Current Electricity Charging Discharging Of Capacitors Question 366
Question: 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
[MP PMT 1994]
Options:
A) 18 : 1
B) 9 : 1
C) 6 : 1
D) 2 : 3
Show Answer
Answer:
Correct Answer: B
Solution:
Current density $ J=\frac{i}{A}=\frac{i}{\pi r^{2}} $
Therefore $ \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 $
BETA
