Question

Chapter 30, Problem 015 GO A square wire loop with 2.0 m sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field as shown in the figure. The loop contains an ideal battery with emf 8 = 15 V. If the magnitude of the field varies with time according to B = 0.040 - 0.82 t, with B in teslas and t in seconds, what are (a) the net emf in the circuit and (b) the direction of the (net) current around the loop? & (a) (b) the tolerance is +/-5%

Answer 1

Ich Magnitude of emf induced is End = 1 a=2m side length of square loop E Eind Ebat = 150 Gemf of battery B=0.040-0.82 G magnetic field magnetic flux throug the loop is $= Bxle) = (0.040--01824)x ► 05 (0.08–1644) . Rate of change of flur is do do. 2² 50–1964.de 1.64 It It Eina = 1.640 at 9 | -1,641 v sence, do is negative, it means flux Hirough the loop is decreasing, so from Lenz's law polarity of Eind should be such that current induced due to Eind

produces a magnetic field which try to increase the decreasing magnetic fileese. So, induced magnetic field should be in the direction out of plane of paper. For this induced current should be counterclockwise. Palarity of Eind is as shown in figure below since, Eind and Ebat are in same sense, so net emf in the circuit is Enet- Ebatt Eind Eind Ebat Enet Fisut 1.645 >Enet = 16.64V 6 So, direction of not current is Counter clockwise" 66