Home
Class 12
PHYSICS
A straight copper-wire of length 1 = 100...

A straight copper-wire of length 1 = 1000 m and corss-sectional area A = 1.0 `m m^(2)` carries a current i=4.5A. The sum of electric forces acting on all free electrons in the given wire is `M xx 10^(6)N`. Find M. Free electron density and resistivity of copper are `8.5xx10^(28)//m^(3)` and `1.7xx10^(-8)Omega`.m respectively.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will follow the outlined steps based on the information provided in the question. ### Step 1: Calculate the number of free electrons per unit volume (n) The formula for the number of free electrons per unit volume (n) is given by: \[ n = \frac{N_A \cdot d}{M} \] Where: - \(N_A\) = Avogadro's number = \(6.022 \times 10^{23} \, \text{mol}^{-1}\) - \(d\) = density of copper = \(8.5 \times 10^{28} \, \text{m}^{-3}\) - \(M\) = atomic mass of copper = \(63.5 \, \text{g/mol} = 63.5 \times 10^{-3} \, \text{kg/mol}\) Now, substituting the values: \[ n = \frac{(6.022 \times 10^{23}) \cdot (8.5 \times 10^{28})}{63.5 \times 10^{-3}} \] Calculating this gives: \[ n \approx 8.49 \times 10^{28} \, \text{electrons/m}^3 \] ### Step 2: Calculate the total electric force acting on all free electrons in the wire The formula for the total electric force (F) acting on all free electrons in the wire is given by: \[ F = n \cdot e \cdot L \cdot \rho \cdot I \] Where: - \(e\) = charge of an electron = \(1.6 \times 10^{-19} \, \text{C}\) - \(L\) = length of the wire = \(1000 \, \text{m}\) - \(\rho\) = resistivity of copper = \(1.7 \times 10^{-8} \, \Omega \cdot \text{m}\) - \(I\) = current = \(4.5 \, \text{A}\) Substituting the values into the formula: \[ F = (8.49 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (1000) \cdot (1.7 \times 10^{-8}) \cdot (4.5) \] Calculating this gives: \[ F \approx 2.30 \times 10^{6} \, \text{N} \] ### Step 3: Relate the force to the given expression According to the problem, the total electric force can also be expressed as: \[ F = M \times 10^{6} \, \text{N} \] From our calculation, we have: \[ 2.30 \times 10^{6} = M \times 10^{6} \] ### Step 4: Solve for M To find \(M\), we can divide both sides by \(10^{6}\): \[ M = 2.30 \] ### Final Answer Thus, the value of \(M\) is: \[ M = 2.30 \]
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE - II LEVEL-I (MAIN) (Straight Objective Type Questions))|14 Videos
  • CURRENT ELECTRICITY

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE - II LEVEL-II (ADVANCED) (More than One correct answer Type Questions))|3 Videos
  • CURRENT ELECTRICITY

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE - I LEVEL-II (ADVANCED) (Matrix Matching Type Questions))|1 Videos
  • CURRENT ELECTRICITY

    AAKASH SERIES|Exercise PROBLEMS (LEVEL-II)|27 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    AAKASH SERIES|Exercise PRACTICE EXERCISEX|42 Videos

Similar Questions

Explore conceptually related problems

A uniform copper wire having a cross sectional area of 1 mm^(2) carries a current of 5 A. Calculate the drift speed of free electrons in it. (Free electron number of density of copper =2 times 10^(28)//m^(3) .)

A straight copper-wire of length 100m and cross-sectional area 1.0mm^(2) carries a current 4.5A . Assuming that one free electron corresponds to each copper atom, find (a) The time it takes an electron to displace from one end of the wire to the other. (b) The sum of electrostatic forces acting on all free electrons in the given wire. Given resistivity of copper is 1.72xx10^(-8)Omega-m and density of copper is 8.96g//cm^(3) .

A current of 1.8 A flows through a wire of cross-sectional area 0.5 mm^(2) ? Find the current density in the wire. If the number density of conduction electrons in the wire is 8.8 xx 10^(28) m^(-3) , find the drift speed of electrons.

A uniform copper wire of length 1m and cross section area 5 xx 10^(-7)m^(2) carries a current of 1A . Assuming that are 8 xx 10^(28) free electron per m^(3) in copper, how long will an electron take to drift from one end of the wire an electron the other. Charge on an electron = 1.6 xx 10^(-19)C

Calculate the electric field in a copper wire of cross-sectional area 2.0mm^(2) carrying a current of 1A .The resistivity of copper =1.7xx10^(-8)(Omega)m .

A sonometer wire of length 1.5m is made of steel. The tension in it produces an elastic strain of 1% . What is the fundamental frequency of steel if density and elasticity of steel are 7.7 xx 10^(3) kg//m^(3) and 2.2 xx 10^(11) N//m^(2) respectively ?

A copper wire has a square cross-section, 2.0 mm on a side. It carries a current of 8 A and the density of free electrons is 8 xx 10^(28)m^(-3) . The drift speed of electrons is equal to

Estimate the average drift velocity of conduction electrons in a copper wire of cross-sectional area 2.5 xx 10^(-7) m^(2) , carrying a current of 2.7 A. Assume the density of conduction electrons to be 9 xx 10^(28) m^(-3) .

A copper rod of length 20cm and cross-sectional area 2mm^(2) is joined with a similar aluminium rod as shown in figure .Find the resistance of the combination between the ends, Resistivity of copper =1.7xx10^(-8)(Omega)m and that of aluminium =2.6xx10^(-8)(Omega)m .

A uniform copper wire of mass 2.23 xx10^(-3) kg carries a current of 1 A when 1.7 V is applied across it. Calculate the length and the area of cross section, if the wire is uniformaly stretched to double its length, calculate the new resistance. Density of copper is 8.92xx10^3 kgm^(-3) and resistivity is 1.7xx10^(-8) Omega m.