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An electric cell of emf $E$ is connected across a copper wire of diameter $d$ and length $l$. The drift velocity of electrons in the wire is $v_{d}$. If the length of the wire is changed to $2 l,$ the new drift velocity of electrons in the copper wire will be
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Verified Answer
The correct answer is:
$v_{d} / 2$
The drift velocity is given as
$$
\begin{aligned}
v_{d} &=\frac{i}{n e A}=\frac{E}{R \times n e A} \\
&=\frac{E \times A}{\rho l \times n e A}=\frac{E}{ln e}
\end{aligned}
$$
When length of wire changed to 2l , the new drift velocity.
$$
\begin{aligned}
v_{d}^{\prime} &=\frac{E}{\rho \times 2 l \times n e} \\
\therefore \quad & \frac{v_{d}^{\prime}}{v_{d}}=\frac{E / \rho 2 \text { lne }}{E / \rho \text { lne }}=\frac{1}{2} \\
\Rightarrow \quad v_{d}^{\prime} &=\frac{v_{d}}{2}
\end{aligned}
$$
$$
\begin{aligned}
v_{d} &=\frac{i}{n e A}=\frac{E}{R \times n e A} \\
&=\frac{E \times A}{\rho l \times n e A}=\frac{E}{ln e}
\end{aligned}
$$
When length of wire changed to 2l , the new drift velocity.
$$
\begin{aligned}
v_{d}^{\prime} &=\frac{E}{\rho \times 2 l \times n e} \\
\therefore \quad & \frac{v_{d}^{\prime}}{v_{d}}=\frac{E / \rho 2 \text { lne }}{E / \rho \text { lne }}=\frac{1}{2} \\
\Rightarrow \quad v_{d}^{\prime} &=\frac{v_{d}}{2}
\end{aligned}
$$
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