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Feliz [49]
3 years ago
14

A length of copper wire carries a current of 14 A, uniformly distributed through its cross section. The wire diameter is 2.5 mm,

and its resistance per unit length is 3.3 /km.
a) What is the energy density of the magnetic field at the surface of the wire in J/m^3 ?

b)Of the electric field?
Physics
1 answer:
gavmur [86]3 years ago
4 0

Answer:

a. ρ_\beta=1.996J/m^3

b. U_E=9.445x10^{-15} J/m^3

Explanation:

a. To find the density of magnetic field given use the gauss law and the equation:

i=14A, d=2.5mm, R=3.3Ω, l=1 km, E_o=8.85x10^{-12}F/m, u_o=4*x10^{-7}H/m

ρ_\beta=\frac{\beta^2}{2*u_o}

ρ_\beta=\frac{1}{2*u_o}*(\frac{u_o*i^2}{2\pi *r})^2

ρ_\beta=\frac{u_o*i^2}{8\pi*r}=\frac{4\pi *10^{-7}H/m*(14A)^2}{8\pi*(1.25x10^{-3}m)^2}

ρ_\beta=1.996J/m^3

b. The electric field can be find using the equation:

U_E=\frac{1}{2}*E_o*E^2

E=(\frac{i*R}{l})^2

U_E=\frac{1}{2}*8.85x10^{-12}*(\frac{14A*3.3}{1000m})^2

U_E=9.445x10^{-15} J/m^3

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