Question

A current of4. 8ampear is flowing ina copper wire of cross section area 3×10-4m2 find the current density in the wire​

Answer 1

Answer:2MM/SEC

Explanation: 1=5A,A=5×10−6m2,n=5×1026/m3

I=neAvd

vd=ineA=55×1026×1.6×10−19×5×10−6

=180m/s=0.0125m/sec

=1.25×10−2m/sec

(b) (i) I=neAvd

vd=ineA=1102210−6×1.6×10−19×0.01×10−4

=116×10−2=0.0625×10−2m/sec

=6.25×10−4m/sec

(ii) j=σE(Ohm's law)

E=jσ=i/A1/ρ=ρiA=1.6×10−8×10.01×10−4

=1.6×10−2volt/m

( c) Volume V=lA=1×10−6m3=10−6m3

Mass = Density × Volume

=5×103×10−6=5×10−3kg=5g

Number of atoms =MassAt.wt× Avogadro's no.

=560×6×1023=5×1022

Number of free electrons =5×1022

Number of free electrons/volume

n=5×102210−6=5×1028

i=neAvd

vd=ineA=165×1028×1.6×10−19×10−6

=2×10−3m/sec

=2mm/sec.