Question

A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the ends. (a) What is the current in amperes in the wire? (b) What is the magnitude of the current density? (c) Calculate the resistivity of the material of which the wire is made.

Answer 1

(a) Current is 2831.93 A

(b)
`8.40A/m^2`

(c)
`\rho =15.52* 10^(-9)ohm-m`

Step-by-step explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

`A=\pi r^2=3.14* 0.00366^2=4.20* 10^(-5)m^2`

Resistance
`R=11.9mohm=11.9* 10^(-3)ohm`

Potential difference V = 33.7 volt

(A) current is equal to

`i=(V)/(R)=(33.7)/(11.9* 10^(-3))=2831.93A`

(B) Current density is equal to

`J=(i)/(A)`

`J=(2831.93)/(4.20* 10^(-5))=8.40A/m^2`

(c) Resistance is equal to

`R=(\rho l)/(A)`

`11.9* 10^(-3)=(\rho * 3.22)/(4.20* 10^(-5))`

`\rho =15.52* 10^(-9)ohm-m`