A metal wire of cross-sectional area 0.5 mm^2 and length 100 m is connected across a battery of e.m.

Q: Why do we use atomic mass in kg/mol instead of g/mol?

Standard SI units are crucial for consistency. Since density is given in $\text{kg/m}^3$, the atomic mass must be converted to $\text{kg/mol}$ to ensure the units cancel out correctly and give $n$ in $\text{m}^{-3}$.

Q: Why is the internal resistance added to the wire's resistance?

The battery and the wire are connected in series. In a series circuit, resistances add up. The internal resistance acts like a small resistor inside the battery itself.

Q: What does 'one conduction electron per atom' mean mathematically?

It means the number density of free electrons ($n$) is exactly equal to the number density of atoms in the metal. If an atom gave two electrons, $n$ would be twice the atom density.

Q: Why is the drift velocity so small (less than a millimeter per second)?

Even though electrical signals travel near the speed of light, the actual physical electrons move very slowly because they constantly collide with the vibrating atoms in the metal lattice. There are just so many of them that even a slow drift creates a large current!

Q: Can I use $\rho = \frac{1}{\sigma}$ directly in the resistance formula?

Absolutely! Resistivity ($\rho$) is the reciprocal of conductivity ($\sigma$). So $R = \frac{\rho L}{A}$ is mathematically identical to $R = \frac{L}{\sigma A}$.

Question

A metal wire of cross-sectional area $0.5 	ext{ mm}^2$ and length $100 	ext{ m}$ is connected across a battery of e.m.f. $2 	ext{ V}$ and internal resistance $1 \ \Omega$. The density, atomic mass and electrical conductivity of the metal are $6.35 	imes 10^3 	ext{ kg m}^{-3}$, $63.5 	ext{ gm/mole}$ and $2 	imes 10^8 	ext{ mho m}^{-1}$, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in $	ext{mm s}^{-1}$) of the electrons in the wire is:
[Take Avogadro's number as $6 	imes 10^{23}$ and charge of the electron as $1.6 	imes 10^{-19} 	ext{ C}$.]

Options:

(A) 0.052

(B) 0.104

(C) 0.208

(D) 0.156

Sigma Percentile · Accepted Answer

### The Setup: A Macroscopic View

Imagine a simple yet fascinating electrical circuit. We have a thick metal wire connected across a battery. The battery acts as the pump, providing an electromotive force (EMF) of $E = 2	ext{ V}$, but it also has its own internal resistance of $r = 1\ \Omega$.

Our ultimate goal is to zoom into the microscopic world of this wire and figure out exactly how fast the electrons are drifting. But before we go microscopic, we need to understand the macroscopic flow of current.

### Diving into the Wire: Calculating Resistance

First, we need to determine the resistance of the metal wire itself. We are given its physical dimensions and its electrical conductivity. The formula connecting these properties is:

$$R = \frac{L}{\sigma A}$$

Let's plug in the given values. The length $L$ is $100	ext{ m}$. The conductivity $\sigma$ is $2 	imes 10^8	ext{ mho m}^{-1}$. The cross-sectional area $A$ is $0.5	ext{ mm}^2$, which we must convert to standard SI units as $0.5 	imes 10^{-6}	ext{ m}^2$.

$$R = \frac{100}{2 	imes 10^8 	imes 0.5 	imes 10^{-6}}$$

Notice how beautifully the denominator simplifies. $2 	imes 0.5 = 1$, and $10^8 	imes 10^{-6} = 10^2 = 100$. So, the resistance of the wire is exactly $1\ \Omega$.

### The Flow of Charge: Ohm's Law

Now that we know the wire's resistance, we can look at the entire circuit. The total resistance opposing the flow of current is the sum of the wire's resistance and the battery's internal resistance.

$$R_{	ext{total}} = R + r = 1\ \Omega + 1\ \Omega = 2\ \Omega$$

Using Ohm's Law, we can find the total macroscopic current $I$ flowing through the circuit:

$$I = \frac{E}{R_{	ext{total}}} = \frac{2	ext{ V}}{2\ \Omega} = 1	ext{ A}$$

### The Microscopic World: Electron Density

To find the drift velocity, we need to know how crowded the electrons are inside the metal. This is called the electron number density, denoted by $n$. The problem states there is **one conduction electron per atom**. This means the number density of electrons is exactly equal to the number density of atoms.

We can calculate the number of atoms per unit volume using the material's density, atomic mass, and Avogadro's number:

$$n = \frac{	ext{Density} 	imes N_A}{	ext{Atomic Mass}}$$

**Crucial Step:** We must ensure all units are in standard SI format. The density is $6.35 	imes 10^3	ext{ kg m}^{-3}$. Avogadro's number $N_A$ is $6 	imes 10^{23}$. The atomic mass is given as $63.5	ext{ gm/mole}$, which must be converted to $0.0635	ext{ kg/mole}$.

$$n = \frac{6.35 	imes 10^3 	imes 6 	imes 10^{23}}{0.0635}$$

$$n = 10^5 	imes 6 	imes 10^{23} = 6 	imes 10^{28}	ext{ m}^{-3}$$

### The Final Sprint: Drift Velocity

We are finally ready to connect the macroscopic current to the microscopic drift velocity using the fundamental relation:

$$I = n e A v_d$$

Rearranging for drift velocity $v_d$:

$$v_d = \frac{I}{n e A}$$

Let's substitute all our hard-earned values into this equation:

$$v_d = \frac{1}{6 	imes 10^{28} 	imes 1.6 	imes 10^{-19} 	imes 0.5 	imes 10^{-6}}$$

$$v_d = \frac{1}{4.8 	imes 10^3} = 2.08 	imes 10^{-4}	ext{ m/s}$$

The question specifically asks for the answer in $	ext{mm s}^{-1}$. To convert from meters to millimeters, we multiply by $1000$ (or $10^3$).

$$v_d = 2.08 	imes 10^{-4} 	imes 10^3 = 0.208	ext{ mm/s}$$

And there we have it! The electrons are drifting at a sluggish pace of $0.208	ext{ mm/s}$, yet they collectively create a substantial current of $1	ext{ A}$.

Select Answer:

Visualized Solution

$E, r, L, A, σ$

$R = \frac{L}{σ A}$

$R = 1 Ω$

$I = \frac{E}{R + r}$

$I = 1 A$

$n = \frac{d \cdot N _{A}}{M}$

$n = 6 \times 1 0^{28} m^{- 3}$

$I = n e A v_{d}$

$v_{d} = 2.08 \times 1 0^{- 4} m/s$

$v_{d} = 0.208 mm/s$

$T ↑ ⟹ v_{d} ↓$

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Setup

Diving into the Wire

The Flow of Charge

The Microscopic World

The Final Sprint

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E,r,L,A,σ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=σAL​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=1 Ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=R+rE​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=1 A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=Md⋅NA​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=6×1028 m−3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I=neAvd​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vd​=2.08×10−4 m/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vd​=0.208 mm/s

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T↑⟹vd​↓