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InterviewSolution
| 1. |
Derivesigma=("ne"^2 tau)/m, where the symbols have their usual meaning. |
| Answer» <html><body><p></p>Solution :Let .n. be the numberdensityof electrons , .L. be the lengthof the <a href="https://interviewquestions.tuteehub.com/tag/conductora-8288920" style="font-weight:bold;" target="_blank" title="Click to know more about CONDUCTORA">CONDUCTORA</a> and .A.be the areaof cross -sectionof the conductor . <br/>Let `v_d` be the drift <a href="https://interviewquestions.tuteehub.com/tag/velocityof-3259016" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITYOF">VELOCITYOF</a> the electrons and Dx be a small length . <br/>Electrons drift in a directionoppositeto the electric field. Number ofelectronsin .Dx.and area of cross -section .A.is equal to (ADx) n. <br/>Charge on the theseelectrons =(nADx)e. <br/> If Dt is the time taken for effectivedisplacementof electronsthen rate of flow charge = `nAe ((Deltax)/(Deltat))` .<br/> By <a href="https://interviewquestions.tuteehub.com/tag/definition-11288" style="font-weight:bold;" target="_blank" title="Click to know more about DEFINITION">DEFINITION</a>,electric current I=rate of flowof charge <br/> i.e.,`I=nAev_d`...(1) <br/>where , average velocityof electrons with whichit drifts againstthe directionof electric field is known as drift velocity`(v_d)`. <br/> Let .a. be the accelerationof electrons. Let .E.be the electricfieldintensity . Force on electrons, <br/> F=ma <br/> i.e., eE=ma <br/> But = `a=v_dt` where .t.is the relaxationtime . <br/> Relaxationtime representsthe average timetaken for two successivecollisionsof electronsand ionsin the lattice. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SPH_PHY_HND_BOK_PUC_QP_JUN_18_E01_028_S01.png" width="80%"/> <br/> Hence ,`eE=mv_dt` or `v_d=(eE)/(mtau)` ...(2) <br/>We alsoknow that electricpotentialdifferencebetween the ends of the conductorV=EL <br/> `v_d =(eV)/(mtauL)` substitutingthis is the expression (1) <br/>We write, `I=nAe ((eV)/(mtauL))` <br/> i.e., `I=((nAe^2)/(mtauL))V` <br/> `R=(mtau L)/(nAe^2)`is called the electricalresistanceof a conductor . and `K=((nAe^2)/(mtauL))` is called the electric conductance of a conductor . the expression `sigma =("<a href="https://interviewquestions.tuteehub.com/tag/ne-1112263" style="font-weight:bold;" target="_blank" title="Click to know more about NE">NE</a>"^2)/(mtau)` is called the electricalconductivityand `1/sigma=(mtau)/("ne"^2)` is called electricalresistivity .</body></html> | |