Current Electricity Charging Discharging Of Capacitors Question 677
Question: The current density varies with radial distance r as $ j=ar^{2} $ , in a cylindrical wire of radius R. The current passing through the wire between radial distance R/3 and R/2 is
Options:
A) $ \frac{65\pi aR^{4}}{2592} $
B) $ \frac{25\pi aR^{4}}{72} $
C) $ \frac{65\pi a^{2}R^{3}}{2938} $
D) $ \frac{81\pi a^{2}R^{4}}{144} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Given; $ J=ar^{2} $
$ i=\int _{1}^{2}{J\times 2\pi rdr}=\int _{R/3}^{R/2}{ar^{2}\times 2\pi rdr} $
$ =2\pi a\int _{R/3}^{R/2}{r^{3}dr}=2\pi a| \frac{r^{4}}{4} | _{R/3}^{R/2} $
$ =\frac{\pi a}{2}[ {{( \frac{R}{2} )}^{2}}-{{( \frac{R}{3} )}^{4}} ] $
$ =\frac{\pi aR^{4}}{2}\times \frac{65}{81\times 16}=\frac{65\pi aR^{4}}{2592} $
BETA
