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(a) Show that an a.c. circuit containing a pure inductor, the voltage is ahead of current by p/2in phase. (b) A horizontal straight wire of length Lextending from east to westis falling with speed v at right angles to the horizontal component of Earth's magnetic field B. (i) Write the expression for the instantaneous value of the e.m.f. induced in the wire. (ii) What is the direction of the e.m.f.? (iii) Which end of the wire is at the higher potential? |
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Answer» Solution :(a) `epsilon=epsilon_(0)SIN omegat` `epsilon-L (di)/(dt)=0` `""(di)/(dt)=(epsilon)/(L)=(epsilon_(0))/(L)sin omegat` `""underset(0)overset(1)int (di)/(dt)dt=(epsi_(0))/(L)underset(0)overset(1)int sin omegat` `""i=(epsi_(0))/(OMEGAL) COS omegat`  `(epsi_(0))/(X_(L))sin(omegat-pi//2),X_(L)=omegaL` (b)  (i) Horizontal component along AB is .....(i) `AL=H=R cos delta ""(i)` vertical component along AD is `AM=V=R sin delta ""(ii)` Square (i) and (ii), and add `H^(2)+V^(2)+R^(2)(cos^(2)delta +sin^(2)delta)=R^(2)` `therefore ""R=sqrt(H^(2)+V^(2))........(III)` Dividing (ii) by (i), we get `(R sin delta)/(R cos delta)=(V)/(H)="Or "tandelta=(V)/(H)` .......(iv) The value of horizontal component H=R cos `delta` is different at different places. At the magnetic poles, `delta=90^(@)` Horizontal component (H) can be measured using a vibration magnetometer and a deflection magnetometer. The value of H at a place on the surfae of earth is the order of `3.2xx10^(-5)`tesla. (ii) The DIRECTION of the e.m.f. from the south to north. (iii) Wire 1is the higher potential than wire 2. |
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