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How many electrons per second enter the positive end of battery #2 for the following circuit:

  1. 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.
  2. A steady current flows through this circuit.
  3. Each battery has an emf of $ 1.3 $ volts
  4. Each wire has a length and diameter of $ 26 \ cm $ and $ 0.0007 \ m $ respectively.
  5. The wire material (metal) contains $ 7 \times 10^{+28} $ mobile electrons per cubic meter.
  6. 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.

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

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

\[ i = nA \mu E \]

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

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

 

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