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A copper wire has a square cross-section 2.3 mm on a side. The wire is 4.0 m long and carries a current of 3.6 A. The density of free electrons is8.5×1028/m3. Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

Short Answer

Expert verified
  1. Current density in the wire isJ=6.8×105A/m2
  2. Electric field in the wire isE=11.69×10-3V/m
  3. Time required by the electronτ=22.22hr

Step by step solution

01

Step 1:

Given data:

A=l2=2.3×10-3m2=5.3×10-6m2L=4.0ml=3.6A

Density of free electrons=8.5×1028/m3

(a) Current density in the wire is:

J=IA=3.6A5.3×10-6m2=6.8×105A/m2

Therefore, the current density in the wire isJ=6.8×105A/m2.

02

Step 2:

(b) Electric field in the wire is:

As the copper resistivity isρ=1.72×10-8Ω·m

So electric field is:

E=ρJ=1.72×10-8Ω·m6.8×105A/m2=0.01169V/m=11.69×10-3V/m

Hence, the electric field in the wire isE=11.69×10-3V/m.

03

Step 3:

(c) Time required to travel the length of the wire:

From the equation,

τ=Lvd

Here vdis drift speed;

vd=Jnq

So, the time required is:

τ=Lvd=L·nqJ=4.0m×8.5×1028×-1.6×10-19C6.8×105A/m2=80000s=1333.33min=22.22hr

Therefore, the time required by the electronτ=22.22hr.

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