Question

27.A 0.50-mm-diameter silver wire carries a 20 mA current.What are (a) the electric field and (b) the electron drift speed in the wire?

Answer 1

Concepts and reason

The concepts used to solve this problem are electron drift velocity and electric field.

Initially, the electric field can be calculated in terms of resistivity and current density. By replacing the current density with the current per unit area cross section, electric field is calculated.

To calculate the flow of free electrons with different velocity and directions in a material drift, velocity is used. Then drift velocity is calculated by dividing the current and electron density, charge, and area.

Fundamentals

Electric field is the strength created around a material when the current or electron flows.

The force experienced by the other charged particle due to the field around a charge particle is called electric field.

$E=pJ $

Here, is the Resistivity of silver, is the Current density, and is the Electric field.

Expression for current density is,

Here, J is the current density, is the current, and A is the area of cross section.

Use the expression for J to rewrite the expression for E.

Free electrons flow in all possible directions with different velocities. The result of the whole random motion of electrons is called drift velocity.

Here, is the drift velocity of the electron, is the charge of an electron, is the electron density, and is the Cross-sectional area.

(a)

The expression for the electric field is,

$6)(O)=2 $

Rewrite the expression in terms of diameter for area.

Here, d is the diameter of the wire.

Convert the value of I from mA to A.

Convert the value of d from mm to m.

Substitute for , for , and for d to find E.

(b)

The expression for the drift velocity is,

Rewrite the expression in terms of diameter.

Substitute for I, for , for , and for d to find .

Ans: Part a

The electric field of silver wire is.

Part b

The drift velocity of silver wire is.