Question

A wire with conductivity s carries current $I$. The current is increasing at the rate $dI/dt.$

a. Show that there is a displacement current in the wire equal to $(\in o/{\sigma }_{})(dI/dt)$

b. Evaluate the displacement current for a copper wire in which the current is increasing at $1.0\times {10}^{6}A/s.$

Short answer

a) The displacement current in the wire is equal to $I_d=\in o{\frac{d\varnothing e_{}}{dt}}_{}$

b) The displacement current in wire in which current is increasing at $1.0\times {10}^{6}A/s$is$I_d=1.48\times {10}^{-13}A$.

Step-by-step solution

Part(a) Step1: Given information

We have given that In the wire the current density $\underset{J}{\to }$is related to the electric field localid="1649944213928" $\underset{J}{\to }=\sigma \underset{E}{\to }$, Integrating over a cross section of the wire , we have $I=\sigma {\varphi }_E$,

so ${\varphi }_E=\frac{I}{\sigma }$

$\frac{d{\varphi }_E}{dt}=\frac{1}{\sigma }\frac{dI}{dt}$

Part(a) Step2: Simplify

The displacement of current $I_d$is:

$I_d=\in o_{}\frac{d{\varphi }_e}{dt}$

$I_d={\frac{\in o_{}}{\sigma }}_{}\frac{dI}{dt}$$\{{\varphi }_E=\frac{I}{\sigma }\}$

Part(b) Step1: Given information

We have given that For copper, the vacuum permittivity is $\in o=8.85\times {10}^{-12}{\frac{F}{m}}_{}$and $\sigma =5.95\times {10}^7\frac{S}{m}$

we calculate values for $I_d$;

Part(b) Step2: Explanation

$I_d={\frac{\in o_{}}{\sigma }}_{}\frac{dI}{dt}$

$I_d=\frac{8.85\times {10}^{-12}}{5.95\times {10}^7}1.0\times {10}^6$

$I_d=1.45\times {10}^{-13}{A}_{}$