# Answer Week 9 - 2 I |—l b l i Answers Week 9(i The...

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Unformatted text preview: 2; I |—l b.) l i Answers Week 9 (i) The elementary generators of [3 must be magnetic dipoles whose “size” is close to ZERO. (ii) 5 ﬁeld lines form closed loops, there is no “beginning” or end. Non-Coulomb g - ﬁeld lines also form closed loops surrounding the region where , the ﬂux of g is changing with time so total ﬂux of g through any closed Non~Coquinb surface will be zero. a = —0.5 V , bottom end of rod is positive because we need a clockwise “current” I. to oppose the increase in ﬂux of Q out of page. We do not need to apply a force to move the rod. Suppose ﬂip time is At , 5 m f in coil due to ﬂip will be ZWFZB , . ‘ ZﬂrzB Ag 8 = ~ , causmg current I = — w = _ At _ AIR At 27rrZB V So A = . Q R FLUX “UP” IS REDUCING SO FLOW OF CHARGE will be counter clockwise T 0 make a generator, rotate coil at angular velocity 0) about y-axis. If so, ﬂux ofﬁ through coil will vary with time and generate an a m f given by g = '— MB = cuABSi (cot) At for a single turn coil of area A. L=— g -—>.V01tixTime a Amp At ' R : Volt ‘ Amp L So —— ——> Time R When there is a current in an Inductor, there is a g - ﬁeld inside it. The energy U B : é—L I 2 gets stored in this 13 - ﬁeld. 9—15 r=1o*4s, Time for current to reach 90 percent of% will be t = 2.3 xlO'4 s, ...
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