The diagram above shows two wires; wire 1 and wire 2. The charge carriers in wir

Question

The diagram above shows two wires; wire 1 and wire 2. The charge carriers in wire 1 (of circular cross section and radius R) have a drift speed down the wire that is not constant across the wire. Instead, the drift speed rises linearly from zero at the circumference (r = R) to v_0 at the center (r = 0), according to v_d(r) = v_0(l-r/R). A second wire (wire 2) has the same radius, the same density of charge carriers and a constant drift speed given by Vd(r) = f*v_0. Evaluate the ratio of the current carried by wire 1 to the current carried by wire 2, when f = 0.600.

Explanation / Answer

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 = 2r×dr

Hence current in the wire will be integration of ne×(2r×dr)×Vd

Or current1 = ne×2{(R2 /2 )- R2/3}Vo = (ne R2)Vo/3

For second wire

Vd = 0.600Vo

Current 2= neR2×0.60Vo

Hence ratio of current = current1/current2 = 5/9

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