Question

The drift velocity of free electrons in a conductor is $v$, when a current. $I$ is flowing in it if both the radius and current are doubled, then drift velocity will be.

• A. $v/4$
• B. $v/2$
• C. $2v$
• D. $4v$

Answer 1

The correct option is B $v/2$

$\text{Step1 : Relation between current and drift velocity}$

We know that,

$I=neAv$

where, $A$ is the cross-sectional area

$n$ is the no. of electrons.

$e$ is the charge of electrons.

$\therefore v=\frac{I}{neA}$ $=\frac{I}{ne\pi r^2}$

$\Rightarrow v\propto \frac{I}{r^2}$ $......\left(1\right)$

$\text{Step-2: New drift velocity}$

Let, New drift velocity $=v^{\prime }$,

New radius, $r^{\prime }=2r$

New current, $I^{\prime }=2I$

From equation $\left(1\right)$

$\frac{v^{\prime }}{v}=\frac{\left(I^{\prime }/r^{\prime 2}\right)}{\left(I/r^2\right)}=\frac{I^{\prime }}{I}\times {\left(\frac{r}{r^{\prime }}\right)}^2$

$\Rightarrow \frac{v^{\prime }}{v}=\frac{2I}{I}\times {\left(\frac{r}{2r}\right)}^2$

$\therefore v^{\prime }=\frac{v}{2}$

$\therefore$ New drift velocity is $\frac{v}{2}$

Hence, Option B is correct.