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EEL3472HA5S

# EEL3472HA5S - H 4-e F*t 2ol0 of pro5lcm 6.l A with an...

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H4-e F*t( 2ol0 pro5lcm 6..l A stationary conducting loop with an internal resistance of 0.5 Cl is placed in a time-varying magnetic field. When the loop is closed, a current of 5 A hows through it. What will the current be if the loop is opened to create a small gap and a 2-O resistor is connected across its open ends? Solution: %.r is independent of the resistance which is in the loop. Therefore, when the loop is intact and the internal resistance is only 0.5 O, Vemf :5 A x 0.5 Cl : 2.5 V.. When the small gap is created, the total resistancein the loop is infinite and the current flow is zero. With a 2-C) resistor in the gap, I : Y"nl (2 Cl + 0.5 0) -- 2.5 v 12.5 C) : I (A)'

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vcmr-- -# : l'1 xZnx l0asin(2zr x l0ar) x l0-7 : 6.9 x l0-3 sin(22 x l0at) (V). (b) , Vcmf 6,9 x l0-i _:_/.-.. r n4.\ _ 1 ao ^:^t.t* ., I rind:4-JJ : # sin(2nx lgar) : l.38sin(2nxl1at) (mA). At t :0, B is a maximum,it points in -i-directibn, and since it varies as cos(2n x lOat), it is decreasing. Hence, the induced current has to be ;: CW when looking down on the loop,as shown in the figure.
pr0blenr 6.6 The square loop shown in Fig. P6.6 is coplanar with a long, straight wire carrying a current 1(l) : 5cos(2zr x lOar) (A)' (a)Determinetheemfinducedacrossasmallgapcreatedintheloop. (b) Determine the direction and magnitude of the current that would flow through a 4-Q resistor connected across the gap. The loop has an intemal resistance of lo. f igure P6.6: Loop coplanar with long wire (Problem 6'6)' Solution: (a) The magnetic field due to the wire is ^ ltol ^ 1t4l B : Q T*: -* zny, where in the plane of the loop, 0 : -i and r : l.The flux passing through

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EEL3472HA5S - H 4-e F*t 2ol0 of pro5lcm 6.l A with an...

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