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The total momentum of electrons in a straight wire of length 5 m carrying a current of 4 A is

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$11.375 \times 10^{- 12} k g m / s$

$227.5 \times 10^{- 12} k g m / s$

$32 \times 10^{- 12} k g m / s$

$61.5 \times 10^{- 12} k g m / s$

answer is A.

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Detailed Solution

$i = n e A v_{d}$

Momentum of electron = $m v_{d}$

Total number of electrons = $n A l$

Total momentum of electrons = $(n A l) m v_{d}$

$= m n A l v_{d} = \frac{m l i}{e}$

$= \frac{9.1 \times 10^{- 31} \times 5 \times 4}{1.6 \times 10^{- 19}}$

$= 113.75 \times 10^{- 12} k g m / s$

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