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A cylindrical wire if radius R = 2 mm is...

A cylindrical wire if radius R = 2 mm is of uniform area of cross-section. The current density through a cross-section varies with radial distance r as `J = ar^(2)`, where `a =3.2 xx 10^(11)A//m^(2)` and r is in metres. What is the current through the outer portion of the wire between radial distances `R//2` and R?

Text Solution

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Consider an element of the cylindrical wire of radius r and thickness dr shown by shaded portion with front view in figure.

Area of cross-section of the element is
`dA = 2 pir dr`
The current through the element is along the wire and elementary area `vec(dA)` is perpendicular to the cross-section of the wire. Both have the same direction. Therefore, current through the element of the wire is
`dI=vec(J).d vec(A)=J dA cos 0^(@) = JdA`
`=ar^(2)xx2 pi rdr =2pi ar^(3)dr`
The current through the outer portion of the wire is
`I = underset(R//2)overset(R)(int) 2 pia r^(3) dr = 2 pia [r^(4)/4]_(R//2)^(R) = 15/32 pi a R^(4)`
` = 15/32 xx 22/7 xx (3.2 xx 10^(11)) xx ( 2 xx 10^(-3)) = 7.54A`
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