)) * I * L
F = (4π × 10⁻⁷ T m/A * 1 A) / (2π * 0.01 m) * 1 A * 1 m
F = 2 × 10⁻⁷ N

(d) If you could somehow remove the stationary positive charges, the electrical repulsion force would be given by Coulomb's law, which states that the force between two charges is given by kQ₁Q₂ / r², where k is Coulomb's constant. In this case, the charges are the number of free electrons times the charge of an electron.

So, F = kQ₁Q₂ / r²
F = (9 × 10⁹ N m²/C² * (n * e)²) / (0.01 m)²
F = 2.16 × 10⁹ N

The electrical repulsion force is then 2.16 × 10⁹ N / 2 × 10⁻⁷ N = 1.08 × 10¹⁶ times greater than the magnetic force.

Question

Sure, let's break down each part of the question:

(a) To find the density of mobile charges in a piece of copper, we need to find the number of free electrons per unit volume. We know that each copper atom contributes one free electron. The number of atoms per unit volume can be found by dividing the density of copper by the atomic mass and then multiplying by Avogadro's number.

So, the number of free electrons per unit volume (n) is given by:

n = δ * NA / M
n = 9 g/cm³ * 6 × 10²³ mol⁻¹ / 64 g/mol
n = 8.4375 × 10²² cm⁻³

The density of mobile charges (ρ) is then given by multiplying the number of free electrons per unit volume by the charge of an electron:

ρ = n * e
ρ = 8.4375 × 10²² cm⁻³ * 1.6 × 10⁻¹⁹ C
ρ = 1.35 × 10⁴ C/cm³

(b) The average electron velocity (v) in a copper wire can be found using the formula for current (I), which is the charge (Q) divided by the time (t). In this case, the charge is the number of free electrons times the charge of an electron, and the time is the length of the wire divided by the velocity.

So, v = Q / (I * t)
v = (n * e) / (I * (L / v))
v = √(I / (n * e * π * (d/2)²))
v = √(1 A / (8.4375 × 10²² cm⁻³ * 1.6 × 10⁻¹⁹ C * π * (0.1 cm)²))
v = 6.93 × 10⁻⁵ cm/s

(c) The force of attraction between two such wires can be found using Ampere's law, which states that the magnetic field (B) around a current-carrying wire is given by μ₀I / (2πr), where μ₀ is the permeability of free space. The force (F) is then given by B * I * L, where L is the length of the wire.

So, F = B * I * L
F = (μ₀I / (2πr)) * I * L
F = (4π × 10⁻⁷ T m/A * 1 A) / (2π * 0.01 m) * 1 A * 1 m
F = 2 × 10⁻⁷ N

(d) If you could somehow remove the stationary positive charges, the electrical repulsion force would be given by Coulomb's law, which states that the force between two charges is given by kQ₁Q₂ / r², where k is Coulomb's constant. In this case, the charges are the number of free electrons times the charge of an electron.

So, F = kQ₁Q₂ / r²
F = (9 × 10⁹ N m²/C² * (n * e)²) / (0.01 m)²
F = 2.16 × 10⁹ N

The electrical repulsion force is then 2.16 × 10⁹ N / 2 × 10⁻⁷ N = 1.08 × 10¹⁶ times greater than the magnetic force.

Knowee AI · Accepted Answer