A copper wire of length 1, m and radius 1, mm | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Current density $J$ is defined as the current per unit cross-sectional area,given by $J = \frac{I}{A} = \frac{I}{\pi r^2}$.Since the copper wire a

Vedclass · Accepted Answer

$9:1$

Vedclass · Answer

(B) Current density $J$ is defined as the current per unit cross-sectional area,given by $J = \frac{I}{A} = \frac{I}{\pi r^2}$.Since the copper wire and the iron wire are connected in series,the same amount of current $I$ flows through both wires,i.e.,$I_{Cu} = I_{Fe} = I$.The ratio of the current density in the copper wire $(J_{Cu})$ to the iron wire $(J_{Fe})$ is given by:$\frac{J_{Cu}}{J_{Fe}} = \frac{I / (\pi r_{Cu}^2)}{I / (\pi r_{Fe}^2)} = \frac{r_{Fe}^2}{r_{Cu}^2}$.Given $r_{Cu} = 1\, mm$ and $r_{Fe} = 3\, mm$,we substitute these values:$\frac{J_{Cu}}{J_{Fe}} = \frac{(3\, mm)^2}{(1\, mm)^2} = \frac{9}{1} = 9:1$.Thus,the ratio of the current density is $9:1$.

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

Similar Questions

The unit of mobility in terms of fundamental units is . . . . . . .

The relaxation time in conductors

Derive the equation of mobility in terms of relaxation time $(\tau)$. Write its unit.

Vedclass Test Series

Exam Paper Generator

Online Exam Module