**This circuit is made up of two wires and two batteries. All components are connected in series such that the positive terminal of battery # 2 is electrically connected to the negative terminal of battery # 1.****A steady current flows through this circuit.****Each battery has an emf of $ 1.3 $ volts****Each wire has a length and diameter of $ 26 \ cm $ and $ 0.0007 \ m $ respectively.****The wire material (metal) contains $ 7 \times 10^{+28} $ mobile electrons per cubic meter.****The electron mobility has a value of $ 5 \times 10^{-5} \ (m/s) (m/V) $**

The aim of this question is to understand the **flow of electrons** in a metallic wire **under the influence of some electric field**.

The electric field is generated by the emf of the batteries. Therefore the **potential gradient formula** of the electric field strength can be used which is defined as:

**\[ E = \dfrac{ \text{ emf of the battery }}{ \text{ length of wire } } \]**

Once the electric field is known, we can easily find the **flow of electrons through a point** in the circuit by using the following formula:

**\[ \boldsymbol{ i = nA \mu E } \]**

Here, $ n $ is the number of electrons per cubic meter, $ A = \pi \bigg ( { \frac{ diameter }{ 2 } } \bigg )^2 $ is the area of cross section of the wire, $ \mu $ is the electron mobility and $ E $ is the electric field strength.

## Expert Answer

**Step (1): Calculating the cross sectional area of the wire:**

\[ A = \pi \bigg ( { \frac{ d }{ 2 } } \bigg )^2\]

\[ A = \pi \bigg ( { \frac{ 0.0007 }{ 2 } \bigg ) }^2 \]

\[ A = 3.85 \times 10^{-7} \ m^2 \]

**Step (1): Calculating the electric field strength:**

\[ E = \dfrac{ \text{ emf of the battery }}{ \text{ length of wire } } \]

\[ E = \dfrac{ 1.3 \ V }{ 26 \ cm } \]

\[ E = 5 V/m \]

**Step (1): Calculating the current flow:**

\[ i = nA \mu E \]

\[ i = \bigg ( 7 \times 10^{+28} \ electrons \ m^{-3} \bigg ) \bigg ( 3.85 \times 10^{-7} \ m^2 \bigg ) \bigg ( 5 \times 10^{-5} \ ( m/s )( m/V ) \bigg ) \bigg ( 5 \ (V/m) \bigg ) \]

\[ i = 6.73 \times 10^{18} electrons/second \]

## Numerical Result

\[ i = 6.73 \times 10^{18} electrons/second \]

## Example

In the same circuit find the number of electrons entering battery # 2 with the following parameters:

**– Each battery has an emf of $ 5 $ volts**

**– Each wire has a length and diameter of $ 5 \ m $ and $ 0.0001 \ m $ respectively.**

\[ A = \pi \bigg ( { \frac{ d }{ 2 } } \bigg )^2 = \pi \bigg ( { \frac{ 0.0001 }{ 2 } \bigg ) }^2 = 2.5 \times 10^{-9} \ m^2\]

\[ E = \dfrac{ \text{ emf of the battery }}{ \text{ length of wire } } = \dfrac{ 5 \ V }{ 5 \ m } = 1 V/m \]

\[ i = nA \mu E \]

\[ i = \bigg ( 7 \times 10^{+28} \ electrons \ m^{-3} \bigg ) \bigg ( 2.5 \times 10^{-9} \ m^2 \bigg ) \bigg ( 5 \times 10^{-5} \ ( m/s )( m/V ) \bigg ) \bigg ( 1 \ (V/m) \bigg ) \]

\[ i = 8.75 \times 10^{15} electrons/second \]