Question

Suppose that a wire leads into another, thinner wire of the same material that has only half the cross-sectional area. In the steady state, the number of electrons per second flowing through the thick wire must be equal to the number of electrons per second flowing through the thin wire. If the electric field \({E_1}\) in the thick wire is \(1 \times 1{0^{ - 2}}\;N/C\), what is the electric field \({E_2}\) in the thinner wire?

Short answer

The electric field in the thinner wire is \(2 \times {10^{ - 2}}\;{\rm{N/C}}\).

Step-by-step solution

Given information

The cross-sectional area of the thick wire is, \({A_1}\).

The cross-sectional area of the thinner wire is, \({A_2} = \frac{{{A_1}}}{2}\).

The number of electrons per second flowing through the thick wire is, \({n_1}\).

The number of electrons per second flowing through the thinner wire is, \({n_2} = {n_1}\).

The electric field in the thick wire is, \({E_1} = 1 \times {10^{ - 2}}\;{\rm{N/C}}\).

The electric field in the thinner wire is, \({E_2}\).

Electron current in a wire

When a certain amount of current is supplied to a conducting wire, the electric current in the wire is carried by the mobile electrons.

The speed of the electrons flowing in the conducting wire changes with the current supplied, the cross-sectional area of the wire, the electric field in the wire, and the mobility of the electrons.

The electric field in the thinner wire

According to the question, the thick wire leads into another thinner wire of the same material so the same current flows through both the wires.

Then, the formula for the current flowing through the thick wire is given by,

\(\begin{array}{c}{i_1} = {i_2}\\{n_1}{A_1}{u_1}{E_1} = {n_2}{A_2}{u_2}{E_2}\end{array}\)

Putting, \({n_1} = {n_2}\), \({A_2} = \frac{{{A_1}}}{2}\), and assuming the same mobility in both wires \({u_1} = {u_2}\),

\(\begin{array}{c}{n_1}{A_1}{u_1}{E_1} = {n_1}\frac{{{A_1}}}{2}{u_1}{E_2}\\{E_2} = 2{E_1}\\{E_2} = 2\left( {1 \times {{10}^{ - 2}}\;{\rm{N/C}}} \right)\\{E_2} = 2 \times {10^{ - 2}}\;{\rm{N/C}}\end{array}\)

Hence, the electric field in the thinner wire is \(2 \times {10^{ - 2}}\;{\rm{N/C}}\).