Question
The total momentum of electrons in a straight wire of length $$1000\,m$$ carrying a current of $$70A$$ is closest to
A.
$$40 \times {10^{ - 8}}N - \sec $$
B.
$$30 \times {10^{ - 8}}N - \sec $$
C.
$$50 \times {10^{ - 8}}N - \sec $$
D.
$$70 \times {10^{ - 8}}N - \sec $$
Answer :
$$40 \times {10^{ - 8}}N - \sec $$
Solution :
No. of electron in the wire $$ = nA\ell \left( {n = {e^ - }{\text{density}}} \right)$$ and momentum $$\left( {nA\ell } \right){m_e}{v_d}$$
$$\eqalign{
& = neA{v_d}\frac{{\ell {m_e}}}{e} = \frac{{I\ell {m_e}}}{e} \cr
& = \frac{{70 \times 1000}}{{1.6 \times {{10}^{ - 19}}}} \times 9.1 \times {10^{ - 31}} \cr
& = 40 \times {10^{ - 8}}N - \sec \cr} $$