Question

If a current­-carrying wire has a cross­-sectional area that gradually becomes smaller along the length of the wire, the drift velocity: a. increases along the length of the wire if the resistance increases too b. decreases along the length of the wire if the resistance decreases too c. decreases along the length of the wire d. increases along the length of the wire e. remains the same along the length of the wire

Answer 1

Concepts and reason

The concept required to solve this problem is Ohms law and combination of resistors in series and parallel.

First calculate the total charge in the conductor by using charge density and volume of the conductor. Calculate the drift velocity of the electrons by using current, cross sectional area of the conductor, number of electrons and charge of the electron and compare with area of the conductor.

Fundamentals

The rate of change of the charge flow in the conductor is called current.

$I = \frac{Q}{t}$

Here, $Q$is the charge and t is the time.

The drift velocity of the electron is,

${v_d} = \frac{L}{t}$

Here, L is the distance moved by electron in a time t.

Rearrange the above equation for time t.

$t = \frac{L}{{{v_d}}}$

The charge density in a conductor is,

$\rho = \frac{Q}{V}$

Here, Q is the charge and V is the volume of the conductor.

Rearrange the above equation for Q.

$Q = \rho V$

The charge density of the conductor is equal to product of number of electrons n and charge of the electron.

$\rho = ne$

Here, n is the number of electrons and e is the charge of the electron.

Substitute $ne$for $\rho$in the equation$Q = \rho V$.

$Q = neV$

Substitute $neV$for$Q$and $\frac{L}{{{v_d}}}$for$t$in the equation$I = \frac{Q}{t}$.

$\begin{array}{c}\\I = \frac{{neV}}{{\frac{L}{{{v_d}}}}}\\\\ = \frac{{neV{v_d}}}{L}\\\end{array}$

The volume of the conductor is,

$V = AL$

Here, A is the cross sectional area of the conductor and L is the length of the conductor.

Substitute AL for V in the equation$I = \frac{{neV{v_d}}}{L}$

$\begin{array}{c}\\I = \frac{{ne\left( {AL} \right){v_d}}}{L}\\\\ = neA{v_d}\\\end{array}$

Rearrange the above equation for drift velocity.

${v_d} = \frac{I}{{neA}}$

The charge density of the conductor is equal to product of number of electrons n and charge of the electron.

$\rho = ne$

Here, n is the number of electrons and e is the charge of the electron.

The total charge is,

$Q = \rho V$

Substitute $ne$for $\rho$in the above equation.

$Q = neV$

The drift velocity is,

${v_d} = \frac{I}{{neA}}$

Here, I is the current passing through the conductor, n is the number of electron in the conductor, e is the charge of the electron and A is the cross sectional area of the conductor.

The drift velocity and the cross sectional area is inversely proportional to the each other.

So, if the cross sectional area is gradually becomes smaller along the length of the wire then the drift velocity is increases along the length of the wire.

Hence, the answer is increases along the length of the wire.

Therefore, the correct option is d.

Ans:

The answer is increases along the length of the wire.