First slide

current density

Question

A copper wire of length 1 m and radius 1 mm is joined in series with an iron wire of length Z 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

Easy

A

18 : 1
B

9 : 1
C

6 : 1
D

2 : 3

Solution

Current density $J = \frac{i}{A} = \frac{i}{{πr}^{2}} \Rightarrow \frac{J_{1}}{J_{2}} = \frac{i_{1}}{i_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}$
But the wires are in series, so they have the same
current, hence $i_{1} = i_{2} . So \frac{J_{1}}{J_{2}} = \frac{r_{2}^{2}}{r_{1}^{2}} = 9 : 1$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App