Question

oSUniPhys1 30.2. P. 035 My Notes -/1 points The magnetic flux through the loop shown in the accompanying figure varies with time according to m9.00e sin(130nt), wheremis in milliwebers. What are the magnitude and direction of the current (in A) through = the 5.00 Q resistor at the following times? 5.0 (a) t 0s magnitude downward through the resistor direction (b) t 2.17 x 10-2 s magnitude A direction downward through the resistor v (c) t 0.5 s magnitude downward through the resistor v direction

Answer 1

= 00 e 4t sin ( 130 t) mWb. converting it to Wb = 0.009 e-4t sin (130 11 t) Wb Induced ent is given by - -dom 1. dt - :E 3.0 (0.005 e 4t sin(13006) dt & 2-12.0096-418 at sin(13016) - + 0.009 e-44 (1307) cos (130 7t) | E = -0.00964+ [ -4 sin (130 11t) + 1301 cos (130mt) | a) At E -os E = -0.009 eº [ 0 + 13011 Cos(o) ] .: E = -3.674 V - By Ohm's law I= - E R -3.674 5 I= -0.735 A ]

Mag field outward through loop mag. flux decreasing To oppose mag, bield, the direction of induced current must be clock wise Thus - downward through resistance. b) t - 2.17 x 10-2 sec € = -0.009 ē 4(2-77109 T-4 sin (1301 x 2.17X107) + 130 n cos (13011 X 2.17X1024 = 2.86 V : I = E 2-86 E I = 0.572 A downward through R c) t= 0.5 sec -=-0.009 en T-4 sio (650) +1301 CUS (6577 E = 0.495 V II = 0.ogg a downward the R

Answer 2

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