(B) The drift velocity $v_d$ is given by the formula $v_d = \frac{eE\tau}{m}$,where $E = \frac{V}{l}$.Substituting $E$,we get $v_d = \frac{eV\tau}{ml}

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(B) The drift velocity $v_d$ is given by the formula $v_d = \frac{eE\tau}{m}$,where $E = \frac{V}{l}$.Substituting $E$,we get $v_d = \frac{eV\tau}{ml}$.Since the potential difference $V$ is constant,$v_d \propto \frac{1}{l}$.If the length $l$ is increased to $4l$,the new drift velocity $v_{d'}$ becomes $v_{d'} = \frac{eV\tau}{m(4l)} = \frac{v_d}{4}$.Therefore,the drift velocity decreases by $4$ times.

Class 12 Physics Current Electricity Current Density, Drift Velocity and Mobility

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$A$ constant potential difference is applied between the ends of a wire. If the length of the wire is elongated to $4$ times its original length,then the drift velocity of electrons will be:

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A
increases $4$ times
B
decreases $4$ times
C
increases $2$ times
D
decreases $2$ times

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Current Electricity

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Current Density, Drift Velocity and Mobility

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$A$ charged particle having a drift velocity of $7.5 \times 10^{-4} \, m s^{-1}$ in an electric field of $3 \times 10^{-10} \, V m^{-1}$ has a mobility in $m^{2} V^{-1} s^{-1}$ of:

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$A$ current $I$ is passing through a wire having two sections $P$ and $Q$ of uniform diameters $d$ and $d/2$ respectively. If the mean drift velocity of electrons in sections $P$ and $Q$ is denoted by $v_P$ and $v_Q$ respectively,then

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$A$ constant potential difference is applied between the ends of a wire. If the length of the wire is elongated to $4$ times its original length,then the drift velocity of electrons will be:

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$A$ charged particle having a drift velocity of $7.5 \times 10^{-4} \, m s^{-1}$ in an electric field of $3 \times 10^{-10} \, V m^{-1}$ has a mobility in $m^{2} V^{-1} s^{-1}$ of:

The drift velocity of electrons in a conducting wire connected to a cell is $V_{d}$. If the length of the wire is doubled and the area of cross-section is halved,then the drift velocity of electrons becomes:

The drift velocity of electrons in a conducting wire is of the order of $1 \, mm/s$. However,when the switch is turned on,the bulb glows almost instantaneously. This is because:

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$A$ current $I$ is passing through a wire having two sections $P$ and $Q$ of uniform diameters $d$ and $d/2$ respectively. If the mean drift velocity of electrons in sections $P$ and $Q$ is denoted by $v_P$ and $v_Q$ respectively,then

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