Question

An aluminium wire carrying a current has diameter 0.84 mm. The electric field in the wire is 0.49$V/m$. What is,
(a) the current carried by the wire?
(b) the potential difference between two points in the wire 12.0 m apart?

(c) the resistance of a 12.0 m length of this wire?

Specific resistance of aluminum is $2.75\times {10}^8\Omega -m$.

Answer 1

Given : $D=0.84$ mm $=8.4\times {10}^{-4}$ m $E=0.49$ V/m $L=12$ m $\rho =2.75\times {10}^{-8}$ $\Omega$ $m$

Cross-section area of the wire $A=\frac{\pi D^2}{4}=\frac{\pi (8.4\times {10}^{-4}{)}^2}{4}=5.54\times {10}^{-7}{m}^2$

(a) : Current in the wire $I=\frac{AE}{\rho }=\frac{5.54\times {10}^{-7}\times 0.49}{2.75\times {10}^{-8}}=9.9$ A

(b) : Potential difference $V=EL=0.49\times 12=5.88$ volts

(c) : Resistance $R=\rho \frac{L}{A}=2.75\times {10}^{-8}\times \frac{12}{5.54\times {10}^{-7}}=0.60\Omega$