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Find the current density of a material with resistivity 20 units and electric field intensity 2000 units.(a) 400(b) 300(c) 200(d) 100I have been asked this question in class test.This intriguing question originated from Real Time Applications in chapter Electrostatic Fields of Electromagnetic Theory

Answer»

The correct answer is (d) 100

The best explanation: The CURRENT density is given by J = σ E, where σ is the CONDUCTIVITY. THUS resistivity ρ = 1/σ. J = E/ρ = 2000/20 = 100 UNITS.



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