Two wires each of radius of cross section $r$ but of different materials are connected together end to end (in series). If the densities of charge carriers in the two wires are in the ration $1 : 4$ , the drift velocity of electrons in the two wires will be in the ratio:

Question

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Answer 1

Let $n_{1}$ and $n_{2}$ be the densities of charge carriers in the two wires, $V_{d_{1}}$ and $V_{d_{2}}$ be the drift velocities of charge carriers in the two wires,
$A$ be the area of cross section,
$e$ is the charge on electron and
$I$ is the current flowing through both the wires
Hence, $I = n_{1} e A V_{d_{1}} = n_{1} e A V_{d_{2}}$
$∴ n_{1} V_{d_{1}} = n_{2} V_{d_{2}}$
$∴ \frac{V _{d_{1}}}{V _{d_{2}}} = \frac{n _{2}}{n _{1}}$
$∴ \frac{V _{d_{1}}}{V _{d_{2}}} = \frac{4}{1}$

Answer 2

Let

n_{1}

and

n_{2}

be the densities of charge carriers in the two wires,

V_{d_{1}}

and

V_{d_{2}}

be the drift velocities of charge carriers in the two wires,

A

be the area of cross section,

e

is the charge on electron and

I

is the current flowing through both the wires
Hence,

I = n_{1} e A V_{d_{1}} = n_{1} e A V_{d_{2}}

∴ n_{1} V_{d_{1}} = n_{2} V_{d_{2}}

∴ \frac{V _{d_{1}}}{V _{d_{2}}} = \frac{n _{2}}{n _{1}}

∴ \frac{V _{d_{1}}}{V _{d_{2}}} = \frac{4}{1}