Question

Consider a block of copper.

a. Find the density of mobile charges in a piece of copper, assuming each atom contributes two free electrons.

b. Calculate the average electron velocity in a copper wire 1.00mm in diameter which carries a current of 1.00A.

c. What is the force of attraction between two such wires, 1.00cm apart?

d. If you could somehow remove the stationary positive ions, what would the

electrical repulsion force be?

e. How many times greater is the electrical force in part (d) than the

magnetic force in part (c)?

Answer 1

(a)

charge density is found as

_e = q / m

where

is mass density, q is charge, m can be written as N / m

where N is avogadro's number and m is molar mass

for copper, m = 64 g / mol

so,

_e = 9 * 1.6e-19 * 6.022e23 / 64

_e = 1.4e4 C/cm³ ( mind the units, please !!! )

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(b)

we find current density

J = I / A

where A = pi * d² / 4 = 7.85e-7 m²

so,

J = 1 / 7.85e-7

J = 1.27e6 A/m²

so,

v = J /

v = 9.1e-3 cm / s ( mind the units )

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(c)

F / L = u_o I1 I2 / 2 r

F / L = 4e-7 * 1 / 2 * 1

F/L = 2e-7 N/cm

____________________________

(d)

electric force is written as

F = k₁₂ / 2d

where is charge density and is equal = I / v

so,

F = k * I₁ * I₂ / 2dv²

this can be written as

F = I₁ * I₂ / 2_o v²d

and

1/_o = u_o c²

sp,

combining above two equations, we have

F = u_o c² I₁ * I₂ / 2 v²d

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(e)

compare it with magnetic force

F / F (mag) = c² / v²

F / F (mag) = 3e10² / 9.1e-3²

F / F (mag) = 1.1e25

so,

electric force is 1.1e25 times greater than magnetic force.