# Current-carrying gold wire has a diameter of 0.83 mm. The electric field in the wire is 0.49 V/m....

## Question:

Current-carrying gold wire has a diameter of 0.83 mm. The electric field in the wire is 0.49 V/m. What is the current carried by the wire?

## Electric Current

The rate at which the electrons experience a shift due to the movement in a circuit is called an electric current. This movement of charge is a result of the potential created at two ends. Current is obtained as a ratio of the value of charge to the time taken for that flow.

## Answer and Explanation:

**Given data**

- The diameter of gold wire is: {eq}D = 0.83\;{\rm{mm}} {/eq}.

- The electric field in wire is: {eq}E = 0.49\;{\rm{V/m}} {/eq}.

The current carried by the wire is given by the following relation.

{eq}I = \dfrac{{E \times a}}{{{\rho _G}}} \cdots\cdots\rm{(I)} {/eq}

Here, the resitivity of gold is {eq}{\rho _G} {/eq} and the area of cross section of wire is {eq}a {/eq}.

The resistivity of gold is {eq}2.44 \times {10^{ - 8}}\;{\rm{\Omega }}{\rm{.m}} {/eq}.

The area of cross section of wire is computed as,

{eq}a = \dfrac{{\pi {D^2}}}{4} {/eq}

Substitute the values in Equation (I).

{eq}\begin{align*} I &= \dfrac{{E \times \left( {\dfrac{{\pi {D^2}}}{4}} \right)}}{{{\rho _G}}}\\ &= \dfrac{{\left( {0.49\;{\rm{V/m}}} \right) \times \left( {\dfrac{{\pi {{\left( {0.83\;{\rm{mm}}\left( {\dfrac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}} \right)} \right)}^2}}}{4}} \right)}}{{2.44 \times {{10}^{ - 8}}\;{\rm{\Omega }}{\rm{.m}}}}\\ &\approx 10.87\;{\rm{A}} \end{align*} {/eq}

Thus, the current carried by the wire is approximately {eq}10.87\;{\rm{A}} {/eq}.

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