Question

3) In the figure below, the magnetic flux through the loop shown increases according to the equation, 0-8-3r where is in webers and is in seconds: wa a) What is the induced voltage across the resistor atr=2 seconds? (5 pes) b) What is the polarity of the voltage across the resistor (Justify your answer (Spes) c) If the area of the entire closed loop is 14 m", what is the magnitude of the magnetic field at 2 seconds? (4 pes) 4) A cylindrical copper cable carries a current of 1200 A. There is a potential difference of 16 my between the two points on the cable that are 24 cm apart. The number of carriers per unit volume of the current in the wire is 6.95X10 a) Find the radius of the cable) b) Find the average current density in the cable (5 pes) c) What is the drift velocity of the current? (5 pes) 5) Given the following circuit: 4.000 8.00 WWW 3.00 A 6.00 24.0 V PE + w 2.00 V a) Find the current through the 2.00 resistor and the voltage V of the battery to the left of this resistor pts) b) Calculate the power delivered to all elements in the circuit and show energy is conserved. 7 pis

Answer 1

Ans. 3) Q = 8+ - 34 -0 the (a) using faraday's law of induction induced voltage aross the resistor = { E- 24 t? - 6t da dt at ta 2 sec E = 24(23) = 6 (2) E = 8.4V b) from we see that flux through the loop increases with time. so according to Lenz's law the induced voltage opposes ein this increase. FR so the polarity of induced voltage should be as shown in the figure

(c) area of he closed loop = A= 14 m2 Blt) = Q/A 14 B&t=2) 8(23) - (2) = 8 t - 3 t² I 14 I Ans. 4) t=2) 64-12 8.71428 T T4 current = I= 1200 A potential difference = V= 16MV = 16x10v distance between the two points : d=24cm a= 0.24 m 28 n = 6.95810 resistance of the cable between the two points =R = V/I now, R=go/A o resistivity of copper At area of cross-section tay

0- 1.68 X108 and A = X 8² Radius OdI лу R= ed -) ed TE AR 1/2 |-68x10*x 0.24 x 1200 KX 16 X (07 r = ra 0.009.8 11.1 m 0.98111 em b) A = a 2 = 0.0007024 m² average curlent density = . ITA = 1200 0.0003024m? - 3196.825 810 Am

c) drift velocity = Vo = Inea est charge on an election Vd 1200 6.95X1029 X 16 X1019 x 0.0001024 7 4 3.56 857 X10 ma en 8n VAA SA JA + V6 +241 th to Io on Ans.5)(a) Kiachhoff's low for O. gines 24 =-Va + 3x4 + 3x8 y voltage drop across ou resistor. 24 =-6 + 36 Voo 12V Q th I

for loop @ we get Vo 24 + 21 = 12 + 24 +21 VE 36 +2I a

If current through on resistor I6 = Vo/6 -1216 -2 A minus sign shows that the grection of It is to the right. hence Is liol + 3 I= 2 +3=5. A V hence V = 36+ 2 36+245 46 V (6) Power delivered to all elements - (3 x 4 +178) +(2*x65+15X2) 182 W - Power supplied = VI 24x Io - 46X 5 - 24x2 230 48 = 182 w so power delivered = power supplied. hence energy iu consured.