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vovangra [49]
3 years ago
11

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.
Physics
1 answer:
Scorpion4ik [409]3 years ago
4 0

Answer:

(a) Current is 2831.93 A

(b) 8.40A/m^2

(c) \rho =15.52\times 10^{-9}ohm-m

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\times 0.00366^2=4.20\times 10^{-5}m^2

Resistance R=11.9mohm=11.9\times 10^{-3}ohm

Potential difference V = 33.7 volt

(A) current is equal to

i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A

(B) Current density is equal to

J=\frac{i}{A}

J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2

(c) Resistance is equal to

R=\frac{\rho l}{A}

11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}

\rho =15.52\times 10^{-9}ohm-m

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