An infinite straight wire carries current I1 = 5.5 A in the positive y-direction
ID: 1557888 • Letter: A
Question
An infinite straight wire carries current I1 = 5.5 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 = 65 cm from the y-axis and moves with velocity v = 13 cm/s in the negaitve x-direction as shown. The wire has length W = 29 cm.
A) 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.
B) What is (t1), the emf induced in the moving wire at t = t1 = 3.3 s? Define the emf to be positive if the potential at point a is higher than that at point b.
C) The wire is now replaced by a conducting rectangular loop as shown. The loop has length L = 45 cm and width W = 29 cm. At time t = 0, the loop moves with velocity v = 13 cm/s with its left end located a distance d = 65 cm from the y-axis. The resistance of the loop is R = 2 . 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.
E) Suppose now that the loop is rotated 90o and moves with velocity v = 13 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 = 65 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 = 65 cm from the y-axis)?
W b a 1Explanation / Answer
a) Let's calculate teh magnetic field due to I1 = ( 4pi x 10^-7 ) (5.5) / ( 2 pi x 65 x 10^-2)
E = vBL = ( 13 x 10^-2 ) ( 0.169 x 10^ -5) ( 0.29)= 0.638 x 10^ -7 V apprx
b) Distance between I1 and wire after 3.1 sec = 65 - ( 13x 3.3) = 22.1 cm
B = ( 4pi x 10*-7) (5.5) / ( 2 pi x 22.1 x 10^-2) =0.4977 x 10^ -5 T
E = ( 13 x 10^-2 ) ( 0.4977 x 10^ -5) (29 x 10^-2) = 1.8764 x 10^-7 V apprx
c)E = vBL
Let's caculate the net magnetic field as both left and right hand sides of loop will have magnetic field in opposite directon B = ( 4pi x 10^-7 x 5.5) / ( 2pi x 0.65 ) - ( 4pi x 10^- 7 x 5.5)/ ( 2 pi x 1.1) = 6.9224 x 10^-7 T
E = ( 13 x 10^-2) ( 6.9224 x 10^-7) ( 0.29) = 26. 1x 10^ -9 V
I = V/R = 13.05 x 10^-9 ohms apprx
e)B ( net ) = ( 2 x 10^-7 ) I2 ( 1/ 0.29 - 1/ 0.94) = 4.768 x 10^-7 ( I2 )
26. 1x 10^ -9 V = 4.768 x 10^-7 ( I2 ) ( 13x 10^-2) ( (0.45)
I2= 0.9357 A apprx