Question

Fig. 3 Problem 5 A square wire loop with sides of length L=2.00 m is perpendicular to a uniform magnetic field pointing out of the page, with half the area of the loop in the field as shown in Fig. 3. The loop contains an ideal battery with emf Ebar=20.0 V. The magnitude of the field varies with time according to the expression B = 0.0420-0.870t, with B in Tesla and t in seconds. (a) Use Faraday's law to determine the magnitude and direction of the induced emf in the circuit of Fig. 7. (b) Determine the direction of the (net) current around the loop. -249-2-----22 - bal

Answer 1

L=2m, Ebot = 20V, B = 0·0420 - 0.870t (a) From Faraday's law Induced ent (E)= -do dt As $ = BA in E = -d (BA) It is E = – AdB dt -

2 As it is given in question that square wire loop is perpendicular to uniform magnetic field with half area of Loop & A = L² and E=-2² dB 2 dt Putting all values in E = -(2) & (0.0420 -0.870+) 2 at E = -2 (-0.0870) {= 1.748 Total emf (& net) = Et Ebat Enet = 8 1074+20 I Enet = 21.7v imetic (6) As magnetic field is field is out of th out of the page decreasing so Emf and and current will be Counterclockwise /