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Current Electricity - Mechanism of Flow of Current

Grade 12ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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In a conductor, free electrons move randomly with thermal velocities of the order of 10510^5 m/sm/s. However, the average thermal velocity is zero, so there is no net flow of charge in any direction.

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When a potential difference is applied across a conductor, an electric field E⃗\vec{E} is established. Electrons experience a force F⃗=−eE⃗\vec{F} = -e\vec{E}, causing them to accelerate momentarily before colliding with positive ions of the lattice.

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Drift Velocity (vdv_d) is defined as the average velocity with which free electrons get drifted towards the positive terminal of the conductor under the influence of an external electric field.

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Relaxation Time (τ\tau) is the average time interval between two successive collisions of an electron with the positive ions in the conductor.

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The Mean Free Path (λ\lambda) is the average distance traveled by an electron between two successive collisions, given by λ=vd×τ\lambda = v_d \times \tau.

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Current density (JJ) is defined as the current per unit area of cross-section, where J⃗=nev⃗d\vec{J} = n e \vec{v}_d.

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Mobility (μ\mu) is the magnitude of drift velocity per unit electric field. It is always positive and depends on the nature of the charge carrier.

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The microscopic form of Ohm's Law is expressed as J⃗=σE⃗\vec{J} = \sigma \vec{E}, where σ\sigma is the electrical conductivity.

📐Formulae

vd=eEτmv_d = \frac{e E \tau}{m}

I=nAevdI = n A e v_d

J=IA=nevdJ = \frac{I}{A} = n e v_d

μ=vdE=eτm\mu = \frac{v_d}{E} = \frac{e \tau}{m}

σ=ne2τm\sigma = \frac{n e^2 \tau}{m}

ρ=1σ=mne2τ\rho = \frac{1}{\sigma} = \frac{m}{n e^2 \tau}

vd=VeτmLv_d = \frac{V e \tau}{m L}

💡Examples

Problem 1:

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0×10−71.0 \times 10^{-7} m2m^2 carrying a current of 1.51.5 AA. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0×1039.0 \times 10^3 kg/m3kg/m^3 and its atomic mass is 63.563.5 uu.

Solution:

First, calculate the number of electrons per unit volume nn: n=6.022×1023×9.0×10363.5×10−3≈8.5×1028 m−3n = \frac{6.022 \times 10^{23} \times 9.0 \times 10^3}{63.5 \times 10^{-3}} \approx 8.5 \times 10^{28} \text{ m}^{-3}. Using the formula I=nAevdI = n A e v_d, we get: vd=InAev_d = \frac{I}{n A e} vd=1.58.5×1028×1.0×10−7×1.6×10−19v_d = \frac{1.5}{8.5 \times 10^{28} \times 1.0 \times 10^{-7} \times 1.6 \times 10^{-19}} vd≈1.1×10−3 m/sv_d \approx 1.1 \times 10^{-3} \text{ m/s} or 1.11.1 mm/smm/s.

Explanation:

Even though the thermal speed is very high, the drift velocity is remarkably small (around 11 mm/smm/s) because of frequent collisions with the heavy metal ions.

Problem 2:

What is the effect on the drift velocity vdv_d of electrons in a metal conductor if the temperature of the conductor is increased, keeping the applied voltage constant?

Solution:

As temperature increases, the thermal vibrations of the metal ions increase. This leads to more frequent collisions, which decreases the relaxation time τ\tau. Since vd=eEτmv_d = \frac{e E \tau}{m}, a decrease in τ\tau leads to a decrease in the drift velocity vdv_d.

Explanation:

Increased temperature causes the ions to vibrate with greater amplitude, making it harder for electrons to pass through, thereby reducing relaxation time and drift velocity.

Mechanism of Flow of Current - Revision Notes & Key Formulas | ICSE Class 12 Physics