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Current Electricity - Ohm’s Law and Drift Velocity

Grade 12CBSEPhysics

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

🔑Concepts

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Ohm's Law states that the current (II) flowing through a conductor is directly proportional to the potential difference (VV) applied across its ends, provided temperature and other physical conditions remain constant: V=IRV = IR. This linear relationship can be visualized on a V−IV-I graph where the slope represents the resistance RR.

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Drift Velocity (vdv_d) is the average velocity attained by free electrons in a conductor due to an applied electric field. In the absence of a field, electrons move randomly with high thermal speeds (approx. 105 m/s10^5 \text{ m/s}), but their net displacement is zero. Under an electric field, they drift slowly (approx. 10−4 m/s10^{-4} \text{ m/s}) against the field direction.

Illustration of an electron drifting inside a conductor against the direction of the applied electric field.
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Relaxation Time (τ\tau) is the average time interval between two successive collisions of a free electron with the positive ions of the lattice. It decreases as temperature increases because thermal vibrations of ions increase, leading to more frequent collisions.

Zig-zag path of an electron showing successive collisions with lattice ions.
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Electrical Resistivity (ρ\rho) is an intrinsic property of a material. While resistance RR depends on geometry (LL and AA), resistivity depends only on the nature of the material and its temperature. It is defined as ρ=RAL\rho = \frac{RA}{L}.

Cylindrical conductor showing length L and cross-sectional area A used to define resistivity.

📐Formulae

I=qt=nAevdI = \frac{q}{t} = nAe v_d

vd=eEτm=eVτmlv_d = \frac{e E \tau}{m} = \frac{e V \tau}{m l}

R=VI=mlne2τAR = \frac{V}{I} = \frac{m l}{n e^2 \tau A}

ρ=mne2τ\rho = \frac{m}{n e^2 \tau}

J⃗=σE⃗\vec{J} = \sigma \vec{E}

σ=1ρ=ne2τm\sigma = \frac{1}{\rho} = \frac{n e^2 \tau}{m}

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

Rt=R0(1+αΔT)R_t = R_0(1 + \alpha \Delta T)

💡Examples

Problem 1:

A potential difference of 3 V3 \text{ V} is applied across a conductor of length 0.1 m0.1 \text{ m}. If the drift velocity of electrons is 7.5×10−4 m/s7.5 \times 10^{-4} \text{ m/s}, calculate the mobility of the electrons.

Solution:

Given: V=3 VV = 3 \text{ V}, l=0.1 ml = 0.1 \text{ m}, vd=7.5×10−4 m/sv_d = 7.5 \times 10^{-4} \text{ m/s}. \nFirst, find the Electric Field: E=Vl=30.1=30 V/mE = \frac{V}{l} = \frac{3}{0.1} = 30 \text{ V/m}. \nMobility μ=vdE=7.5×10−430=2.5×10−5 m2V−1s−1\mu = \frac{v_d}{E} = \frac{7.5 \times 10^{-4}}{30} = 2.5 \times 10^{-5} \text{ m}^2 \text{V}^{-1} \text{s}^{-1}.

Explanation:

Mobility is defined as the drift velocity acquired per unit electric field applied. We first derive the field from the potential and length.

Problem 2:

A wire of resistance RR is stretched to triple its original length. What will be its new resistance, assuming density and resistivity remain constant?

Solution:

Let initial length be ll and area be AA. Volume V=AlV = Al is constant. \nWhen l′=3ll' = 3l, the new area A′A' must satisfy A′l′=Al  ⟹  A′(3l)=Al  ⟹  A′=A3A'l' = Al \implies A'(3l) = Al \implies A' = \frac{A}{3}. \nInitial Resistance R=ρlAR = \rho \frac{l}{A}. \nNew Resistance R′=ρl′A′=ρ3lA/3=9ρlA=9RR' = \rho \frac{l'}{A'} = \rho \frac{3l}{A/3} = 9 \rho \frac{l}{A} = 9R.

Explanation:

Resistance depends on the geometry of the conductor. When a wire is stretched, its length increases and its cross-sectional area decreases such that the total volume remains the same. The resistance increases by the square of the stretching factor.

Problem 3:

Calculate the current density in a copper wire of radius 0.5 mm0.5 \text{ mm} when a current of 2 A2 \text{ A} flows through it. Also, determine the drift velocity if the number density of free electrons is n=8.5×1028 m−3n = 8.5 \times 10^{28} \text{ m}^{-3}.

A wire showing the direction of current flow used for calculating current density.

Solution:

  1. Area of cross-section A=πr2=π×(0.5×10−3)2=7.85×10−7 m2A = \pi r^2 = \pi \times (0.5 \times 10^{-3})^2 = 7.85 \times 10^{-7} \text{ m}^2.
  2. Current density J=IA=27.85×10−7≈2.55×106 A/m2J = \frac{I}{A} = \frac{2}{7.85 \times 10^{-7}} \approx 2.55 \times 10^6 \text{ A/m}^2.
  3. Drift velocity vd=Jne=2.55×106(8.5×1028)(1.6×10−19)v_d = \frac{J}{ne} = \frac{2.55 \times 10^6}{(8.5 \times 10^{28})(1.6 \times 10^{-19})}.
  4. vd≈1.87×10−4 m/sv_d \approx 1.87 \times 10^{-4} \text{ m/s}.

Explanation:

Current density is current per unit area. Drift velocity is then derived from the relationship J=nevdJ = n e v_d.

Problem 4:

A potential difference of 100 V100 \text{ V} is applied across a conductor of length 2 m2 \text{ m} and resistance 10 \Omega10 \text{ \Omega}. If the relaxation time is 2.5×10−14 s2.5 \times 10^{-14} \text{ s}, find the electron mobility μ\mu. (Take me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg})

Circuit diagram with a 10 ohm resistor connected to a 100V source.

Solution:

  1. Mobility μ=eτm\mu = \frac{e \tau}{m}.
  2. μ=(1.6×10−19 C)×(2.5×10−14 s)9.1×10−31 kg\mu = \frac{(1.6 \times 10^{-19} \text{ C}) \times (2.5 \times 10^{-14} \text{ s})}{9.1 \times 10^{-31} \text{ kg}}.
  3. μ=4×10−339.1×10−31≈4.4×10−3 m2V−1s−1\mu = \frac{4 \times 10^{-33}}{9.1 \times 10^{-31}} \approx 4.4 \times 10^{-3} \text{ m}^2\text{V}^{-1}\text{s}^{-1}.

Explanation:

Mobility is defined as the magnitude of drift velocity per unit electric field. It depends only on the charge of the charge of the carrier, relaxation time, and mass.