The document Notes | EduRev is a part of the JEE Course NEET Revision Notes.

All you need of JEE at this link: JEE

**Question 1: ****One end of an aluminum wire whose diameter is 2.5 mm is welded to one end of a copper wire whose diameter is 1.8 mm. The composite wire carries a steady current i of 17 mA.****(a) What is the current density in each wire?****(b) What is the drift speed of the conduction electrons in the copper wire? Assume that, on the average, each copper atom contributes one conduction electron.****Solution:**

(a) We may take the current density as constant with in each wire (except near the junction, where the diameter changes). The cross-sectional area A of the aluminum wire is

A_{Al }= π(d/2)^{2}

= π/4 (2.5×10^{-3} m)^{2}

= 4.91×10^{-6}m^{2}

And the current density is given by the equation J = i/A,

J_{Al} = i/A

= i/A_{Al}

= (17×10^{-3} A)/(4.91×10^{-6} m^{2})

= 3.5×10^{3} A/m^{2}

As you can verify, the cross-sectional area of the copper wire is 2.54×10^{-6} m^{2}, so

J_{Cu} = i/A_{Cu}

= (17×10^{-3} A)/(2.54×10^{-6} m^{2})

=6.7×10^{3}A/m^{2}

From the above observation we conclude that, the current density in aluminum wire would be 3.5×10^{3} A/m^{2} and copper wire would be 6.7×10^{3}A/m^{2}.

(b) We can find the drift speed from equation J = (ne)v_{d} if we first find n, the number of electrons per unit volume. With the given assumption of about one conduction electron per atom, n is the same as the number of atoms per unit volume and can be found from n/N_{A} = ρ/M or [atoms/m^{3}]/[atoms/mol] = [mass/m^{3}]/[mass/mol].

Here ρ is the density of copper, N_{A} is Avogadro’s number, and M is the molar mass of copper. Thus

n= N_{A}ρ/M

= (6.02×10^{23} mol^{-1}) (9.0×10^{3} kg/m^{3})/(64×10^{-3} kg/mol

=8.47×10^{28} electrons/m^{3}

We then have from equation J = (ne)v_{d},

v_{d }= [6.7×10^{3}A/m^{2}]/[(8.47×10^{28} electrons/m^{3})(1.6×10^{-19} C/electron)]

= 4.9×10^{-7}m/s = 1.8 mm/h

From the above observation we conclude that, the drift speed of the conduction electrons in the copper wire would be 1.8 mm/h.

**Question 2: ****A rectangular block of iron has dimensions 1.2 cm×1.2 cm×15 cm.****(a) What is the resistance of the block measured between the two square ends?****(b) What is the resistance between two opposite rectangular faces?****Solution:**

(a) The resistivity of iron at room temperature is 9.68×10^{-8} Ω.m. The area of a square end is (1.2×10^{-2}m)^{2}, or 1.44×10^{-4} m^{2}. From equation R = ρL/A,

R= ρL/A

= [(9.68×10^{-8} Ω.m)(0.15 m)]/ [1.44×10^{-4} m^{2}]

=1.0×10^{-4 }Ω

= 100μΩ

Therefore, the resistance of the block measured between the two square ends would be 100μΩ.

(b) The area of a rectangular face is (1.2×10^{-2}m) (0.15 m), or 1.80×10^{-3} m^{2}. From equation R = ρL/A,

R = ρL/A

= [(9.68×10^{-8 }Ω.m)(1.2×10^{-2}m)]/ [1.80×10^{-3} m^{2}]

= 6.5×10^{-7}Ω

= 0.65 μΩ

From the above observation we conclude that, the resistance between two opposite rectangular faces would be 0.65 μΩ.

**Question 3: (IIT-JEE):-****(a) What is the mean free path time τ between collisions for the conduction electrons in copper?****(b) What is the mean free path λ for these collisions? Assume an effective speed v _{eff} of 1.6×10^{6} m/s.**

(a) We know that,

ρ=m/ne2

The number of conduction electrons per unit volume (n) in copper will be,

n/N

Here ρ is the density of copper, N

n = N

= (6.02×10

=8.47×10

The value of ρ for copper is 1.69×10

So the denominator of the above equation will be,

ne

= 3.66×10

= 3.66×10

Here we converted units as,

C

For the mean free time we then have,

= (9.1×10

= 2.5×10

Therefore, the mean free path time τ between collisions for the conduction electrons in copper would be 2.5×10

(b) We defined the mean free path as being the average distance traversed by a particle between collisions. Here the time between collisions of a free electron is τ and the speed of the electron is v

So, λ = v

= (2.5×10

= 4.0×10

From the above observation, we conclude that the mean free path λ for these collisions would be 40 nm. This is about 150 times the distance between nearest- neighbor atoms in a copper lattice.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

291 docs

### NCERT Exemplars - Current Electricity

- Doc | 20 pages
### NCERT Exemplars - Electromagnetic Induction

- Doc | 33 pages

- Revision Notes - Current Electricity
- Doc | 5 pages
- NCERT Exemplars - Magnetism and Matter
- Doc | 18 pages