Radius of both wires are equal and it is = R
For first wire
Vd = V0 (1 - r/R)
Since drift velocity is variable
Consider a small ring of thickness dr at distance r from the centre
Then the area of the ring will be = 2πr×dr
Hence current in the wire will be integration of ne×(2πr×dr)×Vd
Or current1 = ne×2π{(R2 /2 )- R2/3}Vo = (ne πR2)Vo/3
For second wire
Vd = 0.600Vo
Current 2= neπR2×0.60Vo
Hence ratio of current = current1/current2 = 5/9
The diagram above shows two wires; wire 1 and wire 2. The charge carriers in wire...
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