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Question: Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper...

Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper having 1 mm diameter, then the drift velocity (approx.) will be (Density of copper =9×103 kg m−3= 9 \times 10 ^ { 3 } \mathrm {~kg} \mathrm {~m} ^ { - 3 } and atomic weight = 63)

A

0.3 mm/sec0.3 \mathrm {~mm} / \mathrm { sec }

B

0.1 mm/sec0.1 \mathrm {~mm} / \mathrm { sec }

C

D

0.2 cm/sec0.2 \mathrm {~cm} / \mathrm { sec }

Answer

0.1 mm/sec0.1 \mathrm {~mm} / \mathrm { sec }

Explanation

Solution

Density of Cu=9×103 kg/m3C u = 9 \times 10 ^ { 3 } \mathrm {~kg} / \mathrm { m } ^ { 3 } (mass of 1 m3 of Cu)

6.0 × 1023 atoms has a mass = 63 × 10–3 kg

∴ Number of electrons per m3 are

=6.0×102363×10−3×9×103=8.5×1028= \frac { 6.0 \times 10 ^ { 23 } } { 63 \times 10 ^ { - 3 } } \times 9 \times 10 ^ { 3 } = 8.5 \times 10 ^ { 28 }

Now drift velocity =vd=ineA= v _ { d } = \frac { i } { n e A }

=1.18.5×1028×1.6×10−19×π×(0.5×10−3)2= \frac { 1.1 } { 8.5 \times 10 ^ { 28 } \times 1.6 \times 10 ^ { - 19 } \times \pi \times \left( 0.5 \times 10 ^ { - 3 } \right) ^ { 2 } }

=0.1×10−3 m/sec= 0.1 \times 10 ^ { - 3 } \mathrm {~m} / \mathrm { sec }