Question

Suppose that wire A and wire B are made of different metals and are subjected to the same electric field in two different circuits. Wire B has the 6 times the cross sectional area, 1.3 times as many mobile electrons per cubic centimetre and 4 times the mobility of wire A. In the steady state \({\bf{2 \times 1}}{{\bf{0}}^{{\bf{18}}}}\) electrons enters wire A every second. How many electrons enter wire B every second?

Short answer

The number of electrons enter in wire B per second are \(6.24 \times {10^{19}}\;{\rm{electrons}}/{\rm{s}}\).

Step-by-step solution

Identification of given data

The electric field for circuits of wire A and wire B is\({E_A} = {E_B}\)

The cross sectional area of the wire B is\({A_B} = 6{A_A}\)

The number of electrons per cubic centimetre for wire B is\({n_B} = 1.3{n_A}\)

The mobility of wire B is\({\mu _B} = 4{\mu _A}\)

The number of electrons enters in wire A per second is\({I_A} = 2 \times {10^{18}}\;{\rm{electrons}}/{\rm{s}}\)

The number of electrons entering in wire B per second is calculated by equating the electric field for both metal wires.

Determination of expression to find the electrons enter in wire B per second

The electric field for both metals are same so,

\(\begin{array}{c}{E_A} = {E_B}\\\frac{{{V_A}}}{{{L_A}}} = \frac{{{V_B}}}{{{L_B}}}\\\frac{{{I_A}{R_A}}}{{{L_A}}} = \frac{{{I_B}{R_B}}}{{{L_B}}}\\\frac{{{I_A}}}{{{L_A}}}\left( {\frac{{{L_A}}}{{{\sigma _A}{A_A}}}} \right) = \frac{{{I_B}}}{{{L_B}}}\left( {\frac{{{L_B}}}{{{\sigma _B}{A_B}}}} \right)\end{array}\)

\(\begin{array}{c}\frac{{{I_A}}}{{{L_A}}}\left( {\frac{{{L_A}}}{{{n_A}{\mu _A}{A_A}}}} \right) = \frac{{{I_B}}}{{{L_B}}}\left( {\frac{{{L_B}}}{{{n_B}{\mu _B}{A_B}}}} \right)\\\frac{{{I_A}}}{{{n_A}{\mu _A}{A_A}}} = \frac{{{I_B}}}{{{n_B}{\mu _B}{A_B}}}\end{array}\)

Determination of electrons enter in wire B per second

Substitute all the values in above equation.

\(\begin{array}{c}\frac{{\left( {2 \times {{10}^{18}}\;{\rm{electrons}}/{\rm{s}}} \right)}}{{{n_A}{\mu _A}{A_A}}} = \frac{{{I_B}}}{{\left( {1.3{n_A}} \right)\left( {4{\mu _A}} \right)\left( {6{A_A}} \right)}}\\{I_B} = 6.24 \times {10^{19}}\;{\rm{electrons}}/{\rm{s}}\end{array}\)

Therefore, the number of electrons enter in wire B per second are \(6.24 \times {10^{19}}\;{\rm{electrons}}/{\rm{s}}\).