# For the wire, calculate the following, assuming that the wire carries a current of 4.9 A. (a) the...

## Question:

For the wire , calculate the following, assuming that the wire carries a current of {eq}4.9 \ A {/eq}.

(a) the current density

(b) the electric field in the wire

## Current density:

The ratio of the amount of current that flows within a conductor through unit area is called as current density. This quantity is vector and is represented by unit Ampere per square meter.

Given data

• The current in the wire is {eq}I = 4.9\;{\rm{A}} {/eq}.

The expression for the Electric field is given as,

{eq}E = \dfrac{J}{\sigma }......\left( 1 \right) {/eq}

Here, J is the current density and {eq}\sigma {/eq} is the conductivity.

The expression for the current density is given as,

{eq}J = \dfrac{I}{A}......\left( 2 \right) {/eq}

Here, I is the current and A is the area of cross section.

(a)

Substituting the values in equation 2,

{eq}\begin{align*} J &= \dfrac{{4.9}}{A}\\ J &= \dfrac{{4.9}}{A}\;{\rm{A/}}{{\rm{m}}^{\rm{2}}} \end{align*} {/eq}

Thus, the current density is {eq}\dfrac{{4.9}}{A}\;{\rm{A/}}{{\rm{m}}^{\rm{2}}} {/eq}.

(b)

Substituting the values in equation 1,

{eq}\begin{align*} E &= \dfrac{{\dfrac{{4.9}}{A}}}{\sigma }\\ E &= \dfrac{{4.9}}{{A\sigma }}\;{\rm{N/C}} \end{align*} {/eq}

Thus, the electric field in the wire is {eq}\dfrac{{4.9}}{{A\sigma }}\;{\rm{N/C}} {/eq}.