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What contradiction is found by using ampere circuital law to obtain magnetic field during charging of capacitor ? |
| Answer» <html><body><p></p>Solution :In figure (a) parallel <a href="https://interviewquestions.tuteehub.com/tag/plate-1156364" style="font-weight:bold;" target="_blank" title="Click to know more about PLATE">PLATE</a> capacitor (C ) is shown in its charging state. During charging of capacitor at time t current i(t) flows which is obtained from battery. No current flow inside capacitor.<br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/KPK_AIO_PHY_XII_P1_C08_E01_005_S01.png" width="80%"/><br/>A parallel plate capacitor C, as part of a circuit through which a time dependent current i(t) flows, (a) a loop of radius r, to determine magnetic <a href="https://interviewquestions.tuteehub.com/tag/field-987291" style="font-weight:bold;" target="_blank" title="Click to know more about FIELD">FIELD</a> at a point P on the loop , (b) a pot - shaped surface passing through the interior between the capacitor plates with the loop shown in (a) as its rim , (c ) a tiffinshaped surface with the circular loop as its rim and a flat circular bottom S between the capacitor plates. the circular bottom S between the capacitor plates. The arrows show uniform electric field between the capacitor plates.<br/>At a point P outside capacitor, consider loop of radius r whose centre is on the wire plane is perpendicular to wire.<br/>From symmetry it can be shown that at any point on the loop magnetic field is along circumference of the loop and magnitude is equal of at any point of the loop.<br/>By using ampere circuital law point P outside capacitor,<br/> `oint <a href="https://interviewquestions.tuteehub.com/tag/vec-723433" style="font-weight:bold;" target="_blank" title="Click to know more about VEC">VEC</a>(B).vec(d)l=mu_(0)i(t)` ....(1)<br/> where `Sigma I = i(t)` is total current.<br/>If `|vec(B)|=B`and `oint vec(d)l=2pir`, then<br/>`B(2pi r)=mu_(0)i(t)`...(2)<br/>Now consider another surface shown in figure (b) which do not touch current <a href="https://interviewquestions.tuteehub.com/tag/carrying-909956" style="font-weight:bold;" target="_blank" title="Click to know more about CARRYING">CARRYING</a> conductor bottom of surface lies in region between plates.<br/>Also in figure (c ) consider oval shaped surface which is between plates of capacitor and radius equal to r. In figure (b) and (c ) current do not flow in closed surface considered hence closed surface having radius r by using ampere circuital law.<br/> In `int vec(B).vec(d)l=mu_(0)Sigma I`<br/>`int dl=2pi r`<br/>`Sigma I=mu_(0)i(t)=0`<br/>Thus, we get different value of magnetic field outside and inside which is contradictory. Thus, outside and inside which is contradictory. Thus, outside plates of capacitor we get non - zero magnetic field and inside (between) plates of capacitor magnetic field is zero.<br/>This contradiction is by using ampere circuital law hence it can be said that there is something missing in ampere circuital law.</body></html> | |