Question

5.0 4 A 5.0-cm x 10.0-cm rectangular loop made of 3.0-mm-diameter copper wire (resisitivity 1.7*10-822*m) is placed next to a long, straight wire that carries a 5.0-A current (see figure). The distance initially separating the loop and the straight wire is 30 cm. 30 cm The loop then begins to move away from the straight wire, at a constant speed of 7.5 cm/s. 5.0 cm Determine the average magnitude and direction of the current induced inside the loop during the first 5.0 seconds of its motion. 10.0 cm For simplicity, approximate the magnetic field inside the loop as uniform, with the strength equal to the value at the center of the loop at that time.

Answer 1

21 % Initial magnere field (ar centre of loop), B, = MI Mo= 4T XO? I= current in wire = 5A. 8 = 30+ 25 = 32.5cm = 0.325m distance from Clarme) B = 4 11 8 10 ? x 5 30.70x107T al xo.325 travelled by centre of loop 5-V6=7.5810 * 5 =37.5xion - 7 seme Final field By, Mo I 4TX10 X5 It-time = 5sec = 14.29x10r 8 = 33.5+ 37.5 change in field, dB = B, -B₂ = (30.77 - 14:29)X10? 12 = 70cm 30-70m vzrelowing Distance 2T 82 anxo.7 dB = 16.48x10T Induced enf, e=dd = f(BA) = AdB olt DE A Ecosea of loop = lengthy breadth = 10x10x5702= 504.10 2,2 dt= time assee. 16.48810 volt e-50x154 x 16.488107 Resistance of loop, R=SL A .-7.21 x 10-12 &= resishting = 1.7810-8 Ls toral length of loop - Llox2)+6x)]- 30 kón A area of cross sembn=18? = 3.14x1.5x103) 2 A' = 7.07x106 [= sadims -diamers - 3mm] .R=1.7410 x30x10-2 7.07x10-6 -TO -6 indued current, I'-e 16. 48x10 - 2.286 x 10 A 7.21400-4 As the doop moves away from wire magnent foeld decreases. (B2<B.). Hence indrued tries to field is into the doop (by daw. As right hand rule), indeed is clockwise, which produces into the coil field increase it by Lena's current Current

21 % Initial magnere field (ar centre of loop), B, = MI Mo= 4T XO? I= current in wire = 5A. 8 = 30+ 25 = 32.5cm = 0.325m distance from Clarme) B = 4 11 8 10 ? x 5 30.70x107T al xo.325 travelled by centre of loop 5-V6=7.5810 * 5 =37.5xion - 7 seme Final field By, Mo I 4TX10 X5 It-time = 5sec = 14.29x10r 8 = 33.5+ 37.5 change in field, dB = B, -B₂ = (30.77 - 14:29)X10? 12 = 70cm 30-70m vzrelowing Distance 2T 82 anxo.7 dB = 16.48x10T Induced enf, e=dd = f(BA) = AdB olt DE A Ecosea of loop = lengthy breadth = 10x10x5702= 504.10 2,2 dt= time assee. 16.48810 volt e-50x154 x 16.488107 Resistance of loop, R=SL A .-7.21 x 10-12 &= resishting = 1.7810-8 Ls toral length of loop - Llox2)+6x)]- 30 kón A area of cross sembn=18? = 3.14x1.5x103) 2 A' = 7.07x106 [= sadims -diamers - 3mm] .R=1.7410 x30x10-2 7.07x10-6 -TO -6 indued current, I'-e 16. 48x10 - 2.286 x 10 A 7.21400-4 As the doop moves away from wire magnent foeld decreases. (B2<B.). Hence indrued tries to field is into the doop (by daw. As right hand rule), indeed is clockwise, which produces into the coil field increase it by Lena's current Current