A copper wire of length `1 m` and radius is joined in series with an iron wire of length `2 m` and radius `3 mm` and a current is passed through the wire. The ratio of the current density in the copper and iron wires is

Options

A. `18 : 1`
B. `9 : 1`
C. `6 : 1`
D. `2 : 3`

Answer

B

Explanation

(b) Current density `J = (i)/(A) = (i)/(pi r^(2)) implies (J_(1))/(J_(2)) = (i_(1))/(i_(2)) xx (r_(2)^(2))/(r_(1)^(2))`
But the wires are in series,so they have the same
Curren hence `i_(1) = i_(2)`. So `(J_(1))/(J_(2)) = (r_(2)^(2))/(r_(1)^(2)) = 9 : 1`