The current density in a cylindrical wire of radius 4   mm is 4 × 10 6   A   m - 2 . The current through the outer portion of the wire between radial distances R 2 and R is _____ π   A .

Question

The current density in a cylindrical wire of radius 4 &#160; mm is 4 &#215; 10 6 &#160; A &#160; m - 2 . The current through the outer portion of the wire between radial distances R 2 and R is _____ &#960; &#160; A .

Aicharya · Accepted Answer

Current density is defined as current per unit cross-section area. Therefore,

$i = J A = J \times π (R^{2} - \frac{R^{2}}{4})$

$\Rightarrow i = 4 \times 10^{6} \times π \times 16 \times 10^{- 6} \times \frac{3}{4}$

$\Rightarrow i = 48 π A$

The current density in a cylindrical wire of radius 4   mm is 4 × 10 6   A   m - 2 . The…

Solution

The current density in a cylindrical wire of radius 4 &#160; mm is 4 &#215; 10 6 &#160; A &#160; m - 2 . The…

Solution

The current density in a cylindrical wire of radius 4 mm is 4 × 10 6 A m - 2 . The…