5 એકમ વિદ્યુત ક્ષેત્ર ધરાવતા પદાર્થની અવરોધકતા 4.5 એકમ હોય, તો વહન કરંટ ઘનતા શોધો.
જવાબ અને પગલાવાર ઉકેલ
વહન કરંટ ઘનતા ઓહ્મના નિયમના બિંદુ સ્વરૂપથી વિદ્યુત ક્ષેત્ર સાથે જોડાયેલી છે: J = σE = E/ρ. E = 5 અને ρ = 4.5 સાથે J = 5/4.5 ≈ 1.11.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Concept & Reasoning
Resistivity tells us how strongly a material opposes conduction. Conductivity is its reciprocal, σ = 1/ρ. Therefore a finite electric field produces a conduction current density J = σE. The correct numerical value follows directly from dividing the field intensity by resistivity; multiplying E by ρ would reverse the physical relationship.
How to Solve It in the Exam
Remember: resistivity in denominator, conductivity in numerator.
Important Exam Point
Do a dimensional/physical sense check: larger resistivity must reduce J for the same E.
Quick Trick
Remember: resistivity in denominator, conductivity in numerator.
Common Mistake
Using J = ρE instead of J = E/ρ.
Calculate the conduction current density when the resistivity of a material with an electric field of 5 units is 4.5 units.
Answer & step-by-step solution
Conduction current density is related to electric field by the point form of Ohm’s law: J = σE = E/ρ. With E = 5 and resistivity ρ = 4.5, the current density is 5/4.5 ≈ 1.11 in the corresponding current-density units.
Concept & Reasoning
Resistivity tells us how strongly a material opposes conduction. Conductivity is its reciprocal, σ = 1/ρ. Therefore a finite electric field produces a conduction current density J = σE. The correct numerical value follows directly from dividing the field intensity by resistivity; multiplying E by ρ would reverse the physical relationship.
How to Solve It in the Exam
Remember: resistivity in denominator, conductivity in numerator.
Important Exam Point
Do a dimensional/physical sense check: larger resistivity must reduce J for the same E.
Quick Trick
Remember: resistivity in denominator, conductivity in numerator.
Common Mistake
Using J = ρE instead of J = E/ρ.
5 इकाई विद्युत क्षेत्र वाले पदार्थ की प्रतिरोधकता 4.5 इकाई हो, तो चालन करंट घनत्व ज्ञात कीजिए।
उत्तर और चरणबद्ध हल
चालन करंट घनत्व ओम के नियम के बिंदु रूप से विद्युत क्षेत्र से जुड़ा है: J = σE = E/ρ। E = 5 और ρ = 4.5 के साथ J = 5/4.5 ≈ 1.11।
विस्तृत हल (मूल भाषा में)
Concept & Reasoning
Resistivity tells us how strongly a material opposes conduction. Conductivity is its reciprocal, σ = 1/ρ. Therefore a finite electric field produces a conduction current density J = σE. The correct numerical value follows directly from dividing the field intensity by resistivity; multiplying E by ρ would reverse the physical relationship.
How to Solve It in the Exam
Remember: resistivity in denominator, conductivity in numerator.
Important Exam Point
Do a dimensional/physical sense check: larger resistivity must reduce J for the same E.
Quick Trick
Remember: resistivity in denominator, conductivity in numerator.
Common Mistake
Using J = ρE instead of J = E/ρ.
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