Question

Determine the current density in an aluminium wire of length 2.00m and radius 0.650 cm the potential difference between the ends of the wire is 60.0 v​

Answer 1

Answer:

The current density in an aluminum wire of length 2.00m and radius 0.650 cm the potential difference between the ends of the wire is 60.0 v is $1.132075\times10^{9} A/m^{2}$

Explanation:

Current density in an aluminum wire  , $J=\frac{I}{A}$

         Where ,

              I - current flowing through the wire

              $I=\left(\frac{V}{R}\right)$,     V is the potential difference between the wire and R is the resistance of the wire.  

              A - Area of cross section of the wire

               $\text { A } =\pi r^{2}$  , r is the radius of the given wire.

             

Given:

      Length of the aluminum wire, l =  2.00 m

       Radius of the wire, r = 0.650 cm = 0.0065 m

       Potential difference between the ends of the wire, V = 60.0 v

   We know,

       Resistivity of aluminum , $\rho=2.65 \times 10^{-8} \Omega \mathrm{m}$  

       Resistance of the wire ,    $R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^{2}}$

       Substituting for 'I' and then for 'R' in the equation for 'J'.

              $\begin{aligned}J &=\frac{I}{A} \\&=\left(\frac{V}{R}\right) \frac{1}{(A)} \\&=\frac{V}{\left(\frac{\rho l}{A}\right) A} \\&=\frac{V}{\rho l}\end{aligned} \\ = \frac{60}{2.65\times10^{-8}\times 2}= 1132075471.7= 1.132075\times10^{9} A/m^{2}$