Chapter 30: Q71P (page 900)

A length of copper wire carries a current of 10 A uniformly distributed through its cross section. (a) Calculate the energy density of the magnetic field. (b) Calculate the energy density of the electric field at the surface of the wire. The wire diameter is 2.5 mm, and its resistance per unit length is $3.3 Ω / km$ .

Short Answer

Expert verified

a) Energy density of the magnetic field at the surface of the wire is $1.0 J / m^{3}$ .

b) Energy density of the electric field at the surface of the wire is $4.8 \times 10^{- 15} J / m^{3}$ .

Step by step solution

Given

$I = 10 A d = 2.5 mm = 2.5 \times 10^{- 3} m$

Resistance per unit length $\frac{R}{I} = 3.3 \frac{Ω}{km} = 3.3 \times 10^{- 3} Ω / m$

Understanding the concept

We need to use the formula for the energy density for the magnetic field and the electric field at the surface of the wire. Using these equations and the given data, we can calculate the energy density for both, the electric field and the magnetic field.

Formula:

$u_{B} = \frac{B^{2}}{2 μ_{0}} B = \frac{μ_{0} I}{2 R} u_{E} = \frac{1}{2} ε E^{2}$

Calculate the energy density of the magnetic field at the surface of the wire .

We have,

$u_{B} = \frac{B^{2}}{2 μ_{0}} But, B = \frac{μ_{0} I}{2 R} So u_{B} = \frac{1}{2 μ_{0}} \times (\frac{μ_{0} I}{2 R}) u_{B} = \frac{μ_{0} I^{2}}{2 R^{2}} u_{B} = \frac{4 π \times 10^{- 7} \times 10^{2}}{8 \times (\frac{2.5 \times 10^{- 3}}{2})} u_{B} = 1.0 J / m^{3}$

(b) Calculate Energy density of the electric field at the surface of the wire.

We have,

$u_{E} = \frac{1}{2} ε E^{2} But, E = ρJ u_{E} = \frac{1}{2} ε {(ρJ)}^{2} But, J = I / A u_{E} = \frac{1}{2} ε {(\frac{ρ I}{A})}^{2}$

We multiply and divide inside the bracket by

$u_{E} = \frac{1}{2} ε {(\frac{ρ I I}{A I})}^{2}$

But, we know that,

$R = \frac{ρ I}{A} u_{E} = \frac{ε}{2} {(\frac{R I}{I})}^{2} u_{E} = \frac{8.85 \times 10^{- 12}}{2} \times {(10 \times 3.3 \times 10^{- 3})}^{2} u_{E} = 4.8 \times 10^{- 15} J / m^{3}$