Question

If current of 80 A is passing through a straight conductor of length 10 m, then the total momentum of electrons in the conductor is (mass of electron \( = 9.1 \times 10^{-31} \) kg and charge of electron \( = 1.6 \times 10^{-19} \) C)

• A. \( 910 \times 10^{-9} \, \text{Ns} \)
• B. \( 910 \times 10^{-11} \, \text{Ns} \)
• C. \( 455 \times 10^{-9} \, \text{Ns} \)
• D. \( 455 \times 10^{-11} \, \text{Ns} \)

Hint:

- Current \( I = nAev_d \), where \(n\)=number density of charge carriers, \(A\)=cross-sectional area, \(e\)=charge of a carrier, \(v_d\)=drift velocity. - Total number of charge carriers in length L: \( N = nAL \). - Total momentum of these carriers: \( P = N \cdot (\text{mass of carrier} \cdot v_d) \). - Substitute \( nAv_d = I/e \) into the momentum expression: \( P = (I/e) L m \).

Answer 1

The Correct Option is D

Current \( I = 80 \) A.
Length of conductor \( L = 10 \) m.
Mass of electron \( m_e = 9.
1 \times 10^{-31} \) kg.
Charge of electron \( e = 1.
6 \times 10^{-19} \) C.
(Magnitude) Current \( I = n A e v_d \), where \(n\) is the number density of free electrons, \(A\) is the cross-sectional area, \(e\) is the charge of an electron, and \(v_d\) is the drift velocity of electrons.
The total number of free electrons \(N\) in the conductor of length \(L\) and area \(A\) is \( N = n (AL) \).
The total momentum of electrons \( P_{total} = N \cdot (m_e v_d) \).
\[ P_{total} = (nAL) m_e v_d = (n A e v_d) \frac{L m_e}{e} \] Since \( I = n A e v_d \), \[ P_{total} = I \frac{L m_e}{e} \] Substitute the given values: \[ P_{total} = (80 \, \text{A}) \frac{(10 \, \text{m}) (9.
1 \times 10^{-31} \, \text{kg})}{1.
6 \times 10^{-19} \, \text{C}} \] \[ P_{total} = \frac{80 \times 10 \times 9.
1 \times 10^{-31}}{1.
6 \times 10^{-19}} \, \text{Ns} \] \[ P_{total} = \frac{800 \times 9.
1}{1.
6} \times 10^{-31 - (-19)} = \frac{800 \times 9.
1}{1.
6} \times 10^{-12} \] \[ \frac{800}{1.
6} = \frac{8000}{16} = 500 \] \[ P_{total} = 500 \times 9.
1 \times 10^{-12} = 4550 \times 10^{-12} \, \text{Ns} \] \[ P_{total} = 4.
550 \times 10^3 \times 10^{-12} = 4.
55 \times 10^{-9} \, \text{Ns} \] To match the options format \( X \times 10^{-11} \): \( 4.
55 \times 10^{-9} = 455 \times 10^{-2} \times 10^{-9} = 455 \times 10^{-11} \, \text{Ns} \).
This matches option (4).