Background info An infinite straight wire carries current I1

Background info: An infinite straight wire carries current I₁ = 3.9 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 = 63 cm from the y-axis and moves with velocity v = 16 cm/s in the negaitve x-direction as shown. The wire has length W = 22 cm.

Part1- The wire is now replaced by a conducting rectangular loop as shown. The loop has length L = 50 cm and width W = 22 cm. At time t = 0, the loop moves with velocity v = 16 cm/s with its left end located a distance d = 63 cm from the y-axis. The resistance of the loop is R = 1.9 . 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. Answer: _____ amps

Part 2- Suppose now that the loop is rotated 90o and moves with velocity v = 16 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 = 63 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 = 63 cm from the y-axis)? Answer: ____ amps

Solution

Magnetic field due to a long current carrying conductor = 2kI/r
now, considering the rectangular loop as given in the problem

Due to changing flux, EMF will be induced in the loop
flux through the loop at time t = F
consider an infinitesimal part of the loop of thichness dx, and length = l = 0.5 m
flux through this dF = BdA = 2kI*l*dx/x
at time t, the left end of the loop is at x = 0.63 - 0.16t, right end is at x = 0.63 + 0.22 - 0.16t = 0.85 - 0.16t
integrating flux within limits of x
F = 2kI*l*ln([0.85 - 0.16t]/[0.63 - 0.16t]) = 2*k*3.9*ln([0.85 - 0.16t]/[0.63 - 0.16t]) = 7.8k*ln([0.85 - 0.16t]/[0.63 - 0.16t])
dF/dt = 7.8k*0.5(0.63 - 0.16t)[(0.85 - 0.16t)(-0.16) - (0.63 - 0.16t)(-0.16)]/(0.85 - 0.16t)(0.63 - 0.16t)^2
at t = 0
dF/dt = 7.8*0.5k(0.16)[-(0.85) + (0.63)]/(0.85)(0.63) = -1.2451*10^-7 *0.5 V
but V = IR
I = V/1.9 = 3.27676*10^-8 A

Part 2. When the loop is rotated by 90 deg, i.e length becomes width and width becomes length
flux through this dF = BdA = 2kI*l*dx/x
at time t, the left end of the loop is at x = 0.63 - 0.16t, right end is at x = 0.63 + 0.5 - 0.16t = 0.113 - 0.16t
integrating flux within limits of x
F = 2kI*l*ln([0.113 - 0.16t]/[0.63 - 0.16t]) = 2*0.22*k*3.9*ln([0.113 - 0.16t]/[0.63 - 0.16t]) = 1.716k*ln([0.113 - 0.16t]/[0.63 - 0.16t])
dF/dt = 1.716k*(0.63 - 0.16t)[(0.113 - 0.16t)(-0.16) - (0.63 - 0.16t)(-0.16)]/(0.113 - 0.16t)(0.63 - 0.16t)^2
at t = 0
dF/dt = 1.716*k(0.16)[-(1.13) + (0.63)]/(1.13)(0.63) = -1.9283*10^-8 V
but V = IR
I = V/1.9 = 1.0149*10^-8 A