An infinite straight wire carries current I 1 = 5 A in the positive y-direction
ID: 1520960 • Letter: A
Question
An infinite straight wire carries current I1 = 5 A in the positive y-direction as shown. At time t = 0, a conducting wire, aligned with the y-direction is located a distance d = 53 cm from the y-axis and moves with velocity v = 19 cm/s in the negaitve x-direction as shown. The wire has length W = 12 cm.
1)What is (0), the emf induced in the moving wire at t = 0? Define the emf to be positive if the potential at point a is higher than that at point b.
V
2)What is (t1), the emf induced in the moving wire at t = t1 = 1.9 s? Define the emf to be positive if the potential at point a is higher than that at point b.
V
3)
The wire is now replaced by a conducting rectangular loop as shown. The loop has length L = 51 cm and width W = 12 cm. At time t = 0, the loop moves with velocity v = 19 cm/s with its left end located a distance d = 53 cm from the y-axis. The resistance of the loop is R = 2.9 . What is i(0), the induced current in the loop at time t = 0? Define the current to be positive if it flows in the counter-clockwise direction.A
4)
Suppose the loop now moves in the positive y-direction as shown. What is the direction of the induced current now?(choose correct answer)
A- The current flows counterclockwise
B-The current flows clockwise
C-There is no induced current now
5)
Suppose now that the loop is rotated 90o and moves with velocity v = 19 cm/s in the positive x-direction as shown. What is I2, the current in the infinite wire, if the induced current in the loop at the instant shown (d = 53 cm) is the same as it was in the third part of this problem (i.e., when the left end of loop was at a distance d = 53 cm from the y-axis)?
A
Explanation / Answer
I1=5A
Distance=d=0.53m
Velocity=0.19m/sec
Lenghtvof wire=W=0.12m
Part(1);
Induced emf, E= v×W×B
B=uoI/2pi d
=4(3.14)×10-7(5)/2(3.14)(0.53)
=18.8×10-7T
Induced emf=(0.19)(0.12)(18.8×10-7)
= 0.428×10-7
=4.28×10-8V
Part (2);
Distance between the wire=d'=d-(v×t1)
=0.53 - (0.19×1.9)
=1.18m
B'=uoI/2pi d'
=4(3.14)×10-7(5)/2(3.14)(1.18)
=8.4×10-7T
E'=v×W×B'
=(0.19)(0.12)(8.4×10-7)
=1.9×10-8V
Part(3);
Lenght of the loop=0.51m, width=0.12m , resistance R=2.9 ohm
Induced emf in the left side of the loop=E=4.28×10-8
B''= uoI/2(pi)d"
= (4)(3.14)×10-7(5)/2(3.14)(0.51+0.53)
=9.65×10-7T
Induced emf in the right side= E''=(0.19)(0.12)(9.65×10-7)
=2.20×10-8V
Induced current= i= E"/R
=2.30×10-8/2.9
= 0.75×10-8A
Part(4);
Here velocity is parallel to current, hence induced emf= 0V. So there is no induced current.
(18)