Question

3. A long, straight wire carries a current of I = 9.00 A. If next to the wire is a square loop ABCD with sides 4.00 cm in length and the loop carries a current of 12 = 3.00 A in the direction indicated in Fig. 3.1, (a) determine the directions and magnitudes of the forces exerted on each side of the square and (b) the direction and magnitude of the net force on the square loop ABCD. (c) If the square loop is moving to the right with a speed 10.0 m/s at that moment, what is the magnitude of the induced electromotive force in the loop? If the square loop ABCD is in the position shown in Fig. 3.2, (d) what are the directions and magnitudes of the forces exerted on side AB and side CD? If I in Fig. 3.2 increases, (e) is there an induced current in loop ABCD and, if there is, what is the direction of the induced current? (1) If the square loop in Fig. 3.1 is replaced by a circular loop of diameter 4.00 cm as shown in Fig. 3.3, what is the rent in the loop so that the magnetic field at the center of the loop is exactly zero? 4.00 cm D 4.00 cm 02-0 4.00 cm 4.00 cm B 4.00 cm 4.00 cm Fig. 3.1: For question (a), (b), and (c) Fig. 3.2: For questions (d) and (c) Fig. 3.3: For question (f)

Answer 1

for fig 3.1 force on current carsing element Radley) Illesima) force between two parallel wires С. До XII, X4 gad 44cm .4cm for AB current is same direction i force of attraction blo wire & AB fo Mo (II) (12) 1 a 40(9) (30) x 4 w and 22 (4) Gåx107xgx3 = 54x10 N - For co lih o current is in opposite direction force of repulsion between wire and CD 8cm f: Mo II I2d zad 20d Hax10x9x3xy zou 8. To = 27410-7N

for AD and be I 2 force on AD using Right hand A sulle TEAD a foe force on BC: using right hand sule: & pc sinee value of current and all the poents from AD & BC are at equal distance from bong wire i I FAD I = If Bcl i. Net force on AD & BC = FAD-FBC=0 Mathematical expression for FAD k a da 22 consider small element da at a distance of a from long wire on wire as Magnetic feeld of wire Is at a distance x - Modi From sem 2 Ye rare

. Force at a distance a on element de = I&B sino df = 22 (lot.) dx sime [0=900] (df = Mo II I2 dae ano I ran Net force sem sem Jak spor de M5S SEM celor 12 da WOL 2an . 21 som . 22 yem a 8am 40m 4cm - Mode 22 Elna - lodi I2 I en m 1 = wax 10 xgx3 2ne zal yani FAD = 54 dn 2 x 10-7N ( downward) Similarly foc = 54212X10N (upward) (6) Net force = FAB+ FBct fcd & FAD = 54x10r (right) + sulinexion ( up) +9481o-Ingle) +54 eneo) (down) = 27X107W ( sight words