Microscopic Ohm Law
For linear, homogeneous, isotropic conduction at fixed physical conditions, J equals sigma E. Combining E = V/L and I = JA for a uniform wire gives V = IR.
Why this shows up in the exam
Finding current from conductivity and field · Separating material response from geometry · Deriving a wire resistance
Learn the idea
In an ohmic isotropic material, current density is proportional to electric field. Voltage pushes carriers through a material; conductivity measures how strongly the material responds locally. Geometry enters only when the local law is converted into a whole-wire resistance.
🧠 Memory hook: Local law is J-sigma-E; circuit law is V-I-R.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- J = sigma E — local Ohm law under linear isotropic conditions
- E = rho J — equivalent resistivity form
- I = sigma A V/L — uniform straight wire with a uniform axial field
How to approach it
- 1Decide whether data are local or circuit-level
- 2Use consistent geometry
- 3Verify linear-response assumptions
Common slip-ups that cost marks
- •Treating sigma as a geometric property
- •Using V/L for a nonuniform field without integration
- •Assuming every device is ohmic
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A cell of emf 6 V and internal resistance 1 ohm is connected to a 2 ohm resistor. Find the circuit current.
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