In a straight conductor of uniform cross-section charge q is flowing for time t. Let s be the specific charge of an electron. The momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is:

\frac{q}{t s}

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(\frac{q}{t s})^{2}

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\sqrt{\frac{q}{t s}}

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q t s

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Solution

The correct option is A $\frac{q}{t s}$
$\Rightarrow I = \frac{q}{t}$
Total mass flowing $= m_{e} . n . A = m_{t}$

$V_{d} = \frac{I}{n e A} = \frac{q}{n e A t}$

Momentum $= m_{t} V_{d}$
$= m e n A (\frac{q}{n e A t})$
$= \frac{q}{t (\frac{e}{m e})} = \frac{q}{t s}$

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In a straight conductor of uniform cross-section charge q is flowing for time t. Let s be the specific charge of an electron. The momentum of all the free electrons per unit length of the conductor, due to their drift velocity only is:

The correct option is A qts⇒I=qtTotal mass flowing =me.n.A=mtVd=IneA=qneAtMomentum =mtVd =menA(qneAt) =qt(eme)=qts

The correct option is A $\frac{q}{t s}$
$\Rightarrow I = \frac{q}{t}$
Total mass flowing $= m_{e} . n . A = m_{t}$

$V_{d} = \frac{I}{n e A} = \frac{q}{n e A t}$

Momentum $= m_{t} V_{d}$
$= m e n A (\frac{q}{n e A t})$
$= \frac{q}{t (\frac{e}{m e})} = \frac{q}{t s}$