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Grade 11Electric Current

A straight conductor of uniform cross-section carries a current I. Let S= specific charge of an electron. the momentum of all the free electrons per unit length of the condutor , due to their drift velocities only is .....................(answer in terms of I & S).

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14 Years agoGrade 11
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ApprovedApproved Tutor Answer1 Year ago

To determine the momentum of all the free electrons per unit length of a straight conductor carrying a current \( I \), we need to consider the relationship between current, charge, and drift velocity. Let's break this down step by step.

Understanding Current and Charge

Current (\( I \)) is defined as the flow of electric charge per unit time. Mathematically, it can be expressed as:

I = n \cdot A \cdot q \cdot v_d

  • n = number density of charge carriers (electrons) per unit volume
  • A = cross-sectional area of the conductor
  • q = charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs)
  • v_d = drift velocity of the electrons

Relating Charge Density to Specific Charge

The specific charge (\( S \)) of an electron is defined as the charge per unit mass. For electrons, it can be expressed as:

S = \frac{q}{m_e}

where \( m_e \) is the mass of an electron (approximately \( 9.11 \times 10^{-31} \) kg). This relationship allows us to express the number density of electrons in terms of the specific charge.

Calculating Momentum

The momentum (\( p \)) of a single electron is given by:

p = m_e \cdot v_d

Now, to find the total momentum of all the free electrons per unit length of the conductor, we need to consider the number of electrons in a unit length of the conductor. The number of electrons per unit length (\( N \)) can be calculated as:

N = n \cdot A

Substituting \( n \) from the current equation, we have:

N = \frac{I}{q \cdot v_d} \cdot A

Combining Everything

The total momentum per unit length (\( P \)) of all the electrons can now be expressed as:

P = N \cdot p = N \cdot (m_e \cdot v_d)

Substituting for \( N \), we get:

P = \left(\frac{I}{q \cdot v_d} \cdot A\right) \cdot (m_e \cdot v_d)

Notice that \( v_d \) cancels out:

P = \frac{I \cdot m_e \cdot A}{q}

Expressing in Terms of Specific Charge

Now, substituting \( q \) with \( S \cdot m_e \) gives:

P = \frac{I \cdot m_e \cdot A}{S \cdot m_e} = \frac{I \cdot A}{S}

Final Result

Thus, the momentum of all the free electrons per unit length of the conductor, expressed in terms of the current \( I \) and the specific charge \( S \), is:

P = \frac{I \cdot A}{S}

This equation shows how the momentum of the electrons is directly proportional to the current and the cross-sectional area of the conductor, while inversely proportional to the specific charge of the electrons.