⚡ 10,279 VERIFIED QUESTIONS 333 AUTHENTIC BOARD PAPERS CBSE Board Revision Archive 100% FREE PUBLIC REVISION 37 NCERT CHAPTERS 3,054 PARALLEL SET CLONES ⚡ 10,279 VERIFIED QUESTIONS 333 AUTHENTIC BOARD PAPERS CBSE Board Revision Archive 100% FREE PUBLIC REVISION 37 NCERT CHAPTERS 3,054 PARALLEL SET CLONES
HOME CBSE 12 PHYSICS ELECTRIC CHARGES AND FIELDS
✓ 4 MARKS Case Study
🔥 APPEARED IN 3 BOARD SETS
In a metallic conductor, an electron, moving due to thermal motion, suffers collisions with the heavy fixed ions but after collision, it will emerge out with the same speed but in random directions. If we consider all the electrons, their average velocity will be zero. When an electric field is applied, electrons move with an average velocity, known as drift velocity ($v_d$). The average time between successive collisions is known as relaxation time ($ au$). The magnitude of drift velocity per unit electric field is called mobility ($\mu$). An expression for current through the conductor can be obtained in terms of drift velocity, number of electrons per unit volume ($n$), electronic charge ($- ext{e}$), and the cross-sectional area ($ ext{A}$) of the conductor. This expression leads to an expression between current density ($\vec{j}$) and the electric field ($\vec{ ext{E}}$). Hence, an expression for resistivity ($ ho$) of a metal is obtained. This expression helps us to understand increase in resistivity of a metal with increase in its temperature, in terms of change in the relaxation time ($ au$) and change in the number density of electrons ($n$).
YearPaper CodeSetPosition
CBSE 2025Series 55-7Set 1Question 29
CBSE 2025Series 55-7Set 2Question 30
CBSE 2025Series 55-7Set 3Question 29
  • Step 1: Stating the fundamental formula / law statement (+1 Mark)
  • Step 2: Intermediate mathematical substitution or proof steps (+1 Mark)
  • Step 3: Correct final value or conclusion with mandatory SI units (+1 Mark)
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