Chapter 21, Problem 85GP

To determine

(a) To determine:

To find the large value for electric and magnetic fields.

Answer to Problem 85GP

Solution:

The energy density for each field is $3.6\times {10}^9$.

Explanation of Solution

Given info:

$\text{V/m}$ is volt per meter, ${\text{U}}_{\text{B}}$ is energy volume of magnetic field for B and E is electric field.

Formula used:

${\text{U}}_{\text{B}}\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{B}}^{\text{2}}}{{\text{μ}}_{\text{0}}}$

Calculation:

Magnetic field $B=2T$

Electric filed ${\text{E=1×10}}^{\text{4}}\text{V/m}$

Magnetic energy density ${\text{U}}_{\text{B}}\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{B}}^{\text{2}}}{{\text{μ}}_{\text{0}}}$

$\begin{array}{l}{\text{U}}_{\text{B}}\text{=}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{1}}{{\text{4π×10}}^{\text{-7}}}{\text{×2}}^{\text{2}}{\text{J/m}}^{\text{3}}\\ {\text{U}}_{\text{B}}\text{=1}{\text{.6×10}}^{\text{6}}{\text{J/m}}^{\text{3}}\end{array}$

Electric energy density ${\text{U}}_{\text{E}}\text{=}\frac{\text{1}}{\text{2}}{\text{ε}}_{\text{0}}{\text{E}}^{\text{2}}$

$\begin{array}{l}\text{=}\frac{\text{1}}{\text{2}}\text{×8}{\text{.85×10}}^{\text{-12}}\text{×}\left({\text{1×10}}^{\text{4}}\right){\text{J/m}}^{\text{3}}\\ \text{=4}{\text{.4×10}}^{\text{-4}}{\text{J/m}}^{\text{3}}\end{array}$

The ratio

$\begin{array}{l}\frac{{\text{U}}_{\text{B}}}{{\text{U}}_{\text{E}}}\text{=}\frac{\text{1}{\text{.592×10}}^{\text{6}}{\text{J/m}}^{\text{3}}}{\text{4}{\text{.425×10}}^{\text{-4}}{\text{J/m}}^{\text{3}}}\\ \text{=3}{\text{.6×10}}^{\text{9}}\end{array}$

To determine

(b) To determine:

To find the large value for electric and magnetic fields.

Answer to Problem 85GP

Solution:

a)The energy density for each field is $3.6\times {10}^9$.

b)The magnitude electric field would be needed to produce the same energy density as the 2.0 T magnetic is $6\times {10}^8\text{ V/m}$.

Explanation of Solution

Given info:

$\text{V/m}$ is volt per meter, ${\text{U}}_{\text{B}}$ is energy volume of magnetic field for B and E is electric field.

Formula used:

$\frac{\text{1}}{\text{2}}{\text{ε}}_{\text{0}}{\text{E}}^{\text{2}}\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{B}}^{\text{2}}}{{\text{μ}}_{\text{0}}}$

Calculation:

The electric field required to deliver the some vitality thickness as a 2T attractive field is

$\begin{array}{l}\frac{\text{1}}{\text{2}}{\text{ε}}_{\text{0}}{\text{E}}^{\text{2}}\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{B}}^{\text{2}}}{{\text{μ}}_{\text{0}}}\\ {\text{E}}^{\text{2}}\text{=}\frac{{\text{B}}^{\text{2}}}{{\text{ε}}_{\text{0}}{\text{μ}}_{\text{0}}}\\ {\text{E}}^{\text{2}}{\text{=c}}^{\text{2}}{\text{B}}^{\text{2}}\\ \text{E=}\left({\text{3×10}}^{\text{8}}\right)\left(\text{2}\right)\text{ V/m}\\ {\text{E=6×10}}^{\text{8}}\text{V/m}\end{array}$